NDA Math Sequences and Series

Let a be the first term and k be the common ratio of a G.P.
Then, 3rd term =t3=ak3–1=p
⇒ak2=p....(i)
8th term =t8=ak8–1=q
ak7=q...(ii)
and, 13th term=t13=ak13–1=r
⇒ak12=r....(iii)
From (ii), divided by (i), we get

ak7

ak2

=

q

p

⇒k5=

q

p

...(iv)
From (iii), divided by (ii), we get
‌‌

ak12

ak7

=‌‌

r

q

⇒k5=‌‌

r

q

....(v)
From (i) and (ii), we get

q

p

=

r

p

⇒q2=pqr