Question

phone numbers are in arithmetic progression the sum of first and last terms is 8 and the product of both middle terms is 15 the last number of the series​

Answer 1

Answer:11

Step-by-step explanation:

Let the terms be

a+3d,a+d,a-d+a-3da+3d,a+d,a−d+a−3d

It is given that

T_{1}+T_{4}=8T

1

​

+T

4

​

=8

Or

a+3d+(a-3d)=8a+3d+(a−3d)=8

2a=82a=8

a=4a=4 ...(i)

And

T_{2}\times T_{3}=15T

2

​

×T

3

​

=15

(a+d)(a-d)=15(a+d)(a−d)=15

a^{2}-d^{2}=15a

2

−d

2

=15

Now from i a=4a=4.

Hence

16-d^{2}=1516−d

2

=15

d^{2}=1d

2

=1

d=1d=1

Therefore the numbers are

7,5,3,17,5,3,1

Hence the least number of the sequence is 11.