Question

The numbers 1, 5, 25 can be three terms (not necessarily consecutive) of-

• A. At least one A.P.
• B. At least one G.P.
• C. Infinite number of A.P's
• D. Infinite number of G.P.'s

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Infinite number of A.P's
B Infinite number of G.P.'s
C At least one A.P.
D At least one G.P.

Let 1, 5, 25 be the $p^{th},q^{th},r^{th}$ terms of an A.P. with common difference d, then $(q-p)d=(5-1)$ and $(r-p)d=25-1$
$\Rightarrow \frac{q-p}{1}=\frac{r-p}{6}=k\left(say\right)$
$\Rightarrow q=p+k=rp+6k$ where k is any natural number.
Let 1, 5, 25 be the $p^{th},q^{th},r^{th}$ terms of a G.P., with common ratio R, then $R^{q-p}=5,R^{r-p}=25$
$\therefore r-p=2q-2p\Rightarrow r+p-2q=0$
There exist infinitely many triplets of natural numbers satisfying this relation.