Question

The fifth, eight and eleventh terms of a
geometric progression are p, q and r
respectively. Show that : q2 = pr.​

Answer 1

• Answer:

$a {R}^{n - 1}$,

this is the form of any general term of GP where 'a'is first term of GP

And R= general difference between two terms of GP

then p= a R^4

q = aR^7

r =aR^1

Now, p÷q=aR^4÷aR^7=R^-3 ......1

q÷r=aR^7÷aR^10= R^-3.....2

1&2 are equal so,

p÷q= q÷r

q^2=pr......proved