Question

Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and

4
th term is 14.
Please answer me but it should be correct.
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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

Answer 2

Step-by-step explanation:

sum is 12 and the sum of 3rdand4th term is 14.

Let four consecutive terms be

a -d ,a,a+d, a+2d

Answer :

... Fourth term =a+2d=1+2(4)=1+8=9.