Statistics
Variance of Arithmetic Progression
nta_pyq_2025_apr
Grade 11
Question:
The variance of the numbers $8, 21, 34, 47, \ldots, 320$ is
Step-by-Step Solution
Key Concept: The sequence is an AP with $a = 8$, $d = 13$. Find $n$ using the last term, then compute mean and variance using $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$.
$8 + (n-1)13 = 320 \Rightarrow 13n = 325 \Rightarrow n = 25$. Mean $= \frac{25/2 \cdot (8+320)}{25} = 164$. Variance $= \frac{8^2 + 21^2 + \cdots + 320^2}{25} - 164^2 = 8788$.
Correct Answer: 8788