Question

How many terms of an arithmetic progression must be taken for their sum to be equal to $91$, if its third term is $9$ and the difference between the seventh and the second term is $20$?

Answer 1

Third term of $AP,a_3=9$

$\Rightarrow a_1+2d=9-\left(i\right)$

Difference between $7th\&2nd$ term,

$a_7-a_2=20\Rightarrow (a_1+6d)-(a_1+d)=20\Rightarrow a_1+6d-a_1-d=20\Rightarrow 5d=20\Rightarrow d=4From\left(i\right),a_1+2\times 4=9\Rightarrow a_1=9-8=1$

$\therefore$Sum of n terms of $AP=91$

$S_n=91\Rightarrow \frac{n}{2}[2a_1+(n-1\left)d\right]=91Now,a=1,d=4\Rightarrow \frac{n}{2}[2\times 1+(n-1)\times 4]=91\Rightarrow n[2+(n-1\left)4\right]=182\Rightarrow n[2+4n-4]=182\Rightarrow 2n+4n^2-4n=182\Rightarrow 4n^2-182-2n=0\Rightarrow 2n^2-91-n=0\Rightarrow 2n^2-n=91\Rightarrow n(2n-1)=7\times 13$

Equating both sides we get, $n=7$

$\therefore$ We need $7$ terms to get a sum of $91$