# What is the sum of a 36-term arithmetic sequence where the first term is 11 and the last term is 151?

Feb 5, 2016

2916

#### Explanation:

For an arithmetic sequence.

a , a+d , a + 2d , a+3d +........

the nth term = a + (n-1)d

where a is the first and d , the common difference

for n = 36 : 11 + 35d = 151 → 35d =140 → d = 4

the sum to n terms for an arithmetic sequence is :

${S}_{n} = \frac{n}{2} \left[2 a + \left(n - 1\right) d\right]$

$\Rightarrow {S}_{36} = \frac{36}{2} \left[\left(2 \times 11\right) + \left(35 \times 4\right)\right]$

= 18 [ 22 +140 ] = 2916