Question

If the sum of first 4 terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

OR

Find the sum of the first 30 positive integers divisible by 6.

Answer 1

Let a and d respectively be the first term and the common difference of the given A. P

The sum of first four terms
$S_4=40$
$\Rightarrow \frac{4}{2}\{2a+(4-1\left)d\right\}=40$
$\Rightarrow 2a+3d=20\cdots \cdots \left(1\right)$
The sum of first 14 terms
$S_{14}=280$
$\Rightarrow \frac{14}{2}\{2a+(14-1\left)d\right\}=280$
$\Rightarrow 2a+13d=40\cdots \cdots \left(2\right)$
Subtracting equation (1) from equation (2)
(2a+13d)-(2a+3d) = 40 - 20
$\Rightarrow$ 10d = 20
$\Rightarrow$ d = 2
Substituting d = 2 in equation (1),
Substituting d = 2 in equation (1),
2a + 3 $\times$ 2 = 20
$\Rightarrow$ 2a = 20 - 6 = 14
$\Rightarrow a\frac{14}{2}=7$
$\therefore$ Sum of first n terms, $S_n=\frac{n}{2}\{2a+(n-1\left)d\right\}$
$=\frac{n}{2}\{2\times 7+(n-1)\times 2\}$
$=\frac{n}{2}\{14+2n-2\}$
$=\frac{n}{2}(2n+12)$
$=\frac{n}{2}\times (n+6)$
$=n(n+6)$
$=n^2+6n$

OR

The first 30 integers divisible by 6 are 6 , 12, 18 ......180

Sum of first 30 integers
= 6 + 12 + 18 +.... + 180

$=\frac{30}{2}(6+180)\left[S_{n=\frac{n}{2}(a+1)}\right]$
$=15\times 186$
$=2790$