In an AP:

(i) given a = 5, d = 3, an = 50, find n and Sn .

(ii) given a = 7, a13 = 35, find d and S13.

(iii) given a12 = 37, d = 3, find a and S12.

(iv) given a3 = 15, S10 = 125, find d and a10.

(v) given d = 5, S9 = 75, find a and a9 .

(vi) given a = 2, d = 8, Sn = 90, find n and an .

(vii) given a = 8, an = 62, Sn = 210, find n and d.

(viii) given an = 4, d = 2, Sn = –14, find n and a.

(ix) given a = 3, n = 8, S = 192, find d.

(x) given l = 28, S = 144, and there are total 9 terms. Find a.

Question

In an AP:

(i) given a = 5, d = 3, an = 50, find n and Sn .

(ii) given a = 7, a13 = 35, find d and S13.

(iii) given a12 = 37, d = 3, find a and S12.

(iv) given a3 = 15, S10 = 125, find d and a10.

(v) given d = 5, S9 = 75, find a and a9 .

(vi) given a = 2, d = 8, Sn = 90, find n and an .

(vii) given a = 8, an = 62, Sn = 210, find n and d.

(viii) given an = 4, d = 2, Sn = –14, find n and a.

(ix) given a = 3, n = 8, S = 192, find d.

(x) given l = 28, S = 144, and there are total 9 terms. Find a.

Tasneem · Accepted Answer

(i) Given, $a = 5, d = 3, a_{n} = 50$

$a_{n}= a + (n -1)d$

∴ $50 = 5 + (n -1)3$

$45 = (n -1)3$

$15 = n - 1$

n = 16

$S_{n}=\frac{n}{2}[a+a_{n}]$

$S_{16}=\frac{16}{2}[5+50]$ $=440$

(ii)Given, $a = 7, a_{13}= 35$

$a_{n}= a + (n -1)d$

$herefore a_{13} = a + (13 -1) d$

35 = 7 + 12 d

$35 - 7 = 12d$

28 = 12d

$d=\frac{7}{3}$

$S_{13}=\frac{12}{2}[7+35] =273$

(iii) Given, $a_{12}$ = 37, d = 3

a₁₂ = a + (12 − 1)3

37 = a + 33

a = 4

$S_{13}=\frac{12}{2}[4+37] =246$

(iv) Given, $a_{3}= 15, S_{10} = 125$

$a_{3} = a + (3 -1)d$

$15 = a + 2d....... (i)$

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

$S_{10}=\frac{10}{2}[2a+(10-1)d]$

$25=2a+9d........(ii)$

On multiplying equation (1) by 2, we obtain

30 = 2a + 4d (iii)

On subtracting equation (iii) from (ii), we obtain

$-5 = 5d$

d = −1

From equation (i),

$15 = a + 2(-1)$

$15 = a - 2$

a = 17

$a_{10} = a + (10 -1)d$

$a_{10}= 17 + (9) (-1)$

$17 - 9 = 8$

(v)Given, $d = 5, S_{9} = 75$

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

$S_{9}=\frac{9}{2}[2a+(9-1)5]$

$75=\frac{9}{2}[2a+40]$

$25 = 3(a + 20)$

25 = 3a + 60

$3a = 25 - 60$

$a=-\frac{35}{5}$

$a_{n}= a + (n -1)d$

$a_{9} = a + (9 - 1) (5)$

$=\frac{-35}{3}+8(5)$

$=\frac{85}{3}$

(vi) Given, $a = 2, d = 8, S_{n} = 90$

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

$90= \frac{n}{2}[4+(n-1)8]$

$90 = n [2 + (n - 1)4]$

$90 = n [2 + 4n - 4]$

$90 = n (4n - 2) = 4n^{2} -2n$

$4n^{2} - 2n - 90 = 0$

$4n^{2} - 20n + 18n - 90 = 0$

$4n (n - 5) + 18 (n - 5) = 0$

$(n -5) (4n + 18) = 0$

Either n - 5 = 0 or 4n + 18 = 0

n = 5 or $n= -\frac{18}{4}=-\frac{9}{2}$

However, n can neither be negative nor fractional.

Therefore, n = 5

$a_{n}= a + (n -1)d$

$a_{5} = 2 + (5 - 1)8$

= 2 + (4) (8)

= 2 + 32 = 34

(vii) Given, $a = 8, a_{n} = 62, S_{n} = 210$

$S_{n}=\frac{n}{2}[a+a_{n}]$

$210=\frac{n}{2}[8+62]$

n = 6

$a_{n}= a + (n -1)d$

$62 = 8 + (6 - 1)d$

62 - 8 = 5d

54 = 5d

$d=\frac{54}{5}$

(viii) Given, $a_{n} = 4, d = 2, S_{n}= -14$

$a_{n}= a + (n -1)d$

4 = a + (n - 1)2

4 = a + 2n - 2

a + 2n = 6

a = 6 - 2n.............. (i)

$S_{n}=\frac{n}{2}[a+a_{n}]$

$-14=\frac{n}{2}[a+4]$

$-28 = n (a + 4)$

$-28 = n (6 - 2n + 4)$ {From equation (i)}

$-28 = n (- 2n + 10)$

$-28 = - 2n^{2} + 10n$

$2n^{2} -10n - 28 = 0$

$n^{2}- 5n -14 = 0$

$n^{2} - 7n + 2n - 14 = 0$

$n (n - 7) + 2(n - 7) = 0$

$(n - 7) (n + 2) = 0$

Either n=- 7 = 0 or n + 2 = 0

n = 7 or n = -2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we obtain

$a = 6 - 2n$

$a = 6 - 2(7)$

= 6 - 14

= -8

(ix)Given, $a = 3, n = 8, S = 192$

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

$192=\frac{8}{2}[2(3)+(8-1)d]$

192 = 4 [6 + 7d]

48 = 6 + 7d

42 = 7d

d = 6

(x)Given, l = 28, S = 144 and there are total of 9 terms.

$S_{n}=\frac{n}{2}[a+l]$

$144=\frac{9}{2}[a+28]$

$(16) imes (2) = a + 28$

32 = a + 28

a = 4

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