Class 10 Mathematics Chapter 5 Question Answer | Arithmetic Progressions | English Medium

Class 10 Mathematics Chapter 5 Question Answer | Arithmetic Progressions | English Medium | ASSEB

Welcome to HSLC Guru! This complete guide covers Class 10 Mathematics Chapter 5 — Arithmetic Progressions for the ASSEB (SEBA) English medium syllabus. The chapter teaches students how to recognise an arithmetic pattern in a sequence of numbers, how to find any term of such a sequence, and how to add a long list of numbers in seconds using compact formulas. We have explained the definition, the meaning of common difference, the $n^{th}$ term formula, and the sum formula in simple language, then walked through every problem of Exercises 5.1, 5.2, 5.3 and 5.4 with neat step-by-step solutions. Extra HOTS, MCQs, application problems and a glossary are added at the end so you can revise quickly before the HSLC examination.

Chapter Summary

An Arithmetic Progression (AP) is a list of numbers in which each term, except the first, is obtained by adding a fixed number to the term before it. This fixed number is called the common difference and is denoted by $d$. The first term is denoted by $a$ (or $a_1$). A general AP looks like:

$$a,\ a+d,\ a+2d,\ a+3d,\ \dots,\ a+(n-1)d$$

If we know $a$ and $d$, we can write the entire AP. The common difference $d$ may be positive, negative or zero. When $d$ is positive the AP is increasing (e.g. $2, 5, 8, 11, \dots$); when $d$ is negative the AP is decreasing (e.g. $10, 7, 4, 1, \dots$); and when $d=0$ all terms are equal (e.g. $7, 7, 7, \dots$).

An AP that has a fixed last term $l$ and only a finite number of terms is called a finite AP; otherwise it is an infinite AP. To check whether a given list is an AP, find $a_2-a_1, a_3-a_2, a_4-a_3, \dots$ — if all these differences are equal, the list is an AP.

Key Formulas at a Glance

These are the most important formulas used throughout the chapter. Memorise them carefully.

1. nth term (general term):

$$a_n = a + (n-1)d$$

2. nth term from the end (when last term $l$ is known):

$$a_n^{end} = l – (n-1)d$$

3. Sum of first $n$ terms:

$$S_n = \frac{n}{2}\big[2a + (n-1)d\big]$$

4. Sum when first term $a$ and last term $l$ are known:

$$S_n = \frac{n}{2}(a + l)$$

5. Relation between $S_n$ and $a_n$:

$$a_n = S_n – S_{n-1}$$

6. Sum of first $n$ natural numbers:

$$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$$

7. Sum of first $n$ odd natural numbers:

$$1 + 3 + 5 + \dots + (2n-1) = n^2$$

8. Sum of first $n$ even natural numbers:

$$2 + 4 + 6 + \dots + 2n = n(n+1)$$

9. Three terms in AP are conveniently written as $a-d,\ a,\ a+d$ (sum $=3a$).

10. Five terms in AP are written as $a-2d,\ a-d,\ a,\ a+d,\ a+2d$ (sum $=5a$).

Exercise 5.1 — Identifying Arithmetic Progressions

Q1. In which of the following situations does the list of numbers involved make an AP, and why?

(i) The taxi fare after each km is Rs 15 for the first km and Rs 8 for each additional km.

Answer: Fares are $15, 23, 31, 39, \dots$. Differences: $23-15=8,\ 31-23=8,\ 39-31=8$. Difference is constant, so it is an AP with $a=15,\ d=8$.

(ii) The amount of air present in a cylinder when a vacuum pump removes $\tfrac{1}{4}$ of the air remaining each time.

Answer: If $V$ is initial air, then amounts are $V, \tfrac{3V}{4}, \tfrac{9V}{16}, \dots$. Successive differences are $-\tfrac{V}{4}, -\tfrac{3V}{16}, \dots$ which are not equal. Not an AP.

(iii) Cost of digging a well for the first metre is Rs 150 and rises by Rs 50 for each subsequent metre.

Answer: Costs are $150, 200, 250, 300, \dots$. Common difference $=50$. It is an AP.

(iv) Amount of money in account when Rs 10000 is deposited at $8\%$ compound interest per year.

Answer: Amounts are $10000(1.08), 10000(1.08)^2, 10000(1.08)^3, \dots$. The differences are not constant. Not an AP.

Q2. Write the first four terms of the AP when $a$ and $d$ are given.

Answer:

Case	$a$	$d$	First four terms
(i)	$10$	$10$	$10, 20, 30, 40$
(ii)	$-2$	$0$	$-2, -2, -2, -2$
(iii)	$4$	$-3$	$4, 1, -2, -5$
(iv)	$-1$	$\tfrac{1}{2}$	$-1, -\tfrac{1}{2}, 0, \tfrac{1}{2}$
(v)	$-1.25$	$-0.25$	$-1.25, -1.50, -1.75, -2.00$

Q3. For the following APs, write the first term $a$ and the common difference $d$.

(i) $3, 1, -1, -3, \dots$ Answer: $a = 3,\ d = 1-3 = -2$.

(ii) $-5, -1, 3, 7, \dots$ Answer: $a = -5,\ d = -1-(-5) = 4$.

(iii) $\tfrac{1}{3}, \tfrac{5}{3}, \tfrac{9}{3}, \tfrac{13}{3}, \dots$ Answer: $a = \tfrac{1}{3},\ d = \tfrac{4}{3}$.

(iv) $0.6, 1.7, 2.8, 3.9, \dots$ Answer: $a = 0.6,\ d = 1.1$.

Q4. Which of the following are APs? If they form an AP, find the common difference $d$ and write three more terms.

(i) $2, 4, 8, 16, \dots$ — Differences $2, 4, 8$ not equal. Answer: Not an AP.

(ii) $2, \tfrac{5}{2}, 3, \tfrac{7}{2}, \dots$ — $d = \tfrac{1}{2}$. Answer: AP. Next three terms: $4, \tfrac{9}{2}, 5$.

(iii) $-1.2, -3.2, -5.2, -7.2, \dots$ — $d = -2$. Answer: AP. Next: $-9.2, -11.2, -13.2$.

(iv) $-10, -6, -2, 2, \dots$ — $d = 4$. Answer: AP. Next: $6, 10, 14$.

(v) $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots$ — $d = \sqrt{2}$. Answer: AP. Next: $3+4\sqrt{2},\ 3+5\sqrt{2},\ 3+6\sqrt{2}$.

(vi) $0.2, 0.22, 0.222, 0.2222, \dots$ — Differences not equal. Answer: Not an AP.

(vii) $0, -4, -8, -12, \dots$ — $d=-4$. Answer: AP. Next: $-16, -20, -24$.

(viii) $-\tfrac{1}{2}, -\tfrac{1}{2}, -\tfrac{1}{2}, \dots$ — $d=0$. Answer: AP. Next three terms all $-\tfrac{1}{2}$.

(ix) $1, 3, 9, 27, \dots$ — Not constant difference. Answer: Not an AP.

(x) $a, 2a, 3a, 4a, \dots$ — $d = a$. Answer: AP. Next: $5a, 6a, 7a$.

(xi) $a, a^2, a^3, a^4, \dots$ — Not AP unless $a=1$. Answer: Not an AP.

(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots = \sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}$ — $d=\sqrt{2}$. Answer: AP. Next: $5\sqrt{2}, 6\sqrt{2}, 7\sqrt{2}$.

(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$ — Differences not equal. Answer: Not an AP.

(xiv) $1^2, 3^2, 5^2, 7^2, \dots = 1, 9, 25, 49, \dots$ — Differences $8, 16, 24$ not equal. Answer: Not an AP.

(xv) $1^2, 5^2, 7^2, 73, \dots = 1, 25, 49, 73$ — $d = 24$. Answer: AP. Next: $97, 121, 145$.

Exercise 5.2 — The nth Term of an AP

Q1. Fill in the blanks in the following table given $a, d, n$ and $a_n$ where $a_n = a+(n-1)d$.

Answer:

Case	$a$	$d$	$n$	$a_n$
(i)	$7$	$3$	$8$	$28$
(ii)	$-18$	$2$	$10$	$0$
(iii)	$-3$	$-2$	$18$	$-37$ (so $a_n=-5$ given)
(iv)	$-18.9$	$2.5$	$10$	$3.6$
(v)	$3.5$	$0$	$105$	$3.5$

Q2. Choose the correct option.

(i) 30th term of $10, 7, 4, \dots$ is:

Answer: $a=10, d=-3$. $a_{30}=10+29(-3)=10-87=-77$. Option (C) $-77$.

(ii) 11th term of $-3, -\tfrac{1}{2}, 2, \dots$ is:

Answer: $a=-3, d=\tfrac{5}{2}$. $a_{11}=-3+10(\tfrac{5}{2})=-3+25=22$. Option (B) $22$.

Q3. Find missing terms in each AP.

(i) $2, \_\_, 26$. Answer: Middle term $=\tfrac{2+26}{2}=14$.

(ii) $\_\_, 13, \_\_, 3$. Answer: $a+d=13, a+3d=3 \Rightarrow 2d=-10, d=-5, a=18$. AP: $18, 13, 8, 3$.

(iii) $5, \_\_, \_\_, 9\tfrac{1}{2}$. Answer: $a=5, a+3d=\tfrac{19}{2} \Rightarrow d=\tfrac{3}{2}$. Missing terms $\tfrac{13}{2}, 8$.

(iv) $-4, \_\_, \_\_, \_\_, \_\_, 6$. Answer: $a=-4, a+5d=6 \Rightarrow d=2$. Missing: $-2, 0, 2, 4$.

(v) $\_\_, 38, \_\_, \_\_, \_\_, -22$. Answer: $a+d=38, a+5d=-22 \Rightarrow 4d=-60, d=-15, a=53$. Missing: $53, 23, 8, -7$.

Q4. Which term of $3, 8, 13, 18, \dots$ is $78$?

Answer: $a=3, d=5, a_n=78$. $78=3+(n-1)5 \Rightarrow 75=5(n-1) \Rightarrow n=16$. So $78$ is the $16^{th}$ term.

Q5. Find the number of terms in each AP.

(i) $7, 13, 19, \dots, 205$. $a=7, d=6, a_n=205 \Rightarrow 205=7+(n-1)6 \Rightarrow n=34$. Answer: 34 terms.

(ii) $18, 15\tfrac{1}{2}, 13, \dots, -47$. $a=18, d=-\tfrac{5}{2}$. $-47=18+(n-1)(-\tfrac{5}{2}) \Rightarrow n=27$. Answer: 27 terms.

Q6. Check whether $-150$ is a term of the AP $11, 8, 5, 2, \dots$

Answer: $a=11, d=-3$. If $a_n=-150$, then $-150=11+(n-1)(-3) \Rightarrow n=\tfrac{164}{3}$, not a whole number. So $-150$ is NOT a term.

Q7. Find the 31st term of an AP whose 11th term is $38$ and 16th term is $73$.

Answer: $a+10d=38,\ a+15d=73 \Rightarrow 5d=35, d=7, a=-32$. $a_{31}=-32+30(7)=178$.

Q8. An AP consists of $50$ terms of which 3rd term is $12$ and last term is $106$. Find the 29th term.

Answer: $a+2d=12,\ a+49d=106 \Rightarrow 47d=94, d=2, a=8$. $a_{29}=8+28(2)=64$.

Q9. If 3rd and 9th terms of an AP are $4$ and $-8$ respectively, which term is zero?

Answer: $a+2d=4, a+8d=-8 \Rightarrow 6d=-12, d=-2, a=8$. Set $a_n=0: 0=8+(n-1)(-2)\Rightarrow n=5$. 5th term is zero.

Q10. The 17th term of an AP exceeds its 10th term by $7$. Find the common difference.

Answer: $a_{17}-a_{10}=7d=7 \Rightarrow d=1$.

Q11. Which term of the AP $3, 15, 27, 39, \dots$ will be $132$ more than its 54th term?

Answer: $d=12,\ 132=12k\Rightarrow k=11$. Required term $=54+11=65$. 65th term.

Q12. Two APs have the same common difference. The difference between their 100th terms is $100$. Find the difference between their 1000th terms.

Answer: Difference of $n^{th}$ terms $=a_1-a_2$, independent of $n$. Difference of 1000th terms $=100$.

Q13. How many three-digit numbers are divisible by $7$?

Answer: AP: $105, 112, \dots, 994$. $a=105, d=7$. $994=105+(n-1)7 \Rightarrow n=128$. 128 numbers.

Q14. How many multiples of $4$ lie between $10$ and $250$?

Answer: AP: $12, 16, \dots, 248$. $a=12, d=4$. $248=12+(n-1)4 \Rightarrow n=60$. 60 multiples.

Q15. For what value of $n$ are the $n^{th}$ terms of $63, 65, 67, \dots$ and $3, 10, 17, \dots$ equal?

Answer: $63+(n-1)2 = 3+(n-1)7 \Rightarrow 60 = 5(n-1) \Rightarrow n=13$.

Q16. Find the AP whose 3rd term is $16$ and 7th term exceeds 5th term by $12$.

Answer: $a_7-a_5=2d=12 \Rightarrow d=6$. $a+2d=16 \Rightarrow a=4$. AP: $4, 10, 16, 22, \dots$.

Q17. Find the 20th term from the last term of the AP $3, 8, 13, \dots, 253$.

Answer: Use $l-(n-1)d = 253-19(5) = 253-95 = 158$.

Q18. The sum of 4th and 8th terms of an AP is $24$ and the sum of 6th and 10th terms is $44$. Find the first three terms.

Answer: $(a+3d)+(a+7d)=2a+10d=24$; $(a+5d)+(a+9d)=2a+14d=44$. Subtract: $4d=20 \Rightarrow d=5, a=-13$. AP: $-13, -8, -3, \dots$.

Q19. Subba Rao’s salary started at Rs $5000$ in $1995$ with annual increments of Rs $200$. In which year did his salary reach Rs $7000$?

Answer: $7000=5000+(n-1)(200) \Rightarrow n-1=10 \Rightarrow n=11$. Year $=1995+10=2005$.

Q20. Ramkali saved Rs $5$ in week 1 and increased her weekly savings by Rs $1.75$. In which week did her savings become Rs $20.75$?

Answer: $20.75 = 5 + (n-1)(1.75) \Rightarrow 15.75 = 1.75(n-1) \Rightarrow n=10$. 10th week.

Exercise 5.3 — Sum of First n Terms of an AP

Q1. Find the sum of the following APs.

(i) $2, 7, 12, \dots$ to $10$ terms. $a=2, d=5$. $S_{10}=\tfrac{10}{2}[4+9(5)]=5(49)=245$.

(ii) $-37, -33, -29, \dots$ to $12$ terms. $a=-37, d=4$. $S_{12}=6[-74+44]=6(-30)=-180$.

(iii) $0.6, 1.7, 2.8, \dots$ to $100$ terms. $a=0.6, d=1.1$. $S_{100}=50[1.2+99(1.1)]=50(110.1)=5505$.

(iv) $\tfrac{1}{15}, \tfrac{1}{12}, \tfrac{1}{10}, \dots$ to $11$ terms. $a=\tfrac{1}{15}, d=\tfrac{1}{60}$. $S_{11}=\tfrac{11}{2}[\tfrac{2}{15}+\tfrac{10}{60}]=\tfrac{11}{2}\cdot\tfrac{18}{60}=\tfrac{33}{20}$.

Q2. Find the sums.

(i) $7+10\tfrac{1}{2}+14+\dots+84$. $a=7, d=\tfrac{7}{2}$. $84=7+(n-1)\tfrac{7}{2} \Rightarrow n=23$. $S_{23}=\tfrac{23}{2}(7+84)=\tfrac{23\times 91}{2}=1046\tfrac{1}{2}$.

(ii) $34+32+30+\dots+10$. $a=34, d=-2, l=10$. $10=34+(n-1)(-2) \Rightarrow n=13$. $S=\tfrac{13}{2}(34+10)=286$.

(iii) $-5+(-8)+(-11)+\dots+(-230)$. $a=-5, d=-3$. $-230=-5+(n-1)(-3) \Rightarrow n=76$. $S=\tfrac{76}{2}(-5-230)=38(-235)=-8930$.

Q3. In an AP:

(i) $a=5, d=3, a_n=50$. Find $n$ and $S_n$.

Answer: $50=5+(n-1)3 \Rightarrow n=16$. $S_{16}=\tfrac{16}{2}(5+50)=440$.

(ii) $a=7, a_{13}=35$. Find $d$ and $S_{13}$.

Answer: $35=7+12d \Rightarrow d=\tfrac{7}{3}$. $S_{13}=\tfrac{13}{2}(7+35)=273$.

(iii) $a_{12}=37, d=3$. Find $a$ and $S_{12}$.

Answer: $37=a+11(3) \Rightarrow a=4$. $S_{12}=\tfrac{12}{2}(4+37)=246$.

(iv) $a_3=15, S_{10}=125$. Find $d$ and $a_{10}$.

Answer: $a+2d=15$; $\tfrac{10}{2}(2a+9d)=125 \Rightarrow 2a+9d=25$. Solve: $d=-1, a=17$. $a_{10}=17+9(-1)=8$.

(v) $d=5, S_9=75$. Find $a$ and $a_9$.

Answer: $\tfrac{9}{2}(2a+8\cdot 5)=75 \Rightarrow 2a+40=\tfrac{50}{3} \Rightarrow a=-\tfrac{35}{3}$. $a_9=a+8d=-\tfrac{35}{3}+40=\tfrac{85}{3}$.

(vi) $a=2, d=8, S_n=90$. Find $n$ and $a_n$.

Answer: $\tfrac{n}{2}[4+8(n-1)]=90 \Rightarrow n(4n)=180 \Rightarrow n=5$ (taking +ve root after solving). $a_5=2+4(8)=34$.

(vii) $a=8, a_n=62, S_n=210$. Find $n$ and $d$.

Answer: $S_n=\tfrac{n}{2}(8+62)=35n=210 \Rightarrow n=6$. $62=8+5d \Rightarrow d=\tfrac{54}{5}$.

(viii) $a_n=4, d=2, S_n=-14$. Find $n$ and $a$.

Answer: $4=a+(n-1)2$; $-14=\tfrac{n}{2}(a+4)$. Solving gives $n=7, a=-8$.

(ix) $a=3, n=8, S=192$. Find $d$.

Answer: $192=\tfrac{8}{2}[6+7d]=4(6+7d) \Rightarrow 7d=42 \Rightarrow d=6$.

(x) $l=28, S=144, n=9$. Find $a$.

Answer: $144=\tfrac{9}{2}(a+28) \Rightarrow a+28=32 \Rightarrow a=4$.

Q4. How many terms of $9, 17, 25, \dots$ must be taken to give a sum of $636$?

Answer: $a=9, d=8, S_n=636$. $636=\tfrac{n}{2}[18+8(n-1)]=n(4n+5) \Rightarrow 4n^2+5n-636=0$. Solving, $n=12$.

Q5. The first term of an AP is $5$, the last term is $45$, and the sum is $400$. Find the number of terms and the common difference.

Answer: $400=\tfrac{n}{2}(5+45)=25n \Rightarrow n=16$. $45=5+15d \Rightarrow d=\tfrac{40}{15}=\tfrac{8}{3}$.

Q6. The first and last terms of an AP are $17$ and $350$. If the common difference is $9$, how many terms are there and what is the sum?

Answer: $350=17+(n-1)9 \Rightarrow n=38$. $S_{38}=\tfrac{38}{2}(17+350)=19(367)=6973$.

Q7. Find the sum of the first $22$ terms of an AP in which $d=7$ and 22nd term is $149$.

Answer: $149=a+21(7) \Rightarrow a=2$. $S_{22}=\tfrac{22}{2}(2+149)=11(151)=1661$.

Q8. Find the sum of the first $51$ terms of an AP whose second and third terms are $14$ and $18$.

Answer: $d=4, a=10$. $S_{51}=\tfrac{51}{2}[20+50(4)]=\tfrac{51}{2}(220)=5610$.

Q9. If the sum of first $7$ terms of an AP is $49$ and that of $17$ terms is $289$, find the sum of first $n$ terms.

Answer: $\tfrac{7}{2}(2a+6d)=49 \Rightarrow a+3d=7$. $\tfrac{17}{2}(2a+16d)=289 \Rightarrow a+8d=17$. So $d=2, a=1$. $S_n=\tfrac{n}{2}[2+2(n-1)]=n^2$.

Q10. Show that $a_1, a_2, \dots, a_n$ form an AP where $a_n$ is given by:

(i) $a_n=3+4n$. Answer: $a_{n+1}-a_n=4$, constant. AP. $S_{15}=\tfrac{15}{2}(7+63)=525$.

(ii) $a_n=9-5n$. Answer: Difference $-5$. AP. $S_{15}=\tfrac{15}{2}(4+(-66))=-465$.

Q11. If the sum of first $n$ terms of an AP is $4n-n^2$, what is the first term, sum of first two terms, second term, third term, 10th term and $n^{th}$ term?

Answer: $S_1=3=a$. $S_2=4=a_1+a_2 \Rightarrow a_2=1$. $d=-2$. $a_3=-1, a_{10}=3+9(-2)=-15$. $a_n=a+(n-1)d=5-2n$.

Q12. Find the sum of first $40$ positive integers divisible by $6$.

Answer: $6, 12, 18, \dots$, $a=6, d=6$. $S_{40}=\tfrac{40}{2}[12+39(6)]=20(246)=4920$.

Q13. Find the sum of first $15$ multiples of $8$.

Answer: $a=8, d=8$. $S_{15}=\tfrac{15}{2}[16+14(8)]=\tfrac{15}{2}(128)=960$.

Q14. Find the sum of odd numbers between $0$ and $50$.

Answer: $1, 3, 5, \dots, 49$. $a=1, d=2$. $49=1+(n-1)2 \Rightarrow n=25$. $S_{25}=\tfrac{25}{2}(1+49)=625$.

Q15. A contract on construction job specifies penalty: Rs $200$ for first day, Rs $250$ for second, Rs $300$ for third and so on. If contractor delays by $30$ days, find total penalty.

Answer: $a=200, d=50, n=30$. $S_{30}=\tfrac{30}{2}[400+29(50)]=15(1850)=27750$. Total penalty Rs $27750$.

Q16. A sum of Rs $700$ is to be used as cash prizes for $7$ students with each prize Rs $20$ less than its preceding prize. Find the value of each prize.

Answer: $S_7=700, n=7, d=-20$. $700=\tfrac{7}{2}[2a+6(-20)] \Rightarrow 200=2a-120 \Rightarrow a=160$. Prizes: Rs $160, 140, 120, 100, 80, 60, 40$.

Q17. In a school, students plant trees: each section of class I plants $1$ tree, class II plants $2$, … class XII plants $12$. There are $3$ sections of each class. Total trees planted?

Answer: Trees per class $=3, 6, 9, \dots, 36$. $a=3, d=3, n=12$. $S_{12}=\tfrac{12}{2}(3+36)=234$.

Q18. A spiral is made of successive semicircles with radii $0.5, 1.0, 1.5, 2.0, \dots$ cm. Find the total length of the spiral made up of thirteen consecutive semicircles. Take $\pi=\tfrac{22}{7}$.

Answer: Lengths $\pi(0.5), \pi(1.0), \pi(1.5), \dots$. Total $=\pi(0.5+1.0+\dots+6.5)$. Sum $=\tfrac{13}{2}(0.5+6.5)=\tfrac{13}{2}(7)=45.5$. Total length $=\tfrac{22}{7}\times 45.5=143$ cm.

Q19. $200$ logs are stacked: $20$ in bottom row, $19$ in next, $18$ in next and so on. In how many rows are the $200$ logs placed and how many logs in the top row?

Answer: $a=20, d=-1, S_n=200$. $200=\tfrac{n}{2}[40-(n-1)] \Rightarrow n^2-41n+400=0 \Rightarrow n=16$ or $25$. $n=25$ rejected since 25th term would be negative. So $n=16$. Top row: $a_{16}=20-15=5$ logs.

Q20. In a potato race, a bucket is placed at the start, and potatoes are kept $5, 8, 11, \dots$ m apart. The competitor starts from the bucket, picks up nearest potato, runs back, drops, returns for next, and so on. If there are ten potatoes, find total distance.

Answer: Distances run for each potato $=2(5), 2(8), 2(11), \dots$ for $10$ potatoes. AP: $10, 16, 22, \dots$, $a=10, d=6, n=10$. $S_{10}=\tfrac{10}{2}[20+9(6)]=5(74)=370$ m.

Exercise 5.4 — Optional Higher-Order Problems

Q1. Which term of the AP $121, 117, 113, \dots$ is its first negative term?

Answer: $a=121, d=-4$. Need $a_n<0$: $121+(n-1)(-4)<0 \Rightarrow n>31.25$. So $n=32$. 32nd term is the first negative term.

Q2. The sum of the third and seventh terms of an AP is $6$ and their product is $8$. Find the sum of the first sixteen terms of the AP.

Answer: Let $a_3=a+2d, a_7=a+6d$. Sum $=2a+8d=6 \Rightarrow a+4d=3$. Product $=8 \Rightarrow (3-2d)(3+2d)=8 \Rightarrow 9-4d^2=8 \Rightarrow d=\pm\tfrac{1}{2}$. If $d=\tfrac{1}{2}, a=1$, then $S_{16}=\tfrac{16}{2}(2+15\cdot\tfrac{1}{2})=8(\tfrac{19}{2})=76$. If $d=-\tfrac{1}{2}, a=5$, then $S_{16}=8(10-\tfrac{15}{2})=20$.

Q3. A ladder has rungs $25$ cm apart. The rungs decrease uniformly in length from $45$ cm at the bottom to $25$ cm at the top. If the top and bottom rungs are $2.5$ m apart, what is the length of the wood required for the rungs?

Answer: Number of rungs $=\tfrac{250}{25}+1=11$. Lengths form an AP with $a=45, l=25, n=11$. Total length $=S_{11}=\tfrac{11}{2}(45+25)=385$ cm.

Q4. Houses on one side of a street are numbered $1, 2, 3, \dots, 49$. Show there is a value $x$ such that the sum of numbers of houses preceding $x$ equals the sum of numbers following $x$. Find $x$.

Answer: Sum of houses before $x = \tfrac{(x-1)x}{2}$. Sum after $x = \tfrac{(49)(50)}{2}-\tfrac{x(x+1)}{2}$. Equating: $x^2=\tfrac{49\cdot 50}{2}=1225 \Rightarrow x=35$. $x=35$.

Q5. A small terrace has $15$ steps, each $\tfrac{1}{2}$ m high, $50$ cm tread, and total length $50$ m. Volumes of concrete required for steps form an AP. Find total concrete required.

Answer: Volume of $n^{th}$ step $=50 \times \tfrac{1}{2} \times \tfrac{n}{2}=\tfrac{25n}{2}$ m³. AP: $\tfrac{25}{2}, 25, \tfrac{75}{2}, \dots$, $a=\tfrac{25}{2}, d=\tfrac{25}{2}, n=15$. $S_{15}=\tfrac{15}{2}[25+14(\tfrac{25}{2})]=\tfrac{15}{2}(200)=1500$ m³… actually careful: $S_{15}=\tfrac{15}{2}\cdot\tfrac{25}{2}\cdot 16=750$ m³. Total concrete $=750$ m³.

Additional Important Questions and HOTS

Q1. Find the sum: $1+2+3+\dots+100$.

Answer: Use $S_n=\tfrac{n(n+1)}{2}=\tfrac{100\cdot 101}{2}=5050$.

Q2. Find the sum of all two-digit numbers.

Answer: AP: $10, 11, \dots, 99$. $n=90$. $S_{90}=\tfrac{90}{2}(10+99)=45(109)=4905$. (For two-digit multiples of $10$ etc., adapt accordingly.) For all natural numbers from $10$ to $99$ the answer is $4905$. For the sum $10+11+\dots+99$ alternatively, two-digit even numbers $10+12+\dots+98$ sum is $2430$, and two-digit odd numbers sum is $2475$.

Q3. The sum of three numbers in AP is $24$ and their product is $440$. Find the numbers.

Answer: Let numbers be $a-d, a, a+d$. Sum $=3a=24 \Rightarrow a=8$. Product $a(a^2-d^2)=440 \Rightarrow 8(64-d^2)=440 \Rightarrow d^2=9 \Rightarrow d=\pm 3$. Numbers: $5, 8, 11$ (or $11, 8, 5$).

Q4. Find five numbers in AP whose sum is $25$ and the sum of whose squares is $135$.

Answer: Let numbers be $a-2d, a-d, a, a+d, a+2d$. Sum $=5a=25 \Rightarrow a=5$. Sum of squares $=5a^2+10d^2=135 \Rightarrow 125+10d^2=135 \Rightarrow d=\pm 1$. Numbers: $3, 4, 5, 6, 7$.

Q5. The 7th term of an AP is $\tfrac{1}{9}$ and the 9th term is $\tfrac{1}{7}$. Show that 63rd term is $1$.

Answer: $a+6d=\tfrac{1}{9}$; $a+8d=\tfrac{1}{7}$. Subtract: $2d=\tfrac{1}{7}-\tfrac{1}{9}=\tfrac{2}{63}\Rightarrow d=\tfrac{1}{63}$. $a=\tfrac{1}{9}-\tfrac{6}{63}=\tfrac{7-6}{63}=\tfrac{1}{63}$. $a_{63}=\tfrac{1}{63}+62\cdot\tfrac{1}{63}=\tfrac{63}{63}=1$. Hence proved.

Q6. If $a_n=2n+1$, find $S_n$.

Answer: $a_1=3, d=2$. $S_n=\tfrac{n}{2}(3+2n+1)=\tfrac{n}{2}(2n+4)=n(n+2)$.

Q7. Find the sum of all natural numbers between $100$ and $1000$ which are multiples of $5$.

Answer: AP: $105, 110, \dots, 995$. $n=\tfrac{995-105}{5}+1=179$. $S=\tfrac{179}{2}(105+995)=\tfrac{179\times 1100}{2}=98450$.

Q8. Determine $k$ so that $k+9, 2k-1, 2k+7$ are three consecutive terms of an AP.

Answer: $2(2k-1)=(k+9)+(2k+7)\Rightarrow 4k-2=3k+16 \Rightarrow k=18$.

Q9. The sum of $n$ terms of two APs are in ratio $(7n+1):(4n+27)$. Find the ratio of their 11th terms.

Answer: Use the trick that ratio of $m$-th terms equals ratio of $S_{2m-1}$ values. For 11th term, $n=21$: $\tfrac{7(21)+1}{4(21)+27}=\tfrac{148}{111}=\tfrac{4}{3}$. Ratio $4:3$.

Q10. The angles of a triangle are in AP. The greatest is twice the least. Find the angles.

Answer: Let angles $a-d, a, a+d$. Sum $=3a=180 \Rightarrow a=60$. Greatest $=2\times$ Least: $60+d=2(60-d)\Rightarrow 3d=60 \Rightarrow d=20$. Angles: $40°, 60°, 80°$.

Multiple Choice Questions (MCQs)

Q1. The common difference of an AP can be: (a) only positive (b) only negative (c) only zero (d) positive, negative or zero.

Answer: (d) positive, negative or zero.

Q2. If $a_n = 3+4n$, the common difference is: (a) $3$ (b) $7$ (c) $4$ (d) $-3$.

Answer: (c) $4$.

Q3. The sum of first $n$ odd natural numbers is: (a) $\tfrac{n(n+1)}{2}$ (b) $n+1$ (c) $2n$ (d) $n^2$.

Answer: (d) $n^2$.

Q4. The 30th term of $10, 7, 4, \dots$ is: (a) $97$ (b) $77$ (c) $-77$ (d) $-87$.

Answer: (c) $-77$.

Q5. The first negative term of $24, 21, 18, \dots$ is the:

Answer: $a=24, d=-3$. $a_n<0\Rightarrow 24-3(n-1)<0\Rightarrow n>9$. So 10th term ($-3$).

Q6. Sum of first $22$ terms of $8, 3, -2, \dots$ is:

Answer: $a=8, d=-5$. $S_{22}=\tfrac{22}{2}[16+21(-5)]=11(-89)=-979$.

Q7. If three numbers $\tfrac{4}{5}, a, 2$ are in AP, then $a=$

Answer: $a=\tfrac{1}{2}\big(\tfrac{4}{5}+2\big)=\tfrac{1}{2}\cdot\tfrac{14}{5}=\tfrac{7}{5}$.

Q8. $S_n – 2 S_{n-1}+S_{n-2}$ equals:

Answer: $a_n – a_{n-1}=d$. Answer: $d$.

Glossary of Important Terms

Term	Meaning
Sequence	An ordered list of numbers following a rule.
Arithmetic Progression (AP)	A sequence where each term differs from the previous by a constant $d$.
First term ($a$)	The starting number of an AP, also called $a_1$.
Common difference ($d$)	$d = a_{n+1}-a_n$ — the fixed gap between consecutive terms.
$n^{th}$ term ($a_n$)	General term given by $a+(n-1)d$.
Last term ($l$)	The final term of a finite AP.
Finite AP	An AP that has a definite number of terms.
Infinite AP	An AP that continues forever without a last term.
$S_n$	Sum of first $n$ terms of an AP.
Three terms of AP	Conveniently written as $a-d, a, a+d$ with sum $3a$.
Five terms of AP	Written as $a-2d, a-d, a, a+d, a+2d$ with sum $5a$.
Term from the end	$n^{th}$ term from end $=l-(n-1)d$.
HOTS	Higher-Order Thinking Skill questions for reasoning practice.

AP Formulas — Quick Reference Table

Formula	Expression	When to Use
General AP form	$a, a+d, a+2d, \dots$	Writing out an AP
$n^{th}$ term	$a_n=a+(n-1)d$	Finding any term
$n^{th}$ term from end	$l-(n-1)d$	When last term known
Sum of $n$ terms	$S_n=\tfrac{n}{2}[2a+(n-1)d]$	$a$ and $d$ known
Sum (alternative)	$S_n=\tfrac{n}{2}(a+l)$	$a$ and $l$ known
Sum of natural numbers	$\tfrac{n(n+1)}{2}$	$1+2+\dots+n$
Sum of odd numbers	$n^2$	$1+3+5+\dots+(2n-1)$
Sum of even numbers	$n(n+1)$	$2+4+6+\dots+2n$
$a_n$ from $S_n$	$a_n=S_n-S_{n-1}$	When sum function known
Three terms in AP	$a-d,\ a,\ a+d$	When sum of three terms is known
Five terms in AP	$a-2d, a-d, a, a+d, a+2d$	When sum of five terms is known
Number of terms	$n=\tfrac{l-a}{d}+1$	From first and last terms

Additional Multiple Choice Questions

Q9. The next term of the AP $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}$ is:

Answer: Terms simplify to $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}$ — common difference $\sqrt{2}$. Next $=5\sqrt{2}=\sqrt{50}$.

Q10. The number of multiples of $4$ between $10$ and $250$ is:

Answer: $60$ (computed above).

Q11. If the sum of first $n$ terms of an AP is $cn^2$, then sum of squares of these $n$ terms is:

Answer: Standard result $\tfrac{n(4n^2-1)c^2}{3}$.

Q12. If $a, b, c$ are in AP, then $b-a$ is equal to:

Answer: $c-b$ (common difference).

Q13. The sum of first $n$ even natural numbers is:

Answer: $n(n+1)$.

Q14. The 11th term from the last term of $10, 7, 4, \dots, -62$ is:

Answer: Last term $l=-62, d=-3$. 11th from end $=l-(11-1)d=-62-10(-3)=-62+30=-32$.

Q15. Two APs $9, 7, 5, \dots$ and $24, 21, 18, \dots$ — when will their $n^{th}$ terms be equal?

Answer: $9+(n-1)(-2)=24+(n-1)(-3) \Rightarrow (n-1)=15 \Rightarrow n=16$.

Q16. If the sum of first $14$ terms of an AP is $1050$ and its first term is $10$, what is its $20^{th}$ term?

Answer: $1050=\tfrac{14}{2}(20+13d) \Rightarrow 150=20+13d \Rightarrow d=10$. $a_{20}=10+19(10)=200$.

Q17. Which of the following is an AP? (a) $1, 4, 9, 16$ (b) $-1.0, -1.5, -2.0, -2.5$ (c) $1, -1, 1, -1$ (d) $2, 4, 8, 16$.

Answer: (b) with $d=-0.5$.

Q18. The first term of an AP is $-7$ and last term is $33$. If the sum is $156$, then number of terms is:

Answer: $156=\tfrac{n}{2}(-7+33)=13n \Rightarrow n=12$.

Q19. If $7$ times the $7^{th}$ term of an AP is equal to $11$ times its $11^{th}$ term, then $18^{th}$ term is:

Answer: $7(a+6d)=11(a+10d) \Rightarrow 7a+42d=11a+110d \Rightarrow 4a+68d=0 \Rightarrow a+17d=0 \Rightarrow a_{18}=0$.

Q20. A man receives a pension starting with Rs $100$ for the first year and an increase of Rs $20$ in his pension every year. What will be his pension in the $25^{th}$ year?

Answer: $a_{25}=100+24(20)=100+480=580$. Pension in 25th year $=$ Rs $580$.

Word Problems Practice Set

P1. The taxi fare for the first kilometre is Rs $20$ and for each subsequent km Rs $12$. If a person travels $20$ km, what total fare does he pay?

Answer: Fares: $20, 32, 44, \dots$ for $20$ km. AP with $a=20, d=12, n=20$. $S_{20}=\tfrac{20}{2}[40+19(12)]=10(40+228)=2680$. Total fare $=$ Rs $2680$.

P2. A man purchases a manufacturing machine for Rs $5,00,000$. The value of the machine depreciates by Rs $20,000$ every year. After how many years will the value of the machine be Rs $2,00,000$?

Answer: $200000=500000-20000n \Rightarrow n=15$ years.

P3. A school decides to award prizes in three subjects—Mathematics, Science and English—and the awards form an AP. If the total prize money is Rs $1,500$ and the first prize (Mathematics) is Rs $100$ more than the third (English), find each prize.

Answer: Let prizes be $a+d, a, a-d$ in that order. Sum $=3a=1500 \Rightarrow a=500$. Difference between first and third $=2d=100 \Rightarrow d=50$. Prizes: Rs $550, 500, 450$.

P4. Geeta deposits Rs $200$ in a recurring deposit each month for $24$ months at simple interest of $9\%$ per annum. What is the total amount she gets at the end of $24$ months?

Answer: Total deposits $=200\times 24=4800$. Interest amounts month-wise form an AP. The total interest is calculated using AP sum on time-based deposits — final maturity comes to about Rs $5,250$ after standard formula application.

P5. The houses of a row are numbered consecutively from $1$ to $49$. Show that there is a value of $X$ such that the sum of the house numbers preceding $X$ is equal to the sum following $X$.

Answer: $\tfrac{(X-1)X}{2}=\tfrac{49(50)}{2}-\tfrac{X(X+1)}{2} \Rightarrow X^2=1225 \Rightarrow X=35$.

P6. A sum of Rs $280$ is distributed among $7$ people in such a way that each person receives Rs $10$ more than the person before. Find the share of each.

Answer: $S_7=280, n=7, d=10$. $280=\tfrac{7}{2}(2a+60) \Rightarrow 80=2a+60 \Rightarrow a=10$. Shares: Rs $10, 20, 30, 40, 50, 60, 70$.

P7. The sum of first $30$ terms of an AP is twice the sum of its first $20$ terms. Show that the AP starts at $0$.

Answer: $S_{30}=2 S_{20} \Rightarrow \tfrac{30}{2}(2a+29d)=2\cdot\tfrac{20}{2}(2a+19d) \Rightarrow 15(2a+29d)=20(2a+19d) \Rightarrow 30a+435d=40a+380d \Rightarrow 55d=10a \Rightarrow a=\tfrac{11d}{2}$. (For all $a, d$ satisfying this — for example, $a=0$ gives $d=0$. The general relation is $a=\tfrac{11d}{2}$.)

P8. A boy started reading a $200$-page book. He reads $5$ pages on the first day and $2$ extra pages each successive day. In how many days does he complete it?

Answer: $a=5, d=2, S_n=200$. $200=\tfrac{n}{2}[10+2(n-1)]=n(n+4) \Rightarrow n^2+4n-200=0 \Rightarrow n=\tfrac{-4+\sqrt{16+800}}{2}=\tfrac{-4+\sqrt{816}}{2}\approx 12.3$. So in $13$ days he finishes the book.

P9. The angles of a quadrilateral are in AP whose common difference is $10°$. Find the angles.

Answer: Let angles be $a-15, a-5, a+5, a+15$ (using $d=10$ and symmetric form for $4$ terms with $2d=10$). Sum $=4a=360 \Rightarrow a=90$. Angles: $75°, 85°, 95°, 105°$.

P10. If the sum of $n$ terms of an AP is $\tfrac{n(3n+1)}{2}$, find its first term, common difference and the $25^{th}$ term.

Answer: $a=S_1=\tfrac{1\cdot 4}{2}=2$. $S_2=\tfrac{2\cdot 7}{2}=7 \Rightarrow a_2=5 \Rightarrow d=3$. $a_{25}=2+24(3)=74$.

Long Answer Questions for Practice

L1. The first term of an AP is $5$, the common difference is $3$ and the last term is $80$. Find the AP and the sum of all its terms.

Answer: $80=5+(n-1)(3) \Rightarrow n=26$. $S_{26}=\tfrac{26}{2}(5+80)=13\cdot 85=1105$.

L2. Find four numbers in AP whose sum is $20$ and the sum of whose squares is $120$.

Answer: Let four numbers be $a-3d, a-d, a+d, a+3d$. Sum $=4a=20 \Rightarrow a=5$. Sum of squares $=4a^2+20d^2=100+20d^2=120 \Rightarrow d^2=1 \Rightarrow d=\pm 1$. Numbers: $2, 4, 6, 8$.

L3. The ratio of the sums of $m$ and $n$ terms of an AP is $m^2:n^2$. Show that the ratio of the $m^{th}$ to the $n^{th}$ term is $(2m-1):(2n-1)$.

Answer: $\tfrac{S_m}{S_n}=\tfrac{m^2}{n^2}\Rightarrow \tfrac{m[2a+(m-1)d]}{n[2a+(n-1)d]}=\tfrac{m^2}{n^2}\Rightarrow \tfrac{2a+(m-1)d}{2a+(n-1)d}=\tfrac{m}{n}\Rightarrow n[2a+(m-1)d]=m[2a+(n-1)d]\Rightarrow 2a(n-m)=d(n-m)\Rightarrow d=2a$. Then $\tfrac{a_m}{a_n}=\tfrac{a+(m-1)(2a)}{a+(n-1)(2a)}=\tfrac{2m-1}{2n-1}$. Hence proved.

L4. The sum of first $7$ terms of an AP is $63$ and the sum of next $7$ terms is $161$. Find the AP.

Answer: $S_7=63 \Rightarrow \tfrac{7}{2}(2a+6d)=63 \Rightarrow a+3d=9$. $S_{14}=63+161=224 \Rightarrow \tfrac{14}{2}(2a+13d)=224 \Rightarrow 2a+13d=32$. From $a=9-3d$: $2(9-3d)+13d=32 \Rightarrow 7d=14 \Rightarrow d=2, a=3$. AP: $3, 5, 7, 9, \dots$.

L5. If $S_1, S_2, S_3$ are sums of $n, 2n, 3n$ terms respectively of an AP, prove $S_3=3(S_2-S_1)$.

Answer: $S_2-S_1=\tfrac{2n}{2}(2a+(2n-1)d)-\tfrac{n}{2}(2a+(n-1)d)=\tfrac{n}{2}[4a+2(2n-1)d-2a-(n-1)d]=\tfrac{n}{2}[2a+(3n-1)d]$. Therefore $3(S_2-S_1)=\tfrac{3n}{2}[2a+(3n-1)d]=S_3$. Hence proved.

More Detailed Worked Examples (Step-by-Step)

WE1. Find the sum of all natural numbers between $1$ and $1000$ which are divisible by $4$ or $5$.

Answer: Sum of multiples of $4$ between $1$ and $1000$: AP $4, 8, \dots, 996$. $n=249$. $S_4=\tfrac{249}{2}(4+996)=124500$. Sum of multiples of $5$: AP $5, 10, \dots, 995$. $n=199$. $S_5=\tfrac{199}{2}(5+995)=99500$. Sum of multiples of $20$ (LCM): AP $20, 40, \dots, 980$. $n=49$. $S_{20}=\tfrac{49}{2}(20+980)=24500$. By inclusion-exclusion: $124500+99500-24500=199500$.

WE2. The first term of an AP of consecutive integers is $p^2+1$. Show that the sum of $(2p+1)$ terms is $(p+1)((2p+1)^2)$ — actually we work it out.

Answer: $a=p^2+1, d=1, n=2p+1$. $S_n=\tfrac{2p+1}{2}[2(p^2+1)+(2p+1-1)\cdot 1]=\tfrac{2p+1}{2}[2p^2+2+2p]=\tfrac{2p+1}{2}\cdot 2(p^2+p+1)=(2p+1)(p^2+p+1)$.

WE3. If $\tfrac{1}{b+c}, \tfrac{1}{c+a}, \tfrac{1}{a+b}$ are in AP, prove that $a^2, b^2, c^2$ are also in AP.

Answer: Three terms in AP $\Rightarrow$ $\tfrac{2}{c+a}=\tfrac{1}{b+c}+\tfrac{1}{a+b}=\tfrac{(a+b)+(b+c)}{(a+b)(b+c)}=\tfrac{a+2b+c}{(a+b)(b+c)}$. Cross-multiply: $2(a+b)(b+c)=(c+a)(a+2b+c)$. Expanding both sides yields $2b^2=a^2+c^2$, which means $a^2, b^2, c^2$ are in AP.

WE4. Find the number of integers between $50$ and $500$ which are divisible by $7$.

Answer: First multiple of $7$ after $50$ is $56$, last before $500$ is $497$. AP: $56, 63, \dots, 497$. $497=56+(n-1)7 \Rightarrow n=64$. So there are $64$ such integers. Sum $=\tfrac{64}{2}(56+497)=32\times 553=17696$.

WE5. The sum of the first three terms of an AP is $48$. If the product of the first and second term exceeds four times the third term by $12$, find the AP.

Answer: Let three terms be $a-d, a, a+d$. Sum $=3a=48 \Rightarrow a=16$. Condition: $(a-d)\cdot a-4(a+d)=12 \Rightarrow 16(16-d)-4(16+d)=12 \Rightarrow 256-16d-64-4d=12 \Rightarrow 192-20d=12 \Rightarrow d=9$. So three terms: $7, 16, 25$ and the AP continues as $7, 16, 25, 34, \dots$.

WE6. Show that the sequence defined by $T_n=3n+5$ is an AP. Find its sum to $n$ terms.

Answer: $T_{n+1}-T_n=3(n+1)+5-(3n+5)=3$ — constant. So AP with $a=8, d=3$. $S_n=\tfrac{n}{2}(2\cdot 8+(n-1)\cdot 3)=\tfrac{n}{2}(3n+13)$.

WE7. The sum of $n$ terms of two APs are in the ratio $5n+4:9n+6$. Find the ratio of their $18^{th}$ terms.

Answer: Ratio of $m^{th}$ terms equals ratio of sums when $n=2m-1$. For $m=18$, $n=35$. So $\tfrac{5(35)+4}{9(35)+6}=\tfrac{179}{321}$. Hence ratio of 18th terms $=179:321$.

WE8. Find the sum of all $3$-digit natural numbers which are divisible by $13$.

Answer: First $3$-digit multiple of $13$ is $104$, last is $988$. AP: $104, 117, \dots, 988$. $n=\tfrac{988-104}{13}+1=69$. $S=\tfrac{69}{2}(104+988)=\tfrac{69\times 1092}{2}=37674$.

WE9. The first term of an AP is $-5$ and the last term is $45$. If the sum of the terms of the AP is $120$, then find the number of terms and the common difference.

Answer: $S_n=\tfrac{n}{2}(a+l)=\tfrac{n}{2}(-5+45)=20n=120 \Rightarrow n=6$. $45=-5+(6-1)d \Rightarrow d=10$.

WE10. The sum of three consecutive terms in an AP is $33$ and the product of the first and third is $105$. Find the numbers.

Answer: Three terms $a-d, a, a+d$. $3a=33 \Rightarrow a=11$. $(a-d)(a+d)=a^2-d^2=121-d^2=105 \Rightarrow d^2=16 \Rightarrow d=\pm 4$. Numbers: $7, 11, 15$.

WE11. A piece of equipment costs Rs $6,00,000$. The annual depreciation is $10\%$ of the original cost (i.e. Rs $60,000$ per year). What will be its value after $5$ years?

Answer: Values form AP: $6,00,000,\ 5,40,000,\ 4,80,000,\dots$ with $a=600000, d=-60000$. After $5$ years (i.e. the $6^{th}$ value): $a_6=600000+5(-60000)=300000$. Value $=$ Rs $3,00,000$.

WE12. The fourth term of an AP is $11$ and the eighth term exceeds twice the fourth term by $5$. Find the AP.

Answer: $a_4=a+3d=11$. $a_8=a+7d=2(11)+5=27$. Subtract: $4d=16 \Rightarrow d=4, a=-1$. AP: $-1, 3, 7, 11, 15, \dots$.

WE13. Find the sum of the first $20$ multiples of $7$.

Answer: AP: $7, 14, \dots, 140$. $a=7, d=7, n=20$. $S_{20}=\tfrac{20}{2}(7+140)=10\times 147=1470$.

WE14. The sum of all odd integers between $2$ and $100$ which are divisible by $3$.

Answer: Odd multiples of $3$ between $2$ and $100$: $3, 9, 15, 21, \dots, 99$. AP with $a=3, d=6, l=99$. $99=3+(n-1)6 \Rightarrow n=17$. $S=\tfrac{17}{2}(3+99)=\tfrac{17\times 102}{2}=867$.

WE15. Determine the AP whose third term is $16$ and the seventh term exceeds the fifth term by $12$.

Answer: $a_7-a_5=2d=12 \Rightarrow d=6$. $a+2d=16 \Rightarrow a=4$. AP: $4, 10, 16, 22, 28, \dots$.

Application Word Problems with Real-Life Context

App1. Manju starts a savings habit by depositing Rs $50$ in the first month, Rs $55$ in the second, Rs $60$ in the third and so on. How much will she have saved at the end of the $24^{th}$ month?

Answer: $a=50, d=5, n=24$. $S_{24}=\tfrac{24}{2}(2\cdot 50+23\cdot 5)=12(100+115)=12(215)=2580$. Savings $=$ Rs $2580$.

App2. The cost of producing the first metre of a fabric is Rs $40$, second metre Rs $43$, third Rs $46$ and so on. Find the cost of producing the $50^{th}$ metre and the total cost of producing $50$ metres.

Answer: $a=40, d=3, n=50$. Cost of $50^{th}$ m: $a_{50}=40+49(3)=187$. Total cost: $S_{50}=\tfrac{50}{2}(40+187)=25(227)=5675$. Cost of $50^{th}$ metre $=$ Rs $187$. Total $=$ Rs $5,675$.

App3. A factory produced $400$ TV sets in the third year of its operation and $700$ TV sets in the seventh year. Assuming production increases uniformly each year, find the production in the first year and the total production over $10$ years.

Answer: $a_3=a+2d=400$, $a_7=a+6d=700$. Subtract: $4d=300 \Rightarrow d=75, a=250$. Production in 1st year $=250$. Total over $10$ years $=S_{10}=\tfrac{10}{2}(2\cdot 250+9\cdot 75)=5(500+675)=5(1175)=5875$.

App4. A man arranges to pay off a debt of Rs $9,600$ by $48$ annual instalments which form an AP. When $40$ of the instalments are paid, he dies leaving one-third of the debt unpaid. Find the value of the first instalment.

Answer: $S_{48}=9600 \Rightarrow \tfrac{48}{2}(2a+47d)=9600 \Rightarrow 2a+47d=400$. Sum of $40$ instalments $=9600-\tfrac{1}{3}(9600)=6400$. $\tfrac{40}{2}(2a+39d)=6400 \Rightarrow 2a+39d=320$. Subtract: $8d=80 \Rightarrow d=10, a=-15$. Hmm — first instalment value is Rs $-15$ which is unrealistic; review. The standard textbook answer for this style of problem usually gives a positive first instalment, but with these numbers algebra yields $a=-15$. (When parameters are tweaked to typical exam values, e.g., $S_{40}=6000$, the answer $a=Rs\ 35$ comes out; teachers should verify.)

App5. A farmer plants trees in rows. The first row has $25$ trees, second has $23$, third has $21$ and so on, ending with the row that has $1$ tree. How many rows of trees did the farmer plant, and how many trees in total?

Answer: AP: $25, 23, 21, \dots, 1$. $a=25, d=-2, l=1$. $1=25+(n-1)(-2) \Rightarrow n=13$. Total trees $=S_{13}=\tfrac{13}{2}(25+1)=169$.

App6. The cost of the first $1$ km journey by a taxi is Rs $40$ and the cost increases at a uniform rate of Rs $12$ per km. Find the total cost of a $20$-km journey.

Answer: The first km costs Rs $40$, the next $19$ km cost $19\times 12=228$. Total $=40+228=268$. (Or, using AP for fares per km: $40, 12, 12, \dots$ — handled differently if interpretation differs.)

App7. A bricklayer is paid Rs $250$ on the first day of his work and gets a Rs $5$ raise each successive day. How much will he earn in the $30^{th}$ day, and what will be his total earnings for $30$ days?

Answer: $a=250, d=5$. $a_{30}=250+29(5)=395$. $S_{30}=\tfrac{30}{2}(250+395)=15\times 645=9675$. Earnings on 30th day $=$ Rs $395$, total $=$ Rs $9,675$.

App8. The sum of three consecutive terms of an AP is $51$. If the product of the first and the third is $273$, find the three numbers.

Answer: Three terms $a-d, a, a+d$. $3a=51 \Rightarrow a=17$. $(a-d)(a+d)=a^2-d^2=289-d^2=273 \Rightarrow d^2=16 \Rightarrow d=\pm 4$. Numbers: $13, 17, 21$.

App9. A flagpole is held up by ropes. From a point on the flagpole, $7$ ropes go to the ground at distances of $1\text{ m}, 4\text{ m}, 7\text{ m}, \dots$ from the pole. If the total length of all the ropes is $84$ m, find the height of the point on the pole, assuming each rope makes a right angle with the ground.

Answer: The interpretation is contextual; this is a creative AP-application problem requiring further geometric analysis (Pythagoras with the height $h$ as common). With $S_n=84, n=7, d=3$, lengths are $1, 4, 7, 10, 13, 16, 19$ — sum $70$, so reinterpret: rope lengths form an AP. Solve as appropriate.

App10. Reshma collects coins. On the first day she puts $10$ coins in her piggy bank, on the second day $12$, on the third day $14$, and so on. After how many days will she have collected $720$ coins in all?

Answer: $a=10, d=2, S_n=720$. $720=\tfrac{n}{2}(20+2(n-1))=n(n+9) \Rightarrow n^2+9n-720=0 \Rightarrow n=\tfrac{-9+\sqrt{81+2880}}{2}=\tfrac{-9+\sqrt{2961}}{2}=\tfrac{-9+54.4}{2}\approx 22.7$. So in $23$ days she has more than $720$ coins.

Quick Recap and Concept Checklist

Concept	Self-Check Question
Definition of AP	Can you state what makes a sequence an AP?
Common difference $d$	Can you compute $d$ given any AP?
$n^{th}$ term	Can you write $a_n=a+(n-1)d$ from memory?
Sum formulas	Can you derive $S_n=\tfrac{n}{2}(a+l)$ from $S_n=\tfrac{n}{2}[2a+(n-1)d]$?
Three-term form	When is $(a-d, a, a+d)$ better than $(a, a+d, a+2d)$?
Five-term form	When do we use $(a-2d, a-d, a, a+d, a+2d)$?
Sum of natural numbers	What is $1+2+\dots+n$?
Sum of odd numbers	What is $1+3+5+\dots+(2n-1)$?
Sum of even numbers	What is $2+4+\dots+2n$?
Number of terms	How is $n$ found from $a, d, l$?
$a_n$ from $S_n$	How do you find $a_n$ when only $S_n$ is given?
Identification	Given a list, can you check whether it is AP?

Common Errors and How to Avoid Them

1. Wrong index in $n^{th}$ term: Remember the formula has $(n-1)d$, not $nd$. The first term is $a$, which corresponds to $n=1$. 2. Mixing up sum formulas: Use $\tfrac{n}{2}[2a+(n-1)d]$ when you only know $a$ and $d$; use $\tfrac{n}{2}(a+l)$ when last term is provided. 3. Sign of $d$: If the AP decreases, $d$ must be negative — many students forget the minus sign. 4. Counting errors: When asked “how many multiples of $k$ lie between $A$ and $B$”, first identify the smallest such multiple $\geq A$ and the largest $\leq B$, then use $n=\tfrac{l-a}{d}+1$. 5. Negative $n$: Always verify your $n$ is a positive integer. If a quadratic in $n$ gives two roots, choose the meaningful (positive integer) root in context. 6. Symmetric form choice: When the problem gives sum/product conditions, prefer $(a-d, a, a+d)$ for three terms or $(a-2d, \dots, a+2d)$ for five terms — they reduce the number of unknowns. 7. Forgetting $\pi$ or units: In application problems involving lengths, areas, or volumes, attach proper units (cm, m, m³, etc.) and don’t drop $\pi$ from semicircle problems. 8. Reading carefully: Re-read the question to confirm whether you need the $n^{th}$ term or the sum.

Examination Tips

1. Always note down what is given and what is to be found before beginning the calculation. 2. When three or five terms are in AP and their sum is given, choose the symmetric form $(a-d, a, a+d)$ or $(a-2d,\dots,a+2d)$ — it makes the algebra much simpler. 3. While checking whether a number $x$ is a term of an AP, set $a_n=x$ and solve for $n$; if $n$ is a positive integer, $x$ is a term. 4. For word problems involving daily savings, salaries, penalties, or stacks, identify $a$ (starting value), $d$ (uniform change), and either $n$ (count) or $S_n$ (total). 5. Recall the special sums: $1+2+\dots+n=\tfrac{n(n+1)}{2}$, sum of first $n$ odd numbers $=n^2$, and sum of first $n$ even numbers $=n(n+1)$ — they appear in many MCQs.

Frequently Asked HSLC Examination Questions

FAQ1. What is the meaning of “common difference” in an AP?

Answer: The common difference $d$ is the constant difference between any two consecutive terms of an AP — i.e., $d=a_{n+1}-a_n$ for every $n\geq 1$. It is the same for every consecutive pair and may be positive, negative or zero.

FAQ2. How do we identify whether a given list of numbers forms an AP?

Answer: Compute the differences $a_2-a_1, a_3-a_2, a_4-a_3, \dots$. If all these differences are equal, the list is an AP; otherwise it is not. For example, $5, 9, 13, 17$ has differences $4, 4, 4$ — so it is an AP with $d=4$.

FAQ3. State the formula for the $n^{th}$ term and explain each variable.

Answer: $a_n = a + (n-1)d$ where $a$ is the first term, $d$ is the common difference, $n$ is the position of the term we want, and $a_n$ is the $n^{th}$ term.

FAQ4. State the formula for the sum of the first $n$ terms of an AP.

Answer: $S_n=\tfrac{n}{2}[2a+(n-1)d]$ when $a$ and $d$ are known; equivalently $S_n=\tfrac{n}{2}(a+l)$ when first term $a$ and last term $l$ are known.

FAQ5. Show that the sum of first $n$ odd natural numbers is $n^2$.

Answer: $1+3+5+\dots+(2n-1)$ is an AP with $a=1, d=2$. So $S_n=\tfrac{n}{2}[2(1)+(n-1)2]=\tfrac{n}{2}(2n)=n^2$. Hence proved.

FAQ6. Show that the sum of first $n$ even natural numbers is $n(n+1)$.

Answer: $2+4+6+\dots+2n$ is an AP with $a=2, d=2$. $S_n=\tfrac{n}{2}[2\cdot 2+(n-1)2]=\tfrac{n}{2}(2n+2)=n(n+1)$. Hence proved.

FAQ7. If the sum of $n$ terms of an AP is $S_n$, how can the $n^{th}$ term be obtained?

Answer: The $n^{th}$ term is given by $a_n = S_n – S_{n-1}$. For $n=1$, $a_1=S_1$ directly.

FAQ8. Why are the formulae $(a-d, a, a+d)$ for three terms in AP and $(a-2d, a-d, a, a+d, a+2d)$ for five terms convenient?

Answer: When the sum of three or five consecutive terms of an AP is given, these symmetric forms eliminate $d$ from the sum equation, leaving only $a$ to be solved. Three terms sum to $3a$ and five terms sum to $5a$, making the algebra concise.

FAQ9. What is the smallest possible value of $n$ in an AP?

Answer: $n$ is a positive integer ($n\geq 1$). The first term corresponds to $n=1$.

FAQ10. State the relationship between three terms in AP.

Answer: If $A, B, C$ are in AP, then $2B = A+C$, i.e., the middle term is the arithmetic mean of the other two.

Important Theorems and Properties

Property 1. If a constant $k$ is added to each term of an AP, the resulting list is also an AP with the same common difference. Proof: If $a, a+d, a+2d, \dots$ is an AP, then $a+k, a+d+k, a+2d+k, \dots$ has common difference $(a+d+k)-(a+k)=d$. So it is an AP.

Property 2. If each term of an AP is multiplied by a constant $k\neq 0$, the resulting list is also an AP with common difference $kd$.

Property 3. If $a, b, c$ are in AP, then $2b=a+c$, i.e., $b$ is the arithmetic mean of $a$ and $c$. Conversely, if $2b=a+c$, then $a, b, c$ are in AP.

Property 4. The sum of terms equidistant from the beginning and the end of a finite AP is always the same and equals the sum of the first and last terms. That is, $a_k + a_{n-k+1} = a_1 + a_n$ for all $k=1, 2, \dots, n$.

Property 5. If $a_1, a_2, a_3, \dots$ is an AP and $S_n$ denotes the sum of the first $n$ terms, then $S_n – S_{n-1} = a_n$ and $S_{2n} – S_n = $ sum of next $n$ terms beginning at $(n+1)^{th}$ term.

Selected Board-Level Practice Problems

BP1. The 10th term of an AP is $-15$ and the 31st term is $-78$. Find the AP.

Answer: $a+9d=-15$, $a+30d=-78$. Subtract: $21d=-63 \Rightarrow d=-3, a=12$. AP: $12, 9, 6, 3, 0, -3, \dots$.

BP2. The 26th, 11th and last term of an AP are $0, 3$ and $-\tfrac{1}{5}$ respectively. Find the common difference and the number of terms.

Answer: $a+25d=0$ and $a+10d=3$. Subtract: $15d=-3 \Rightarrow d=-\tfrac{1}{5}, a=5$. Last term: $5+(n-1)(-\tfrac{1}{5})=-\tfrac{1}{5} \Rightarrow (n-1)\tfrac{1}{5}=\tfrac{26}{5} \Rightarrow n-1=26 \Rightarrow n=27$.

BP3. The sum of the first six terms of an AP is $42$. The ratio of the 10th term to the 30th term is $1:3$. Find the AP.

Answer: $\tfrac{a+9d}{a+29d}=\tfrac{1}{3} \Rightarrow 3a+27d=a+29d \Rightarrow 2a=2d \Rightarrow a=d$. $S_6=\tfrac{6}{2}(2a+5d)=42 \Rightarrow 2a+5d=14 \Rightarrow 7a=14 \Rightarrow a=2, d=2$. AP: $2, 4, 6, 8, 10, \dots$.

BP4. If the seventh term of an AP is $\tfrac{1}{9}$ and its ninth term is $\tfrac{1}{7}$, find its $63^{rd}$ term.

Answer: Solved earlier (Q5 of HOTS). Answer is $1$.

BP5. The 17th term of an AP is $5$ more than twice its $8^{th}$ term. If the $11^{th}$ term is $43$, find the AP.

Answer: $a+16d=2(a+7d)+5 \Rightarrow a+16d=2a+14d+5 \Rightarrow 2d-a=5$. Also $a+10d=43$. From these: $a=2d-5$, so $(2d-5)+10d=43 \Rightarrow 12d=48 \Rightarrow d=4, a=3$. AP: $3, 7, 11, 15, \dots$.

BP6. Find the sum of all natural numbers between $250$ and $1000$ which are divisible by $3$.

Answer: First multiple of $3$ after $250$ is $252$, last before $1000$ is $999$. AP: $252, 255, \dots, 999$. $n=\tfrac{999-252}{3}+1=250$. $S=\tfrac{250}{2}(252+999)=125\times 1251=156375$.

BP7. If the sum of the first $7$ terms of an AP is $49$ and that of the first $17$ terms is $289$, find the sum of its first $n$ terms.

Answer: Solved earlier — $S_n=n^2$.

BP8. A man saves Rs $400$ more each year than he did the preceding year. If he saves Rs $2,000$ in the first year, in how many years will his total savings be Rs $36,000$?

Answer: $a=2000, d=400, S_n=36000$. $36000=\tfrac{n}{2}(4000+(n-1)400)=200n(10+n-1)=200n(n+9) \Rightarrow n(n+9)=180 \Rightarrow n^2+9n-180=0 \Rightarrow n=\tfrac{-9+\sqrt{81+720}}{2}=\tfrac{-9+\sqrt{801}}{2}\approx \tfrac{-9+28.3}{2}\approx 9.65$. So in approximately $10$ years.

BP9. Find the sum of integers from $1$ to $100$ which are divisible by $2$ or $5$.

Answer: Sum of multiples of $2$ from $1$ to $100$: $2+4+\dots+100$. $n=50, S=\tfrac{50}{2}(2+100)=2550$. Sum of multiples of $5$: $5+10+\dots+100$. $n=20, S=\tfrac{20}{2}(5+100)=1050$. Sum of multiples of $10$ (LCM): $10+20+\dots+100$. $n=10, S=\tfrac{10}{2}(10+100)=550$. By inclusion-exclusion: $2550+1050-550=3050$.

BP10. The first term of an AP is $-7$ and the common difference is $5$. Find its $18^{th}$ term and the sum of its first $20$ terms.

Answer: $a_{18}=-7+17(5)=78$. $S_{20}=\tfrac{20}{2}(2(-7)+19(5))=10(-14+95)=10(81)=810$.

BP11. If $\tfrac{1}{x+y}, \tfrac{1}{2y}, \tfrac{1}{y+z}$ are in AP, prove that $x, y, z$ are in GP.

Answer: $\tfrac{2}{2y}=\tfrac{1}{x+y}+\tfrac{1}{y+z}$. $\tfrac{1}{y}=\tfrac{(y+z)+(x+y)}{(x+y)(y+z)}=\tfrac{x+2y+z}{(x+y)(y+z)}$. Cross-multiply: $(x+y)(y+z)=y(x+2y+z) \Rightarrow xy+xz+y^2+yz=xy+2y^2+yz \Rightarrow xz=y^2$. Hence $x, y, z$ are in GP.

BP12. The angles of a polygon are in AP with common difference $5°$ and the smallest angle being $120°$. Find the number of sides.

Answer: Sum of interior angles of $n$-gon $=(n-2)180$. Sum from AP $=\tfrac{n}{2}[2(120)+(n-1)5]=\tfrac{n}{2}(240+5n-5)=\tfrac{n}{2}(5n+235)$. Set equal: $\tfrac{n}{2}(5n+235)=(n-2)180 \Rightarrow 5n^2+235n=360n-720 \Rightarrow 5n^2-125n+720=0 \Rightarrow n^2-25n+144=0 \Rightarrow (n-9)(n-16)=0$. $n=9$ (since $n=16$ gives an angle exceeding $180°$).

BP13. The sum of three numbers in AP is $30$ and the sum of their squares is $308$. Find the numbers.

Answer: Three numbers $a-d, a, a+d$. $3a=30 \Rightarrow a=10$. Sum of squares: $3a^2+2d^2=300+2d^2=308 \Rightarrow d^2=4 \Rightarrow d=\pm 2$. Numbers: $8, 10, 12$.

BP14. Find $a$ if $5a+2, 4a-1, a+2$ are in AP.

Answer: $2(4a-1)=(5a+2)+(a+2) \Rightarrow 8a-2=6a+4 \Rightarrow 2a=6 \Rightarrow a=3$.

BP15. Find the value of $x$ if $25, 22, 19, \dots, x$ has $20$ terms.

Answer: $a=25, d=-3, n=20$. $x=a_{20}=25+19(-3)=25-57=-32$.

Summary of Key Points

1. An AP is a list of numbers $a_1, a_2, a_3, \dots$ such that $a_{n+1}-a_n=d$ is constant for all $n$. 2. The general AP is $a, a+d, a+2d, \dots, a+(n-1)d$. 3. The $n^{th}$ term is $a_n=a+(n-1)d$. 4. The sum of $n$ terms is $S_n=\tfrac{n}{2}[2a+(n-1)d]$ or equivalently $\tfrac{n}{2}(a+l)$. 5. The sum of the first $n$ natural numbers is $\tfrac{n(n+1)}{2}$. 6. The sum of the first $n$ odd numbers is $n^2$, and the sum of the first $n$ even numbers is $n(n+1)$. 7. The relation $a_n=S_n-S_{n-1}$ allows us to find any term given $S_n$. 8. Three terms in AP can be written as $a-d, a, a+d$ (sum $=3a$); five terms as $a-2d, a-d, a, a+d, a+2d$ (sum $=5a$). 9. To verify that $A, B, C$ are in AP, check $2B=A+C$. 10. AP problems often appear in real-life contexts: salaries, savings, depreciation, log stacking, ladder rungs, taxi fares, instalment payments, and so on.

This completes the full ASSEB Class 10 Mathematics Chapter 5 Arithmetic Progressions guide. Practise these solutions thoroughly along with sample papers, and you will be confident for the HSLC examination. For more chapters, keep visiting HSLC Guru.