# Question 17 Exercise 4.2

Solutions of Question 17 of Exercise 4.2 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

There are $n$ arithmetic means between 5 and 32 such that the ratio of the 3rd and 7th means is $7: 13$, find the value of $n$.

Let $A_1, A_2, A_3, \ldots, A_n$ be $n$ arithmetic means between 5 and 32. Then $5, A_1, A_2, A_3, \ldots, A_n, 32$ are in A.P, where $$a_1=5 \text{ and } a_{n+2}=32.$$ We know $a_n=a_1+(n-1) d$, replacing $n$ by $n+2$, we get
\begin{align}a_{n+2}&=a_1+(n+2-1) d \\ & =a_1+(n+1) d \\ \implies 32&=5+(n+1)d \\ \implies (n+1)d&=32-5\\ \implies n+1&=\dfrac{27}{d}\\ \implies n&=\dfrac{27}{d}-1 ---(i)\end{align} Also, we have given $A_3:A_7=7:13$, where $$A_3=a_4=a_1+3d=5+3d$$ and $$A_7=a_8=a_1+7d=5+7d.$$ Thus we have \begin{align}&\dfrac{A_3}{A_7}=\dfrac{7}{13} \\ \implies &\dfrac{5+3d}{5+7 d}=\dfrac{7}{13} \\ \implies &13(5+3d)=7(5+7d) \\ \implies &65+39d=35+49d\\ \implies &39d-49d=35-65 \\ \implies &-10d=-30 \\ \implies &d=3.\end{align} Putting value of $d$ in (i), we get \begin{align}&n=\dfrac{27}{3}-1\\ \implies &n=8.\end{align}