Sequences & Series
Infinite GP — Two Sum Conditions
nta_pyq_2024_apr
Grade 11
Question:
Let $a,ar,ar^2,\ldots$ be an infinite G.P. If $\sum_{n=0}^{\infty}ar^n=57$ and $\sum_{n=0}^{\infty}a^3r^{3n}=9747$, then $a+18r$ is equal to
Step-by-Step Solution
Key Concept: $\frac{a}{1-r}=57$ and $\frac{a^3}{1-r^3}=9747$. Divide: $\frac{a^3/(1-r^3)}{a/(1-r)}=\frac{9747}{57}=171\Rightarrow a^2\cdot\frac{1-r}{1-r^3}=171\Rightarrow\frac{a^2}{1+r+r^2}=171$.
$a=19$, $r=2/3$. $a+18r=31$.
Correct Answer: 3