Question

If the first term of an ap is 2 and common difference is 4, then sum of its first 40 terms is *

Answer 1

Answer:

3200

Step-by-step explanation:

a=2,d=4,n=40

formula:

Sn=(n/2)*[2a+(n-1)d]

solution:

S40 = 40/2 [2*2 + (40-1) 4]

S40 = 20 * [4 + 39*4]

S40 = 20 * 160

S40 = 3200

Answer 2

The sum of its first 40 terms is 3200.

Explanation:

The sum of the first n term is given by :-

$S_n=\dfrac{n}{2}(2a+(n-1)d)$ , where a = first term

d= common difference

As per given , we have

a= 2

d= 4

n= 40

Then, the sum of first 40 terms will be :-

$S_{40}=\dfrac{40}{2}(2(2)+(40-1)4)\\\\=(20)(4+156)\\\\=(20)(160)=3200$

Hence, the sum of its first 40 terms is 3200.

# Learn more :

If the sum first 10 term of an ap is 4 times the sum of its first five terms, the ratio of the first term to the common difference is

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