Sequences & Series
AP with Even Number of Terms — Odd and Even Position Sums
nta_pyq_2025_apr
Grade 11
Question:
Let an AP have $2k$ terms. If the sum of odd-position terms (1st, 3rd, 5th, ...) is 40, the sum of even-position terms is 55, and the last term minus the first term is 27, then $k$ equals
Step-by-Step Solution
Key Concept: Express sums of odd- and even-position terms in terms of $a$, $d$, $k$; their difference gives $kd=15$, and the last-minus-first relation gives $(2k-1)d=27$; divide to find $k$.
The $k$ odd-position terms form an AP with first term $a$ and common difference $2d$:
$S_{\text{odd}}=k\bigl(a+(k-1)d\bigr)=40$.
The $k$ even-position terms form an AP with first term $a+d$ and common difference $2d$:
$S_{\text{even}}=k\bigl(a+kd\bigr)=55$.
$S_{\text{even}}-S_{\text{odd}}=kd=15$.
Last $-$ first: $(2k-1)d=27$.
$\dfrac{(2k-1)d}{kd}=\dfrac{27}{15}=\dfrac{9}{5}\Rightarrow 10k-5=9k\Rightarrow k=5$.
Correct Answer: 2