Question

If the first term of an AP is 2 and common difference is 4, then find the sum of

its first 40 terms​

Answer 1

A=2 , d=4 , n=40

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

Sn=20(4+156)

Sn=20×160

Sn=3200

sum of first 40 terms is = 3200

Answer 2

Given:

The first term of an AP is 2 so, a = 2

The common difference is 4 i.e. d = 4

To find:

The sum of its 40 terms.

The formula used to find the sum of the first nth term in a given AP is

$\boxed{\sf{\pink{S_{n}=\frac{n}{2}[2a+(n-1)d]}}}$

[In which 'n' is the nth term, 'a' is the first term of the AP, 'd' if the common difference]

So,

$S_{40}=\frac{40}{2}[2(2)+(40-1)4]$

$S_{40}=20[4+(39)4]$

$S_{40}=20[4+156]$

$S_{40} = 20 \times [160]$

S₄₀ = 3200

Hence,

The sum of the first 40 terms is 3200.

- - -

More Information:

• When aₙ = nth term of the AP

$\boxed{\red{S_{n}=\frac{n}{2}[a+a_{n}]}}$

Also,

• When 'l' is the last term of an AP of the nth term, then the sum formula is

$\boxed{\purple{\sf S_{n}=\frac{n}{2}[a+l]}}$