Question

find the sum of series 4+12+20+28+ +100terms

Answer 1

Given ,

First term (a) = 4

Common difference (d) = 8

Last term (an) = 100

We know that , the nth term of an AP is given by

$\boxed{ \sf{ a_{n} = a + (n - 1)d }}$

Thus ,

$\implies \tt 100 = 4 + (n - 1)8$

$\implies \tt 96 = (n - 1)8$

$\implies \tt 12 = n - 1$

$\implies \tt n = 13$

Now , the sum of first n terms of an AP is given by

$\boxed { \sf{ S_{n} = \frac{n}{2}(a + a_{n}) }}$

Thus ,

$\implies \tt sum = \frac{13}{2} (4 + 100)$

$\implies \tt sum =13 \times 52$

$\implies \tt sum =676$

★ The sum of given series is 676

Answer 2

Answer:

The answer is attached