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, $n = 12$.