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

Step-by-step explanation:

If four consecutive terms in A.P. are

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

Then, a-d+a+a + d+a + 2d=12.

=>4x+2d=12—(1)

Now, a+d+a+2d=14

=>(2a+3d=14)×2

=>4a+6d=28—(2)

(2) - (1)

4a+6d=28

-4a+(-)2d=(-)12

4d=16

=>d=16/4=4—(3)

Now (3) in (1)

4a+2d=12

=>4a+2(4)=12

=>4a=12-8=4

=>a=4/4=1.

The four consecutive terms are:

-3,1,5 & 9.........