Sequences & Series
Sequence and Series
star_batch_jee_advanced_2025
Grade 11
Question:
If the $(m+1)^{th}$, $(n+1)^{th}$ and $(r+1)^{th}$ terms of an A.P. are in G.P. and $m,n,r$ are in H.P., then the ratio of the common difference to the first term in the A.P. is equal to:
\frac{2}{n}
\frac{1}{n}
-\frac{1}{n}
\frac{2}{-n}
Step-by-Step Solution
Key Concept: The condition $(a+nd)^2 = (a+md)(a+rd)$ holds when $n$ is related to the harmonic mean of $m$ and $r$.
Starting with $(a+nd)^2 = (a+md)(a+rd)$, expand to get $a^2 + 2and + n^2d^2 = a^2 + ard + amd + mrd^2$. Rearranging: $(n^2-mr)d^2 = ad(r+m-2n)$. This gives $\frac{d}{a} = \frac{r+m-2n}{n^2-mr}$. Setting $n = \frac{2mr}{m+r}$ (the harmonic mean condition) simplifies the ratio to $\frac{d}{a} = -\frac{2}{n}$.
Correct Answer: 4