Math, asked by faraz123, 1 year ago

find four consecutive terms in an A.P.whose sum is 88 and the sum of the 1st and the 3rd terms is 40.

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Answered by neetanarain25
4

Answer:


Step-by-step explanation:

term is 40

Ask for details FollowReport by Riyasign 13.11.2018

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mysticd

Mysticd Maths AryaBhatta

Solution :


Let a , d are first term and common


difference of an A.P


Now ,


a , a+d , a+2d, a + 3d are first four


consecutive terms in A.P


Sum of 4 terms = 88 ( given )


=> a+a+d+a+2d+a+3d =88


=> 4a+6d = 88


Divide each term with 2 , we get


=> 2a + 3d = 44 ----( 1 )


Sum of first and 3rd term = 40(given )


=> a + a + 2d = 40


=> 2a + 2d = 40 ---( 2 )


subtract ( 2 ) from ( 1 ) , we get


d = 4


substitute d = 4 in equation (2),


we get


2a + 2×4 = 40


=> 2a = 40 - 8


=> 2a = 32


=> a = 32/2


=> a = 16


Therefore ,


a = 16 , d = 4


Required 4 terms are ,


a = 16 ,


a + d = 16 + 4 = 20 ,


a+2d = 16+2×4 = 24 ,


a+3d = 16+3×4 = 28



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