Question

If $m$ arithmetic means and three geometric means are inserted between $3$ and $243$ such that $4^{\mathrm{th}}$ A.M. is equal to $2^{\mathrm{nd}}$ G.M., then $m$ is equal to

Answer 1

Step 1: Determine the $4^{\mathrm{th}}$ A.M.

The given numbers are $3$ and $243$.

Let the $m$ arithmetic means are $a_1,a_2,a_3,....,a_m$.

Thus, the series $3,a_1,a_2,a_3,....,a_m,243$ is in arithmetic progression.

We know that $n^{th}$term of an A.P. i.e. $t_n=a+\left(n-1\right)d$, where $a,n,d$ are the first term, number of terms and common difference respectively

The first term of arithmetic progression, $a_0=3$

The last term of the given arithmetic progression can be given by, $t_{m+2}=a_0+\left(m+2-1\right)d$, where $d$ is the common difference

$$\begin{gathered} \Rightarrow 243=3+\left(m+1\right)d \\ \Rightarrow d=\frac{240}{m+1} \end{gathered}$$

Therefore, the $5^{\mathrm{th}}$ term, $t_5=a_4=a_0+\left(5-1\right)d$.

$$\begin{gathered} \Rightarrow a_4=3+\frac{4\times 240}{m+1} \\ \Rightarrow a_4=3+\frac{960}{m+1} \end{gathered}$$

Thus, the $4^{\mathrm{th}}$ A.M., $a_4=3+\frac{960}{m+1}$.

Step 2: Determine the $2^{\mathrm{nd}}$ G.M.

The given numbers $3$ and $243$.

Let the $3$ geometric means are $g_1,g_2,g_3$.

Thus, the series $3,g_1,g_2,g_3,243$ is in geometric progression.

We know that $n^{th}$term of a G.P. i.e. $t_n=a.r^{n-1}$, where $a,r$ are the first term and common ratio respectively.

The first term of geometric progression, $a_0=3$

The last term of the geometric progression can be given by, $t_5=a_0{r}^{5-1}$, where $r$ is the common ratio.

$$\begin{gathered} \Rightarrow 243=3r^4 \\ \Rightarrow r^4=81 \\ \Rightarrow r^4=3^4 \\ \Rightarrow r=3 \end{gathered}$$

Therefore, the $3^{\mathrm{rd}}$ term, $t_3=g_2=a_0{r}^{3-1}$.

$$\begin{gathered} \Rightarrow g_2=3{\left(3\right)}^2 \\ \Rightarrow g_2=27 \end{gathered}$$

Thus, the $2^{\mathrm{nd}}$ G.M., $g_2=27$.

Step 3: Compute the value of $m$

It is given that, $4^{\mathrm{th}}$ A.M. is equal to $2^{\mathrm{nd}}$ G.M.

Thus, $a_4=g_2$

$$\begin{gathered} \Rightarrow 3+\frac{960}{m+1}=27 \\ \Rightarrow \frac{960}{m+1}=24 \\ \Rightarrow \frac{1}{m+1}=\frac{24}{960} \\ \Rightarrow m+1=\frac{960}{24} \\ \Rightarrow m+1=40 \\ \Rightarrow m=39 \end{gathered}$$

Hence, the value of $m$ is $39$.