Question

find sum of 15 multiples of 8 in arthmetic progression​

Answer 1

The first 15 multiples of 8 are 8 × 1, 8 × 2, 8 × 3, ... 8 × 15 i.e., 8, 16, 24 ,32 ,40............ 120, which is an AP and here a=8 d= 8 Therefore sum of 1st 15 multiples of sum=n/2 (a+l) sum= 15/2(8+120) =15/2 ×128 = 15 × 64 = 960

The positive integers which are multiple of 8 are:

8,16,24,................up to 15 terms

It forms an arithmetic series. 

Now

first term a = 8

common difference d = 8

number of terms n = 15

Now sum of first 15 terms of AP = (n/2) *{2a + (n-1)*d}

                                               = (15/2)*{2*8 + (15-1)*8}

                                               = (15/2)*(16 + 14*8)

                                               = (15/2)*(16 + 112)

                                               = (15*128)/2

                                               = 15*64                   (when 64 and 2 is divided by 2)

                                               = 960

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Answer 2

Answer:

960

Step-by-step explanation:

Use summation formula.

First term= 8

Last term= 120

Common difference =8

n=15