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Question

Find four consecutive terms in an A.P whose sum is 12 and sum of 3rd and 4rd term is 14.
(Assume the four consecutive terms in A.P. are ad, a,a+d,a+2d)

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Solution

Let the terms be ad,a,a+d,a+2d
Sum=4a+2d=12
2a+d=6------------(1)
Sum of 3rd & 4th term=(a+d)+(a+2d)=14
2a+3d=14-----------(2)
By substracting Equation (1) from (2)
2d=8
d=4
From Equation(1), a=1
So, the terms are (14),1,(1+4),(1+2(4))
=3,1,5,9

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