Question

There are 4 terms in an AP. in which sum of the first and last term is 13. and the product of the middle term is 40. then terms of the given AP is​

Answer 1

Answer:

Let the terms be

a+3d,a+d,a−d+a−3d

It is given that

T

1

+T

4

=8

Or

a+3d+(a−3d)=8

2a=8

a=4 ...(i)

And

T

2

×T

3

=15

(a+d)(a−d)=15

a

2

−d

2

=15

Now from i a=4.

Hence

16−d

2

=15

d

2

=1

d=1

Therefore the numbers are

7,5,3,1

Hence the least number of the sequence is 1.