Question

Let there be an A.P. with first term 'a', common difference 'd'. If $a_n$ denotes its nth term and $S_n$ the sum of first n terms, find.
(i) n and $S_a$, if a=5, d=3 and $a_n$=50.
(ii) n and a, if $a_n$=4, d=2 and $S_n$=-14.
(iii) d, if a=3, n=8 and $S_n$=192.
(iv)a, if $a_n$=28, $S_n$=144 and n=9.
(v) n and d, if a=8, $a_n$=62 and $S_n$=210.
(vi) n and $a_n$, if a = 2, d=8 and $S_n$=90.
(vii) k, if $S_n=3n^2$+5n and $a_k$=164.
(viii) $S_{22}$, if d =22 and $a_{22}$ =149

Answer 1

As we know,
$a_n=a+(n-1)d$
$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

(i)

$n\Rightarrow$ $50=5+(n-1)3$
$45=3(n-1)$
$15=n-1$
$n=16$
$S_a=S_5$

$\Rightarrow \frac{5}{2}(10+(4\left)\right(3\left)\right)=55$

(ii)

$4=a+(n-1)2$
$4=a+2n-2$
$6=a+2n$ ----- (i)

$-14=\frac{n}{2}(2a+(n-1\left)2\right)$

$=n(a+n-1)$

$=n(6-2n+n-1)$

$=n(5-n)$

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

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

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

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

$n=7,-2$
$n$ cannot be negative, so $n=7$

$a=6-14$

$=-8$

(iii)

$S_n=192$

$192=\frac{8}{2}(6+(7\left)d\right)$

$192=4(6+7d)$
$48=6+7d$
$d=6$

(iv)
$a_n=28$
$a+8d=28$ ----- (i)

$S_n=144$
$\frac{9}{2}(2a+8d)=144$
from (i)
$\frac{9}{2}(2a+28-a)=144$
$a+28=32$
$a=4$

Hence , $a=4$

(v)

$a_n=62$
$8+(n-1)d=62$
$(n-1)d=54$ ------ (i)

$S_n=210$
$\frac{n}{2}(16+(n-1\left)d\right)=210$
from (i)

$\frac{n}{2}(16+54=210$
$\frac{n}{2}\left(70\right)=210$
$n=6$
So, $5d=54$
$d=\frac{54}{5}$

$=10.8$

(vi)
$a_n=2+(n-1)8$ ---- (i)

$S_n=90$

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

$n(4+8n-8)=180$

$8n^2-4n-180=0$

$2n^2-n-45=0$

$2n^2-10n+9n-45=0$

$2n(n-5)+9(n-5)=0$

$(2n+9)(n-5)=0$
$n=5,\frac{-9}{2}$
$\because$ n cannot be -ve, so $n=5$
$a_n=2+(n-1)8$

$\Rightarrow 2+32=34$

(vii)

$a_k=S_k-S_k-1$

$164=3k^2+5k-\left(3\right(k-1{)}_2+5(k-1))$

$164=3k^2+5k-\left(3\right(k^2+1-2k)+5(k-1\left)\right)$

$164=3k^2+5k-(3k^2+3-6k+5k-5))$

(viii)

$a_{22}=149$

$a+21d=149$
$a+21\left(22\right)=149$
$a=-462+149$
$a=-313$
$S_n=11\left(2\right(-313)+21(22\left)\right)$

$=11(-616+462)$

$=11(-154)$

$=-1694$