Question

Given $P=3+5+7+....$ ($n$ terms) and $Q=5+8+11+....\left(10\text{terms}\right)$. What is the value of $n$ if $\frac{P}{Q}=7$?

• A. $27$
• B. $35$
• C. $25$
• D. $37$

Answer 1

Answer: (B)

$35$

$P=3+5+7+...$(n terms)

a=3

d=5-3=2

$\therefore P=\frac{n}{2}[2.3+(n-1\left)2\right]=n^2+2n$

$Q=5+8+11+...$(10 terms)

a=5

d=8-5=3

$\therefore Q=\frac{10}{2}[2.5+(10-1\left)3\right]=185$

$\therefore \frac{P}{Q}=7$

$\Rightarrow \frac{n^2+2n}{185}=7$

$\Rightarrow n^2+2n-1295=0$

$\Rightarrow (n-35)(n+37)=0$

$\therefore n=35$