How may terms of an arithmetic progression (A.P) must be taken for their sum to equal $91$, if its third term is $9$ and the difference between the seventh and the second term is $20$?
Let the first term of A.P $= t_1 = a$
Let the common difference of A.P $= d$
Let the numebr of terms required for the sum of terms of the A.P to equal $91$ be $= n$
Sum to $n$ terms of A.P $= S_n = \dfrac{n \left[2a + \left(n - 1\right) d\right]}{2}$
$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right) d$
$\therefore \;$ $2^{nd}$ term of A.P $= t_2 = a + d$
$3^{rd}$ term of A.P $= t_3 = a + 2d$
$7^{th}$ term of A.P $= t_7 = a + 6d$
Given: $\;$ $S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right] = 91$
i.e. $\;$ $n \left[2a + \left(n - 1\right) d\right] = 182$ $\;\;\; \cdots \; (1)$
And $\;$ $t_3 = a + 2d = 9$ $\;\;\; \cdots \; (2)$
And $\;$ $t_7 - t_2 = 20$
i.e. $\;$ $a + 6d - \left(a +d\right) = 20$
i.e. $\;$ $5 d = 20$ $\implies$ $d = 4$
Substituting $d = 4$ in equation $(2)$ gives
$a + 8 = 9$ $\implies$ $a = 1$
Substituting $a = 1, \; d = 4$ in equation $(1)$ gives
$n \left[2 \times 1 + \left(n - 1\right) \times 4\right] = 182$
i.e. $\;$ $n \left(2 + 4n - 4\right) = 182$
i.e. $\;$ $4 n^2 - 2n = 182$
i.e. $\;$ $2n^2 - n - 91 = 0$
i.e. $\;$ $2 n^2 - 14 n + 13 n - 91 = 0$
i.e. $\;$ $2n \left(n - 7\right) + 13 \left(n - 7\right) = 0$
i.e. $\;$ $\left(2n + 13\right) \left(n - 7 = 0\right)$
i.e. $\;$ $n = \dfrac{-13}{2}$ $\;$ or $\;$ $n = 7$
Since the number of terms of an A.P cannot be negative,
$\therefore \;$ the required number of terms of A.P $= n = 7$