Sequences & Series
Sequence and Series
Allen Star Batch
Grade 11
Question:
Let $a_1, a_2, a_3, ..., a_n$ be the first 'n' terms of an A.P. having common difference '$d$ $(d \neq 0)$, then the greatest value of product of two terms equidistant from the extreme terms is:
$a_1 a_n + \frac{d^2(n-1)^2}{4}$ if n is odd
$a_1 a_n + \frac{d^2(n+1)^2}{4}$ if n is odd
$a_1 a_n + \frac{d^2 n(n+2)}{4}$ if n is even
$a_1 a_n - \frac{d^2}{4}n(n-2)$ is n is even
Step-by-Step Solution
Key Concept: Apply difference of squares to arithmetic progression terms and distinguish cases based on parity.
Given $a_n \cdot a_{n-(k-1)} = (a_n + (k-1)d)(a_n - (k-1)d) = a_n^2 - (k-1)^2d^2$. For odd $n$, this equals $a_1 a_n + rac{(n-1)^2d^2}{4}$; for even $n$, this equals $a_1 a_n + rac{d^2n(n-2)}{4}$.
Correct Answer: 1,4