Question

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

Answer 1

Answer:

Assume that the four consecutive terms in A.P. are a – d, a, a + d, a +2d .

It is given that,

Sum of four consecutive terms = 12

Sum of 3rd and 4th term = 14

• Hence the terms are –3, 1, 5 and 9.

Answer 2

Step-by-step explanation:

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