Question

find the sum of the first 20 multiples of 4​

Answer 1

SOLUTION

TO DETERMINE

The sum of first 20 multiples of 4

EVALUATION

First 20 multiples of 4 are 4 , 8 , 12 ,...., 80

This is an arithmetic progression

First term = a = 4

Common Difference = d = 8 - 4 = 4

Number of terms = 20

Hence sum of first 20 multiples of 4

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

$\displaystyle \sf{ = \frac{20}{2} \times \bigg[ (2 \times 4) + (20 - 1) \times 4\bigg] }$

$\displaystyle \sf{ = 10 \times \bigg[ 8+ 76\bigg] }$

$\displaystyle \sf{ = 10 \times 84 }$

$\displaystyle \sf{ = 840 }$

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Answer 2

Answer:

The sum of the first $20$ multiples of $4$ is $840$

Step-by-step explanation:

Given:

To find the sum of the first $20$ multiples of $4$

They are as :

$4,8,12,16,...$

Since the series form an A.P

The first term is $a=4$

The common difference is $d=8-4=4$

We need to find the sum of $20$ terms so $n=20$

The sum of n terms in A.P is $\frac{n}{2}[2a+(n-1)d]$

By substituting the values we get:

$=\frac{20}{2}[2(4)+(20-1)(4)] \\\\=10.[8+(19)(4)]$

$=10(8+76)\\\\=10*84\\\\=840$