Question

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

Answer 1

Let there be \(n\) terms of this A.P.
For this A.P.,
\(a = 9\) and \(d = a_2 − a_1 = 17 − 9 = 8\)
\(S_n =\frac n2 [2a + (n-1)d]\)

\(636 =\frac n2 [2\times 9 + (n-1)8]\)

\(636 =\frac n2 [18 + (n-1)8]\)
\(636 = n[9 + 4n − 4]\)
\(636 = n(4n + 5)\)
\(4n^2 + 5n − 636 = 0\)
\(4n^2 + 53n − 48n − 636 = 0\)
\(n(4n + 53) − 12 (4n + 53) = 0\)
\((4n + 53) (n − 12) = 0\)
Either \(4n + 53 = 0\) or \(n − 12 = 0\)
\(𝑛=−\frac {53}{4}\) or \(n = 12\)
n can not be −\(\frac {53}{4}\) . As the number of terms can neither be negative nor fractional.

Therefore, \(n = 12\).