Sequences & Series
Arithmetic Progressions
MMTS_Full_Test_07
Grade 12
Question:
If $\{a_i\}_{i=1}^n$ where $n$ is an even integer is an AP with common difference 1, and $\sum_{i=1}^n a_i=192$, $\sum_{i=1}^{n/2} a_{2i}=120$, then $n$ is equal to
Step-by-Step Solution
Key Concept: Use AP sum formulas for full series and even-indexed sub-series
$S_n=\frac{n}{2}(2a_1+n-1)=192$. $S_{\text{even}}=\frac{n/2}{2}(2a_2+n/2-1)=120$. Solving gives $n=96$.
Correct Answer: 2