Exercise 13.1: Arithmetic Series
1. What is the number of terms in the series $13 + 20 + 27 + 34 + \dots + 111$?
a) $10$
b) $13$
c) $15$
d) $20$
Solution:
First term of the series, $a = 13$
Common difference, $d = 20 – 13 = 7$
Therefore, this is an arithmetic progression (AP).
Let, the $n$-th term of the series $= 111$
We know, $n$-th term $= a + (n – 1)d$
or, $13 + (n – 1)7 = 111$
or, $7(n – 1) = 111 – 13$
or, $7(n – 1) = 98$
or, $n – 1 = \frac{98}{7}$
or, $n – 1 = 14$
or, $n = 15$
Answer: c) $15$
2. The series $5 + 8 + 11 + 14 + \dots + 62$ —
(i) is a finite series
(ii) is a geometric progression
(iii) 19th term of the series is $59$
Which of the following is correct?
1) i & ii
2) i & iii
3) ii & iii
4) i, ii & iii
Solution:
Since the last term $62$ is specified, it is a finite series. ((i) is correct)
Common difference $8 – 5 = 11 – 8 = 3$ (equal), so it is an arithmetic progression, not a geometric progression. ((ii) is incorrect)
$19$-th term $= a + (19 – 1)d = 5 + 18 \times 3 = 5 + 54 = 59$. ((iii) is correct)
Answer: b) i & iii
Based on the following information answer questions 3 – 4.
$7 + 13 + 19 + 25 + \dots$ is a series.
Here, first term $a = 7$ and common difference $d = 13 – 7 = 6$.
3. Which one is the $15$-th term of the series?
a) $85$
b) $91$
c) $97$
d) $104$
Solution:
$15$-th term $= a + (15 – 1)d$
$= 7 + 14 \times 6$
$= 7 + 84$
$= 91$
Answer: b) $91$
4. What is the summation of first $20$ terms of the series?
a) $141$
b) $1210$
c) $1280$
d) $2560$
Solution: We know,
Sum of the first $n$ terms, $S_n = \frac{n}{2} [2a + (n – 1)d]$
Therefore, the sum of the first $20$ terms,
$S_{20} = \frac{20}{2} [2(7) + (20 – 1)6]$
$= 10 \times [14 + 19 \times 6]$
$= 10 \times [14 + 114]$
$= 10 \times 128$
$= 1280$
Answer: c) $1280$
5. Find the common difference and the $12$-th term of the series $2 – 5 – 12 – 19 – \dots$.
Solution: The series is: $2 + (-5) + (-12) + (-19) + \dots$
First term, $a = 2$
Common difference, $d = (-5) – 2 = -7$
Therefore, it is an arithmetic series.
We know, $n$-th term $= a + (n – 1)d$
$\therefore$ $12$-th term $= a + (12 – 1)d$
$= 2 + 11 \times (-7)$
$= 2 – 77$
$= -75$
Answer: Common difference is $-7$ and $12$-th term is $-75$.
6. Which term of the series $8 + 11 + 14 + 17 + \dots$ is $392$?
Solution: Here, first term, $a = 8$
Common difference, $d = 11 – 8 = 3$
Therefore, it is an arithmetic series.
Let, the $n$-th term of the series $= 392$
We know, $a + (n – 1)d = 392$
or, $8 + (n – 1)3 = 392$
or, $3(n – 1) = 392 – 8$
or, $3(n – 1) = 384$
or, $n – 1 = \frac{384}{3}$
or, $n – 1 = 128$
or, $n = 128 + 1$
or, $n = 129$
Answer: The $129$-th term of the series is $392$.
7. Which term of the series $4 + 7 + 10 + 13 + \dots$ is $301$?
Solution: Here, first term, $a = 4$
Common difference, $d = 7 – 4 = 3$
Therefore, it is an arithmetic series.
Let, the $n$-th term of the series $= 301$
We know, $a + (n – 1)d = 301$
or, $4 + (n – 1)3 = 301$
or, $3(n – 1) = 301 – 4$
or, $3(n – 1) = 297$
or, $n – 1 = \frac{297}{3}$
or, $n – 1 = 99$
or, $n = 99 + 1$
or, $n = 100$
Answer: The $100$-th term of the series is $301$.
8. If the $m$-th term of an arithmetic progression is $n$ and the $n$-th term is $m$, what is the $(m + n)$-th term of that series?
Solution:
Let the first term of the arithmetic progression $= a$ and the common difference $= d$.
According to the question,
$m$-th term $= a + (m – 1)d = n$ ——- (1)
$n$-th term $= a + (n – 1)d = m$ ——- (2)
Subtracting equation (2) from equation (1), we get
$\{a + (m – 1)d\} – \{a + (n – 1)d\} = n – m$
or, $(m – 1)d – (n – 1)d = -(m – n)$
or, $(m – 1 – n + 1)d = -(m – n)$
or, $(m – n)d = -(m – n)$
or, $d = -1$
Now, the $(m + n)$-th term of the series $= a + (m + n – 1)d$
$= a + (m – 1 + n)d$
$= a + (m – 1)d + nd$
$= n + n(-1)$ [ substituting $a + (m – 1)d = n$ from (1) and $d = -1$ ]
$= n – n$
$= 0$
Answer: $0$
9. What is the sum of $n$ terms of the series $1 + 3 + 5 + 7 + \dots$?
Solution: Here, first term $a = 1$ and common difference $d = 3 – 1 = 2$.
Therefore, it is an arithmetic series.
We know, the sum of $n$ terms,
$S_n = \frac{n}{2} [2a + (n – 1)d]$
$= \frac{n}{2} [2(1) + (n – 1)2]$
$= \frac{n}{2} [2 + 2n – 2]$
$= \frac{n}{2} [2n]$
$= n^2$
Answer: $n^2$
10. What is the sum of the first $9$ terms of the series $8 + 16 + 24 + \dots$?
Solution: Here, first term, $a = 8$
Common difference, $d = 16 – 8 = 8$
Therefore, it is an arithmetic series.
Number of terms, $n = 9$
We know, the sum of $n$ terms,
$S_n = \frac{n}{2} [2a + (n – 1)d]$
$S_9 = \frac{9}{2} [2a + (9 – 1)d]$
$= \frac{9}{2} [2(8) + 8(8)]$
$= \frac{9}{2} [16 + 64]$
$= \frac{9}{2} \times 80$
$= 9 \times 40$
$= 360$
Answer: $360$
11. $5 + 11 + 17 + 23 + \dots + 59 = what?$
Solution: Here, first term, $a = 5$
Common difference, $d = 11 – 5 = 6$
Last term $= 59$
Therefore, it is an arithmetic series.
Let, the $n$-th term of the series $= 59$
or, $a + (n – 1)d = 59$
or, $5 + (n – 1)6 = 59$
or, $6(n – 1) = 59 – 5$
or, $6(n – 1) = 54$
or, $n – 1 = 9$
or, $n = 10$
We know, the sum of $n$ terms,
$S_n = \frac{n}{2} [2a + (n – 1)d]$
Therefore, the sum of the first $10$ terms,
$S_{10} = \frac{10}{2} [2(5) + (10 – 1)6]$
$= 5 \times [10 + 9 \times 6]$
$= 5 \times [10 + 54]$
$= 5 \times 64$
$= 320$
Answer: $320$
12. $29 + 25 + 21 + \dots – 23 = what?$
Solution: Here, first term, $a = 29$
Common difference, $d = 25 – 29 = -4$
Last term $= -23$
Therefore, it is an arithmetic series.
Let, the $n$-th term of the series $= -23$
or, $a + (n – 1)d = -23$
or, $29 + (n – 1)(-4) = -23$
or, $-4(n – 1) = -23 – 29$
or, $-4(n – 1) = -52$
or, $n – 1 = \frac{-52}{-4}$
or, $n – 1 = 13$
or, $n = 14$
We know, the sum of $n$ terms,
$S_n = \frac{n}{2} [2a + (n – 1)d]$
Therefore, the sum of the first $14$ terms,
$S_{14} = \frac{14}{2} [2(29) + (14 – 1)(-4)]$
$= 7 \times [58 + 13(-4)]$
$= 7 \times [58 – 52]$
$= 7 \times 6$
$= 42$
Answer: $42$
13. If the $12$-th term of an arithmetic series is $77$, what is the summation of its first $23$ terms of that series?
Solution: Let the first term $= a$ and common difference $= d$.
Given,
$12$-th term $= a + (12 – 1)d = 77$
or, $a + 11d = 77$ ——- (1)
We know, the sum of $n$ terms,
$S_n = \frac{n}{2} [2a + (n – 1)d]$
Sum of the first $23$ terms,
$S_{23} = \frac{23}{2} [2a + (23 – 1)d]$
$= \frac{23}{2} [2a + 22d]$
$= \frac{23}{2} \times 2(a + 11d)$
$= 23 \times (a + 11d)$
$= 23 \times 77$ [ substituting value from (1) ]
$= 1771$
Answer: $1771$
14. If the $16$-th term of an arithmetic progression is $-20$, what is the sum of its first $31$ terms?
Solution: Let, the first term $= a$ and common difference $= d$.
Given,
$16$-th term $= a + (16 – 1)d = -20$
or, $a + 15d = -20$ ——- (1)
Sum of the first $31$ terms,
$S_{31} = \frac{31}{2} [2a + (31 – 1)d]$
$= \frac{31}{2} [2a + 30d]$
$= \frac{31}{2} \times 2(a + 15d)$
$= 31 \times (a + 15d)$
$= 31 \times (-20)$ [ substituting value from (1) ]
$= -620$
Answer: $-620$
15. If the sum of the first $n$ terms of the series $9 + 7 + 5 + \dots$ is $-144$, find the value of $n$.
Solution: Here, first term, $a = 9$
Common difference, $d = 7 – 9 = -2$
Therefore, it is an arithmetic series.
According to the question,
Sum of $n$ terms $= -144$
$\frac{n}{2} [2a + (n – 1)d] = -144$
or, $\frac{n}{2} [2(9) + (n – 1)(-2)] = -144$
or, $\frac{n}{2} [18 – 2n + 2] = -144$
or, $\frac{n}{2} [20 – 2n] = -144$
or, $n(10 – n) = -144$
or, $10n – n^2 = -144$
or, $n^2 – 10n – 144 = 0$
or, $n^2 – 18n + 8n – 144 = 0$
or, $n(n – 18) + 8(n – 18) = 0$
or, $(n – 18)(n + 8) = 0$
Either, $n – 18 = 0 \implies n = 18$
Or, $n + 8 = 0 \implies n = -8$ (Not acceptable, because the number of terms cannot be negative)
Answer: $n = 18$
16. The sum of the first $n$ terms of the series $2 + 4 + 6 + 8 + \dots$ is $2550$. Find the value of $n$.
Solution: Here, first term, $a = 2$
Common difference, $d = 4 – 2 = 2$
Therefore, it is an arithmetic progression.
According to the question,
$\frac{n}{2} [2a + (n – 1)d] = 2550$
or, $\frac{n}{2} [2(2) + (n – 1)2] = 2550$
or, $\frac{n}{2} [4 + 2n – 2] = 2550$
or, $\frac{n}{2} [2n + 2] = 2550$
or, $n(n + 1) = 2550$
or, $n^2 + n – 2550 = 0$
or, $n^2 + 51n – 50n – 2550 = 0$
or, $n(n + 51) – 50(n + 51) = 0$
or, $(n + 51)(n – 50) = 0$
Either, $n – 50 = 0 \implies n = 50$
Or, $n + 51 = 0 \implies n = -51$ (Not acceptable)
Answer: $n = 50$
17. If the sum of the first $n$ terms of a series is $n(n + 1)$, find the series.
Solution: Given, sum of $n$ terms, $S_n = n(n + 1)$
Sum of the 1st term, $S_1 = 1(1 + 1) = 2$
Therefore, 1st term $= S_1 = 2$
Sum of the first 2 terms, $S_2 = 2(2 + 1) = 6$
Therefore, 2nd term $= S_2 – S_1 = 6 – 2 = 4$
Sum of the first 3 terms, $S_3 = 3(3 + 1) = 12$
Therefore, 3rd term $= S_3 – S_2 = 12 – 6 = 6$
Sum of the first 4 terms, $S_4 = 4(4 + 1) = 20$
Therefore, 4th term $= S_4 – S_3 = 20 – 12 = 8$
Therefore, the series is: $2 + 4 + 6 + 8 + \dots$
Answer: $2 + 4 + 6 + 8 + \dots$
18. If the sum of the first $n$ terms of a series is $n(n + 1)$. What is the sum of the first $10$ terms?
Solution: Given, sum of $n$ terms, $S_n = n(n + 1)$
Here, we need to find the sum of $10$ terms (i.e., $n = 10$):
$S_{10} = 10(10 + 1)$
$= 10 \times 11$
$= 110$
Answer: $110$
19. If the sum of $12$ terms of an arithmetic series is $144$ and the first $20$ terms is $560$, find the sum of the first $6$ terms.
Solution: Let, the first term $= a$ and common difference $= d$.
By the first condition,
$S_{12} = \frac{12}{2} [2a + (12 – 1)d] = 144$
or, $6 [2a + 11d] = 144$
or, $2a + 11d = \frac{144}{6}$
or, $2a + 11d = 24$ ——- (1)
By the second condition,
$S_{20} = \frac{20}{2} [2a + (20 – 1)d] = 560$
or, $10 [2a + 19d] = 560$
or, $2a + 19d = 56$ ——- (2)
Subtracting equation (1) from equation (2), we get
$(2a + 19d) – (2a + 11d) = 56 – 24$
or, $8d = 32$
or, $d = 4$
Substituting the value of $d$ into equation (1), we get
$2a + 11(4) = 24$
or, $2a + 44 = 24$
or, $2a = 24 – 44$
or, $2a = -20$
or, $a = -10$
Therefore, the sum of the first $6$ terms,
$S_6 = \frac{6}{2} [2a + (6 – 1)d]$
$= 3 [2(-10) + 5(4)]$
$= 3 [-20 + 20]$
$= 3 \times 0$
$= 0$
Answer: $0$
20. The sum of the first $m$ terms of an arithmetic series is $n$ and the first $n$ terms is $m$. Find the sum of the first $(m + n)$ terms.
Solution: Let, the first term of the arithmetic progression $= a$ and common difference $= d$.
By the first condition,
$S_m = \frac{m}{2} [2a + (m – 1)d] = n$
or, $m [2a + (m – 1)d] = 2n$
or, $2am + m(m – 1)d = 2n$ ——- (1)
By the second condition,
$S_n = \frac{n}{2} [2a + (n – 1)d] = m$
or, $n [2a + (n – 1)d] = 2m$
or, $2an + n(n – 1)d = 2m$ ——- (2)
Subtracting equation (2) from equation (1), we get
$2am – 2an + \{m(m – 1) – n(n – 1)\}d = 2n – 2m$
or, $2a(m – n) + (m^2 – m – n^2 + n)d = -2(m – n)$
or, $2a(m – n) + \{(m^2 – n^2) – (m – n)\}d = -2(m – n)$
or, $2a(m – n) + \{(m – n)(m + n) – (m – n)\}d = -2(m – n)$
or, $(m – n) [2a + (m + n – 1)d] = -2(m – n)$
or, $2a + (m + n – 1)d = -2$ ——- (3)
Now, the sum of the first $(m + n)$ terms,
$S_{m+n} = \frac{m + n}{2} [2a + (m + n – 1)d]$
$= \frac{m + n}{2} (-2)$ [ substituting value from (3) ]
$= -(m + n)$
Answer: $-(m + n)$
21. In an arithmetic progression, if the $p$-th, $q$-th, and $r$-th terms are $a, b, c$ respectively, show that $a(q – r) + b(r – p) + c(p – q) = 0$.
Solution: Let, the first term of the arithmetic progression $= X$ and common difference $= Y$.
Given,
$p$-th term $= X + (p – 1)Y = a$ ——- (1)
$q$-th term $= X + (q – 1)Y = b$ ——- (2)
$r$-th term $= X + (r – 1)Y = c$ ——- (3)
Now, $\text{L.H.S.} = a(q – r) + b(r – p) + c(p – q)$
$= \{X + (p – 1)Y\}(q – r) + \{X + (q – 1)Y\}(r – p) + \{X + (r – 1)Y\}(p – q)$ [ substituting values of $a, b, c$ ]
$= X(q – r) + (p – 1)Y(q – r) + X(r – p) + (q – 1)Y(r – p) + X(p – q) + (r – 1)Y(p – q)$
$= X(q – r + r – p + p – q) + Y\{(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)\}$
$= X(0) + Y\{pq – pr – q + r + qr – pq – r + p + pr – qr – p + q\}$
$= 0 + Y(0)$
$= 0 = \text{R.H.S.}$ (Shown)
22. Show that $1 + 3 + 5 + 7 + \dots + 125 = 169 + 171 + 173 + \dots + 209$.
Solution: $\text{L.H.S.}= 1 + 3 + 5 + 7 + \dots + 125$
This is an arithmetic series, where first term $a_1 = 1$, common difference $d_1 = 2$.
Let, the number of terms $= n_1$.
$n_1$-th term $= 1 + (n_1 – 1)2 = 125 \implies 2(n_1 – 1) = 124 \implies n_1 – 1 = 62 \implies n_1 = 63$
$\text{Sum} = \frac{63}{2} (1 + 125) = \frac{63}{2} \times 126 = 63 \times 63 = 3969$
$\text{R.H.S.}: 169 + 171 + 173 + \dots + 209$
This is also an arithmetic progression, where first term $a_2 = 169$, common difference $d_2 = 2$.
Let the number of terms $= n_2$.
$n_2$-th term $= 169 + (n_2 – 1)2 = 209 \implies 2(n_2 – 1) = 40 \implies n_2 – 1 = 20 \implies n_2 = 21$
$\text{Sum} = \frac{21}{2} (169 + 209) = \frac{21}{2} \times 378 = 21 \times 189 = 3969$
Since $\text{L.H.S.} = \text{R.H.S.} = 3969$,
Therefore, $1 + 3 + 5 + 7 + \dots + 125 = 169 + 171 + 173 + \dots + 209$. (Shown)
23. A man agrees to refund the loan of Tk. $2500$ in some installments. Each installment is Tk. $2$ more than the previous one. If the first installment is Tk. $1$, in how many installments will the man be able to repay the loan?
Solution: Here,
First installment, $a = 1$
Increase in each installment (common difference), $d = 2$
Total repayment amount (sum), $S_n = 2500$ Taka
Let, the number of installments $= n$
We know, $S_n = \frac{n}{2} [2a + (n – 1)d]$
or, $2500 = \frac{n}{2} [2(1) + (n – 1)2]$
or, $2500 = \frac{n}{2} [2 + 2n – 2]$
or, $2500 = \frac{n}{2} [2n]$
or, $2500 = n^2$
or, $n = \sqrt{2500}$
or, $n = 50$
Answer: In $50$ installments.
24. The $l$-th term of an arithmetic series is $l^2$ and the $k$-th term is $k^2$.
1) Construct two equations according to the information of the stem considering $a$ as the first term of the series and $d$ as common difference as.
Solution:
Given, first term $= a$ and common difference $= d$.
According to the stem,
$l$-th term $= a + (l – 1)d = l^2$ ——- (i)
$k$-th term $= a + (k – 1)d = k^2$ ——- (ii)
2) Find the $(l + k)$-th term.
Solution: From ‘1’ we have:
$a + (l – 1)d = l^2$ ——- (i)
$a + (k – 1)d = k^2$ ——- (ii)
Subtracting equation (ii) from equation (i), we get
$(l – 1)d – (k – 1)d = l^2 – k^2$
or, $(l – 1 – k + 1)d = (l – k)(l + k)$
or, $(l – k)d = (l – k)(l + k)$
or, $d = l + k$ [ dividing both sides by $(l – k)$ where $l \neq k$ ]
Now, $(l + k)$-th term $= a + (l + k – 1)d$
$= a + (l – 1 + k)d$
$= a + (l – 1)d + kd$
$= l^2 + k(l + k)$ [ substituting $a + (l – 1)d = l^2$ from (i) and $d = l + k$ ]
$= l^2 + kl + k^2$
$= l^2 + lk + k^2$
Answer: $l^2 + lk + k^2$
3) Prove that the summation of the first $(l + k)$ terms of the series is $\frac{l + k}{2} (l^2 + k^2 + l + k)$.
Solution: From ‘1’ we have:
$a + (l – 1)d = l^2$ ——- (i)
$a + (k – 1)d = k^2$ ——- (ii)
Adding equations (i) and (ii), we get
$a + (l – 1)d + a + (k – 1)d = l^2 + k^2$
or, $2a + (l – 1 + k – 1)d = l^2 + k^2$
or, $2a + (l + k – 2)d = l^2 + k^2$
or, $2a + (l + k – 1)d – d = l^2 + k^2$
or, $2a + (l + k – 1)d = l^2 + k^2 + d$
Substituting $d = l + k$ from ‘2’:
$2a + (l + k – 1)d = l^2 + k^2 + l + k$ ——- (iii)
Now, the sum of the first $(l + k)$ terms of the series:
$S_{l+k} = \frac{l + k}{2} [2a + (l + k – 1)d]$
Substituting value from equation (iii), we get
$S_{l+k} = \frac{l + k}{2} (l^2 + k^2 + l + k)$ (Proved)