Question

sum of six consecutive term of AP is 48 and the product of first and last term is 63​

Answer 1

Answer:

a=8

d=1/3

Step-by-step explanation:

let the six consecutive terms be

a-3d,a-2d,a-d,a+d,a+2d,a+3d

sum of these consecutive terms is

a-3d+a-2d+a-d+a+d+a+2d+a+3d=48

cancelling -3d,+3d,-2d,+2d,-d and +d

= 6a=48

a=8

now product of first and last term is

(a-3d)(a+3d)=63

=a^2-(3d)^2=63

=8^2-9d^2=63

9d^2=63-64

d^2=-1/9

therefore d= root of -1/9=1/3