Question

Find the sum of first 15 multiples of 8.

Answer 1

Sum of n terms of an AP

The sum of first n terms of an AP with first term 'a' and common difference 'd' is given by

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

Sn=n /2 [ a + l ]       (l = last term)

==========================================================

Solution:

The first 15 multiples of 8 are
8, 16, 24, 32….....120
which  are in an A.P., 

a= 8 ,     d = 8,   l(last term)=120, n= 15
S15 = ?

Sn=n /2 [ a + l ]  

 S15 = 15/2 [8+120] = 15/2 ×128 = 15×64 = 960

S15= 960

Hence, the sum of first 15 multiples of 8 is 960.

==========================================================

Hope this will help you....

Answer 2

Solution:

To Find:

=> Sum of first 15 multiples of 8.

Formula used:

$\sf{\implies S_{n}=\dfrac{n}{2}[2a(n-1)d]}$

Now, multiples of 8 are:

=> 8, 16, 24, 32......

Therefore,

=> a = 8

=> d = 16 - 8 = 8

=> Number of terms (n) = 15

Now, put the values in the formula we get,

$\sf{\implies S_{n}=\dfrac{n}{2}[2a(n-1)d]}$

$\sf{\implies S_{15}=\dfrac{15}{2}[2(8)+(15-1)8]}$

$\sf{\implies S_{15}=\dfrac{15}{2}[16+(14)8]}$

$\sf{\implies S_{15}=\dfrac{15}{2}[16+112]}$

$\sf{\implies S_{15}=\dfrac{15}{2}\times 128}$

$\sf{\implies S_{15}=15\times 64}$

$\sf{\implies S_{15}=960}$

Hence, the sum of first 15 multiples of 8 are 960.