Sequence and Series

Concept:If three numbers are in GP, then the square of the middle term equals the product of the first and third terms.Explanation:Let $a = k$, then $b = 3k$, $c = 5k$ from the given condition $a = \frac{b}{3} = \frac{c}{5}$.Then $b - a = 3k - k = 2k$ and $c - a = 5k - k = 4k$.Check GP condition: $(b-a)^2 = (2k)^2 = 4k^2$, and $(a)(c-a) = k \cdot 4k = 4k^2$. So it holds.Now check $a, b, c$: $a=k$, $b=3k$, $c=5k$. For AP: $2b = 6k$, $a+c = 6k$, so they are in AP.For GP: $b^2 = 9k^2$, $ac = 5k^2$, not equal. For HP: reciprocals are $1/k, 1/(3k), 1/(5k)$, which are not in AP.Thus $a, b, c$ are in AP.Answer:AP