Question

The sum of $4$ numbers in GP is $60$ and the AM of the first and last numbers is $18$. Find the first term and the common difference of the GP.

Answer 1

Step 1:Formula used

In the question, it is given that four numbers are in G.P. and their sum is $60$.
Also, it is given that the A.M. of the first and last term is $18$.
First of all, assume the four numbers in G.P. are $\mathrm{a},\mathrm{ar},{\mathrm{ar}}^2,{\mathrm{ar}}^3$
Where $\mathrm{a}$ is the first term of G.P. and $‘\mathrm{r}’$ is the common ratio.
We know that the formula for finding the sum of terms of a G.P. is given by:
$\text{S =}{\displaystyle \frac{\text{a}\left({\text{r}}^{\text{n}}-1\right)}{\text{r - 1}}}$ ; where $\mathrm{r}>1......\left(1\right)$
$\mathrm{n}$ is the number of terms in G.P.
$\mathrm{r}$ is the common ratio.
$a$ is the first term of G.P.

Step 2:Substituting the known values

Putting the values in the equation $\left(1\right)$, we get:
$\Rightarrow$$\mathrm{S}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}$

$\Rightarrow$$60=\frac{\mathrm{a}\left({\mathrm{r}}^4-1\right)}{\mathrm{r}-1}$

$\Rightarrow$$60=\frac{\mathrm{a}\left(\left({\mathrm{r}}^2+1\right)(\mathrm{r}-1)(\mathrm{r}+1)\right)}{\mathrm{r}-1}$

$\Rightarrow$$60=\mathrm{a}\left(\left({\mathrm{r}}^2+1\right)(\mathrm{r}+1)\right)$

$\Rightarrow$$\mathrm{a}\left(\left({\mathrm{r}}^2+1\right)(\mathrm{r}+1)\right)=60.......\left(2\right)$
Now, we arithmetic mean of two numbers $\mathrm{a}$ and $\mathrm{b}$ is given as:
$\text{A.M.}=\frac{\mathrm{a}+\mathrm{b}}{2}.....\left(3\right)$
Putting the values in the equation $\left(3\right)$, we get:
$\Rightarrow$$18=\frac{\mathrm{a}+{\mathrm{ar}}^3}{2}=\frac{\mathrm{a}({\mathrm{r}}^3+1)}{2}.....\left(4\right)$

$\Rightarrow$ $a(r^3+1)=36$

$\Rightarrow$ $\mathrm{a}=\frac{36}{({\mathrm{r}}^3+1)}$

Step 3: Find the value of $\mathrm{r}$

Putting the value of $‘\mathrm{a}’$ in the equation $\left(2\right)$. We get:

$\Rightarrow$$\frac{36}{{\mathrm{r}}^3+1}\left[({\mathrm{r}}^2+1)(\mathrm{r}+1)\right]=60$

$\Rightarrow$$36\left[({\mathrm{r}}^2+1)(\mathrm{r}+1)\right]=60({\mathrm{r}}^3+1)$
On rearranging the terms on both sides, we get:

$\Rightarrow$$60{\mathrm{r}}^3-36{\mathrm{r}}^3-36{\mathrm{r}}^2-36\mathrm{r}+60-36=0$

$\Rightarrow$ $24{\mathrm{r}}^3-36{\mathrm{r}}^2-36\mathrm{r}+24=0$
On solving the above equation, we get:
$\mathrm{r}=2,-1$ and $0.5$
But we have assumed and used the formula for G.P. with $\mathrm{r}>1$.
Therefore,$\mathrm{r}=2.$

Step 4:Determine the integers

Putting the value of $\mathrm{r}$ in the equation $\left(4\right)$, we get:

$\Rightarrow$$\mathrm{a}=\frac{36}{({\mathrm{r}}^3+1)}$

$\Rightarrow$$\mathrm{a}=\frac{36}{2^3+1}$

$\Rightarrow$$\mathrm{a}=\frac{36}{9}$

$\Rightarrow$$\mathrm{a}=4$

So, First-term $=\mathrm{a}=4$
Second term $=\mathrm{ar}=4\times 2=8$
Third term $={\mathrm{ar}}^2=4\times 2^2=16$
Fourth term $={\mathrm{ar}}^3=4\times 2^3=32.$

Therefore, the four integers are $4,8,16,32$.