Question

find sum of first 8 multiples of 3

Answer 1

Hey !

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Multiples of 3 are

3 , 6 , 9 , ....

a = 3 , d = 3 , n =8

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

=> 8/2 { 2 (3) + 7(3) }

=> 4 { 6 + 21 }

=> 4 { 27 }

=> 108.

Sum of first eight multiples of 3 is 108.

# Hope it helps #

Answer 2

Answer: 108

Step-by-step explanation:

First 8 multiples of 3

3, 6, 9, 12, 15, 18, 21, 24

S = 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24

These numbers are in AP  where a = 3, d = 3 and n = 8

$\bf\huge S_{n} = \frac{n}{2} [2a + (n - 1)d]$

$\bf\huge S_{8} = \frac{8}{2} [2(3) + (8 - 1)3]$

S8 = 4(6 + 21)

S8 ⇒ 4 × 27 = 108

Hence

Sum of first 8 multiples of 3 is 108