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 $20$ ?

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Solution

Third term of $A P, a_{3} = 9$
$\Rightarrow a_{1} + 2 d = 9 - (i)$
Difference between $7 t h & 2 n d$ term,
$a_{7} - a_{2} = 20 \Rightarrow (a_{1} + 6 d) - (a_{1} + d) = 20 \Rightarrow a_{1} + 6 d - a_{1} - d = 20 \Rightarrow 5 d = 20 \Rightarrow d = 4 F r o m (i), a_{1} + 2 \times 4 = 9 \Rightarrow a_{1} = 9 - 8 = 1$
$∴$ Sum of n terms of $A P = 91$
$S_{n} = 91 \Rightarrow \frac{n}{2} [2 a_{1} + (n - 1) d] = 91 N o w, a = 1, d = 4 \Rightarrow \frac{n}{2} [2 \times 1 + (n - 1) \times 4] = 91 \Rightarrow n [2 + (n - 1) 4] = 182 \Rightarrow n [2 + 4 n - 4] = 182 \Rightarrow 2 n + 4 n^{2} - 4 n = 182 \Rightarrow 4 n^{2} - 182 - 2 n = 0 \Rightarrow 2 n^{2} - 91 - n = 0 \Rightarrow 2 n^{2} - n = 91 \Rightarrow n (2 n - 1) = 7 \times 13$
Equating both sides we get, $n = 7$
$∴$ We need $7$ terms to get a sum of $91$

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