Question

find our consecutive terms in an A.P. whose sum is 12 and of 3rd and 4th term is 14.​

Answer 1

Answer:

Let the terms be a−d,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 (1−4),1,(1+4),(1+2(4))

=−3,1,5,9

Step-by-step explanation:

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