UPSC NDA I 2019 Mathematics Solved Paper

(C)
Let for AP,
t1 = 1, tp = 5, tq = 25, c .d = d
∴ 1+( p−1 )d = 5 ⇒ ( p−1) d = 4 …(i)
1 + (q−1) d = 25 ⇒ (q−1) d = 24 …(ii)
∴ from (i) and (ii)

q−1

p−1

= 6 ⇒ q−1 = 6 p − 6
⇒ 6 p − q = 5
∴ there are infinite AP possible.
In same ways, we can show that infinite GP are possible.