Question

SECTION:C
(3X8
1. Find the sum of the first 15 multiples of 8.​

Answer 1

Answer:

hope it helps

Explanation:

The multiples of 8 are 8,16,24,32…

These are in an A.P., having first term as 8 and common difference as 8.

Therefore, a=8,d=8, S15=?

Sn=2n[2a+(n−1)d]

S15=215[2(8)+(15−1)8]

=215[16+(14)(8)]

=215×128

=960

Answer 2

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We know that the,

Multiple of 8 are :-

8 , 16 , 24 , 32 , 40, 48 , 56, 64 , 72 , 80 , 88 , 96,----------------,n

According to the question,

The sum of the first 15 multiples of 8.

Using A.P.

We know that the sum of first n terms are :-

Sn = $\dfrac{n}{2} (2a + (n - 1)d)$

Where, n = no. of terms = 15

a = term = 8.

a2 = 16

d = common difference = a2 - a = 16 - 8 = 8

➝ Sn = $\dfrac{15}{2} (2 \times 8 + (15 - 1)8)$

➝ $\dfrac{15}{2} (16 + (14)8)$

➝ $\dfrac{15}{2} (16 + 112 )$

➝ $\dfrac{15}{2} (128 )$

➝ 7.5 × 128

➝ 960