Question

The sum of first 5 terms of an ap is 45 and that of its first 15 terms is 435. Find the sum of first n terms

Answer 1

$S_5 = \frac{5}{2}[2a_1+(5-1)d]$
$45* \frac{2}{5} = [2a_1+4d]$
$18 = 2[a_1+2d]$
$a_3 = 9$                                                          ....(i)

$S_{15} = \frac{15}{2}[2a_1+(15-1)d]$
$435* \frac{2}{15}= [2a_1+14d]$
$58 = 2[a_1+7d]$
$a_8=29$                                                           ...(ii)

On solving from eq.(i)
$a_1 + 2d = 9$
$a_1 = 9-2d$                                                      ...(iii)

Putting $a_1$'s value from eq.(iii) in eq.(ii) we get
$a_8 = 29$
$a_1 + 7d = 29$
$9-2d+7d=29$
$5d = 20$
$d = 4$                                                                ...(iv)

Putting d's value from eq.(iv) in eq.(ii) we get
$a_1 = 9 - 2(4)$
$a_1 = 1$                                                           ...(v)

Using eq.(iv) and eq.(v) we can get

$S_n = \frac{n}{2}[2a_1+(n-1)d]$
$S_n = \frac{n}{2}[2(1)+(n-1)4]$
$S_n = n[1+2(n-1)]$
$S_n = n[1+2n-2]$
$S_n = n[2n-1]$
$S_n = 2n^2-n$