Sequences & Series
Sequence and Series
star_batch_jee_advanced_2025
Grade 11
Question:
Let $S_n, S_{2n}, S_{3n}$ are respectively the sums of first $n$, $2n$, $3n$ terms of an arithmetic progression, then $S_{3n} =$
$2(S_{2n} - S_n)$
$\frac{3}{2}(S_{2n} - S_n)$
$3(S_{2n} - S_n)$
$6(S_{2n} - S_n)$
Step-by-Step Solution
Key Concept: The difference of partial sums of an arithmetic series can be expressed as a linear combination of the sequence parameters and other partial sums.
The difference $S_{2n} - S_n = \frac{2n}{2}[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d] = n[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d]$ simplifies to $\frac{n}{2}[2a + (3n-1)d] - \frac{1}{3}S_{3n}$.
Correct Answer: 3