Question

find the sum of following series: the multiple of 4 from 4 to 100 inclusive​

Answer 1

SOLUTION

TO DETERMINE

Sum of following series: the multiple of 4 from 4 to 100 inclusive

FORMULA TO BE IMPLEMENTED

Sum of first n terms of an arithmetic progression

$= \displaystyle \sf{ \frac{n}{2} \bigg[2a + (n - 1)d \bigg] }$

Where First term = a

Common Difference = d

EVALUATION

Here we have to find the sum of the series: the multiple of 4 from 4 to 100 inclusive

The series is

4, 8, 12, 16,..., 100

It is an arithmetic progression

First term = a = 4

Common Difference = d = 4

Let 100 be the n th term

∴ 4 + 4(n-1) = 100

$\implies \sf{4(n - 1) = 96}$

$\implies \sf{(n - 1) = 24}$

$\implies \sf{n = 25}$

So the required sum of the series

$= \displaystyle \sf{ \frac{n}{2} \bigg[2a + (n - 1)d \bigg] }$

$= \displaystyle \sf{ \frac{25}{2} \bigg[4 + 100 \bigg] }$

$= \displaystyle \sf{ \frac{25}{2} \times 104}$

$= 25 \times 52$

= 1300

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