Question

Find the sum of the first 15 multiples of 9.​

Answer 1

Answer:

Step-by-step explanation:

A.P= 9,18,27,......

a=9 d=9

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

S15=15/2(2(9)+(15-1)9)

=15/2(18+(14)9)

=15/2(18+126)

=15/2(144)

=15×72

=1080

Answer 2

Sum of first 15 multiples of 9 are 1080.

Given:

• First 15 multiples of 9.

To find:

• Find the sum of the first 15 multiples of 9.

Solution:

Formula to be used:

Sum of first n terms of A.P.

$\bf S_n = \frac{n}{2} \left(2a + (n - 1)d \right)$

Here,

a : first term

d: Common difference

n: number of term

Step 1:

Write the multiples of 9.

9, 18, 27, 36...

a : first term: 9

d: Common difference:18-9= 9

n: number of term = 15

Step 2:

Find the sum.

$S_{15} = \frac{15}{2} (2 \times 9 + (15 - 1)9)$

or

$S_{15} = \frac{15 \times \cancel2}{\cancel2} ( 9 + 7 \times 9)$

or

$S_{15} = 15 (9 + 63)$

or

$S_{15} = 15 \times 72$

or

$\bf S_{15} = 1080$

Thus,

Sum of first 15 multiples of 9 are 1080.

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