Question

The fifth term of an A.P is $1$ whereas its $31$st term is $-77$. Find sum of its first fifteen terms. Also find which term of the series will be $-17$ and sum of how many terms will be $20$.

Answer 1

$T_5=a+4d=1,T_{31}=a+30d=-77$,
Solving the given two, we get
$a=13$ and $d=-3$. ..$\left(1\right)$
$S_{15}=\left(n/2\right)[2a+(n-1\left)d\right]$
$=\left(15/2\right)[26+14(-3\left)\right]=-120$.
Let $T_n=-17$. Then $a+(n-1)d=-17$.
$13+(n-1)(-3)=-17$
$\therefore 3n=33$ or $n=11$.
Let $S_n=20.$ Then $\left(n/2\right)[2a+(n-1\left)d\right]=20$.
$n[2\times 13+(n-1\left)\right(-3\left)\right]=40$, by $\left(1\right)$.
$3n^2-29n+40=0$
$\therefore (n-8)(3n-5)=0$,
$\therefore n=8$.
The value of n cannot be fractional.