Sequences & Series
Arithmetico-Geometric Series — Sum to Infinity
nta_pyq_2024_jan
Grade 11
Question:
If $8=3+\dfrac{1}{4}(3+p)+\dfrac{1}{4^2}(3+2p)+\dfrac{1}{4^3}(3+3p)+\cdots\infty$, then the value of $p$ is
Step-by-Step Solution
Key Concept: Split into geometric series and AGP: $\sum_{n=0}^\infty\frac{3}{4^n}+\sum_{n=1}^\infty\frac{np}{4^n}=\frac{3}{1-1/4}+p\cdot\frac{1/4}{(1-1/4)^2}=4+\frac{4p}{9}=8$.
$4+\frac{4p}{9}=8\Rightarrow p=9$.
Correct Answer: 9