How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Let there be n terms of the AP. 9, 17, 25 …

For this A.P.,
First term, a = 9
Common difference, d
= \(a_2 – a_1\) = 17-9 = 8
As, the sum of n terms, is;

\(\Rightarrow S_n = {{n} \over {2}} (2a + (n – 1)d\)
\(\Rightarrow \) 636 = \({{n} \over {2}}\) [2×a+(8-1)×8]
\(\Rightarrow \) 636 = \({{n} \over {2}}\) [18+(n-1)×8]
\(\Rightarrow \) 636 = n [9 +4n -4]
\(\Rightarrow \) 636 = n (4n +5)
\(\Rightarrow \) 4\(n^2\) +5n -636 = 0
\(\Rightarrow \) 4\(n^2\) +53n -48n -636 = 0
\(\Rightarrow \) n (4n + 53)-12 (4n + 53) = 0
\(\Rightarrow\) (4n +53)(n -12) = 0

Either 4n+53 = 0 or n-12 = 0

n = \({{-53} \over {4}}\) or n = 12

n cannot be negative or fraction, therefore, n = 12 only.