Question

If $N$arithmetic means are inserted between $1$and $31$ such that ratio of first and $n^{th}$mean is $3:29,$then what is the value of $N$?

Answer 1

Solve for value of $N.$

The first term is $a=1$ and the $n^{th}$ term is $a_n=31$

Number of A.M.'s $=n$

Total no. of terms in A.P $=$no. of means$+$ first term$+$ last term

$N=n+2$

Let common difference$=d,$ $a_n=a+(N-1)d$

$31=1+(n+2-1)d\Rightarrow 30=(n+1)d$

$d$ $=\frac{30}{n+1}...\left(1\right)$

The first mean$=1+d$

The $n^{th}$mean $=1+nd$

The ratio of first mean to the $n^{th}$ mean is $3:29,$i.e.

$$\begin{gathered} \frac{1+d}{1+nd}\Rightarrow \frac{3}{29} \\ \Rightarrow 29(1+d)=3(1+nd) \\ \Rightarrow 29+29d=3+3nd \\ \Rightarrow 26=3nd-29d \\ \Rightarrow 26=(3n-29)d \\ \Rightarrow d=\frac{26}{3n-29}...\left(2\right) \end{gathered}$$

By solving both the equations$\left(1\right)$ and $\left(2\right)$ we get:

$$\begin{gathered} \frac{30}{n+1}=\frac{26}{3n-29} \\ 30(3n-29)=26(n+1) \\ 90n-870=26n+26 \\ 90n-26n=26+870 \\ 64n=896 \\ n=\frac{896}{64} \\ n=14 \end{gathered}$$

Hence, the value of $N$is $14.$