Question

In an ap the in an ap the sum of first 3 terms is -60 and that of the last 3 are 84 .if there are 15terms what is the sum of the middle 3 terms

Answer 1

The sum of middle 3 terms is 12

Step-by-step explanation:

Let the first term of AP be a and common difference be d

No. of terms, n = 15

First three terms will be

$a, a+d, a+2d$

According to the question

$a+(a+d)+(a+2d)=-60$

$\implies 3a+3d=-60$

$\implies a+d=-20$  ................ (1)

Last three terms of AP

$a+14d, a+13d, a+12d$

According to the question

$(a+12d)+(a+13d)+(a+14d)=84$

$\implies 3a+39d=84$

$\implies a+13d=28$  ................... (2)

From (1) and (2)

12d=48

d = 4

a = -24

The middle three terms will be 7th term, 8th term and 9th term

Sum of middle three terms

$=(a+6d)+(a+7d)+(a+8d)=3a+21d$

$=3(a+7d)$

$=3(-24+7\times 4)$

$=3(-24+28)$

$=12$

Hope this answer is helpful.

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Answer 2

Given :  in an ap the sum of first 3 terms is -60 and that of the last 3 are 84  . There are 15 terms

To find : Sum of the middle 3 Terms

Solution:

This Method to find Sum with out finding  AP  ( a  &  d )

Sum of first three Terms

= a + a+ d + a  + 2d

Sum of Last three term

= L + L - d  + L - 2d

Addling  both

3(a + L )  =  -60 + 84

=> 3(a + L) = 24

=> a + L  = 8

there are total 15 terms  ( odd number of terms)

=> Middle term  M =  (a + L )/2   = 8/2  = 4

Middle three terms

= M -d  , M  , M + d

Sum = 3M =  3(4) = 12

Sum of Middle 3 Terms = 12

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