Question

In an AP:

1. given a = 5, d = 3, an = 50, find n and Sn
2. given a = 7, a13 = 35, find d and S13.
3. given a12 = 37, d = 3, find a and S12.
4. given a3 = 15, S10 = 125, find d and a10.
5. given d = 5, S9 = 75, find a and a9.
6. given a = 2, d = 8, Sn = 90, find n and an .
7. given a = 8, an = 62, Sn = 210, find n and d.
8. given an = 4, d = 2, Sn = –14, find n and a.
9. given a = 3, n = 8, S = 192, find d.
10. given l = 28, S = 144, and there are total 9 terms. Find a.

Answer 1

(i) Given that, $a = 5$, $d = 3$, $a_n = 50$
As $a_n = a + (n − 1)d$,
$∴ 50 = 5 + (n − 1)3$
$45 = (n − 1)3$
$15 = n − 1$
$n = 16$
$S_n =\frac n2[a + a_n]$

$S_{16} = \frac {16}{2}[5 + 50]$
$S_{16} = 8 \times 55$
$S_{16} = 440$

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(ii) Given that, $a = 7$, $a_{13} = 35$
As, $a_n = a + (n − 1) d$
$∴ a_{13} = a + (13 − 1) d$
$35 = 7 + 12 d$
$35 − 7 = 12d$
$28 = 12d$
$d = \frac {28}{12}$

$d = \frac 73$
$S_n = \frac n2[a + a_n]$

$S_{13} =\frac n2[a + a_{13}]$

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

$S_{13} = \frac {13 \times 42}{2}$
$S_{13} = 13 \times 21$
$S_{13} = 273$

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(iii) Given that, $a_{12} = 37$, $d = 3$
As $a_n = a + (n − 1)d$
$a_{12}= a + (12 − 1)3$
$37 = a + 33$
$a = 4$
$S_n = \frac n2[a + a_n]$

$S_{12} = \frac {12}{2}[4 + 37]$
$S_{12}= 6 \times 41$
$S_{12} = 246$

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(iv) Given that, $a_3 = 15$, $S_{10} = 125$
As, $a_n = a + (n − 1)d$
$a_3 = a + (3 − 1)$
$15 = a + 2d$ ……….(i)
$S_n = \frac {n}{2}[2a + (n-1)d]$

$S_{10} = \frac {10}{2}[2a + (10-1)d]$
$125 = 5[2a + 9d]$
$25 = 2a + 9d$ ……….(ii)
On multiplying equation (i) by 2, we obtain
$30 = 2a + 4d$ ………..(iii)
On subtracting equation (iii) from (ii), we obtain
$−5 = 5d$
$d = −1$
From equation (i),
$15 = a + 2(−1)$
$15 = a − 2$
$a = 17$
$a_{10} = a + (10 − 1)d$
$a_{10}= 17 + (9) (−1)$
$a_{10} = 17 − 9 = 8$

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(v) Given that, $d = 5$, $S_9 = 75$
As, $S_n = \frac {n}{2}[2a + (n-1)d]$

$S_9 =\frac 92[2a + (9-1)5]$

$75 = \frac 92(2a + 40)$
$25 = 3(a + 20)$
$25 = 3a + 60$
$3a = 25 − 60$
$a = -\frac {35}{3}$
$a_n = a + (n − 1)d$
$a_9 = a + (9 − 1)5$
$a_9 = -\frac {35}{3} + 8 \times 5$

$a_9 = -\frac {35}{3} + 40$

$a_9 = \frac {-35+120}{3}$

$a_9 = \frac {85}{3}$

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(vi) Given that, $a = 2$, $d = 8$, $S_n = 90$
As, $S_n = \frac {n}{2}[2a + (n-1)d]$

$90 = \frac n2[2 \times 2 + (n-1)8]$

$90 = \frac n2[4 + (n-1)8]$
$90 = n [2 + (n − 1)4]$
$90 = n [2 + 4n − 4]$
$90 = n (4n − 2)$
$90= 4n^2 − 2n$
$4n^2 − 2n − 90 = 0$
$4n^2 − 20n + 18n − 90 = 0$
$4n (n − 5) + 18 (n − 5) = 0$
$(n − 5) (4n + 18) = 0$
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, $n = 5$
$a_n = a + (n − 1)d$
$a_5 = 2 + (5 − 1)8$
$a_5= 2 + 4 \times 8$
$a_5= 2 + 32$
$a_5 = 34$

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(vii) Given that, $a = 8$, $a_n = 62$, $S_n = 210$
$Sn = \frac {n}{2}[a + a_n]$

$210 = \frac n2[8 + 62]$

$210 = \frac n2 \times 70$
$n = 6$
$a_n = a + (n − 1)d$
$62 = 8 + (6 − 1)d$
$62 − 8 = 5d$
$54 = 5d$
$d = \frac {54}{5}$

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(viii) Given that, $a_n = 4, d = 2, S_n = −14$
$a_n = a + (n − 1)d$
$4 = a + (n − 1)2$
$4 = a + 2n − 2$
$a + 2n = 6$
$a = 6 − 2n$ …….(i)
$S_n = \frac n2[a + a_n]$

$-14 = \frac n2[a + 4]$
$−28 = n (a + 4)$
$−28 = n (6 − 2n + 4)$ {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
$a = 6 − 2n$
$a = 6 − 2 \times 7$
$a= 6 − 14$
$a= −8$

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(ix) Given that, $a = 3$, $n = 8$, $S = 192$
$S_n = \frac n2[2a + (n-1)d]$

$192 = \frac 82[2 \times 3 + (8-1)d]$
$192 = 4 [6 + 7d]$
$48 = 6 + 7d$
$42 = 7d$
$d = 6$

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(x) Given that, $l = 28$, $S = 144$ and there are total of $9$ terms.
$S_n =\frac n2(a+l)$

$144 = \frac 92(a+28)$

$16 = \frac 12(a+28)$
$16 × 2 = a + 28$
$32 = a + 28$
$a = 32 - 28$
$a = 4$