QUESTION

# The sum of four numbers in a G.P. is 60 and A.M. between the first and the last is 18. Find the numbers.

Hint: The first step for solving this type of question is to assume the four numbers in G.P. as a, ar, a${{\text{r}}^2}$and a${{\text{r}}^3}$then use the formula for finding the sum of the four numbers in Geometric progression and also find Arithmetic mean of the first and last term to form two equation in two variable.

In the question it is given that four numbers are in G.P. and their sum is 60.
Also it is given that A.M. of first and last term is 18.
First of all assume the four numbers in G.P. are a, ar, a${{\text{r}}^2}$and a${{\text{r}}^3}$.
Where ‘a’ is the first term of G.P. and ‘r’ is the common ratio.
We know that the formula for finding the sum of terms of a G.P. is given by:
${\text{S = }}\dfrac{{{\text{a}}\left( {{{\text{r}}^{\text{n}}} - 1} \right)}}{{{\text{r - 1}}}}$ ; where r>1. (1)
n is the numbers of terms in G.P.
r is the common ratio.
A is the first term of G.P.
Putting the values in equation 1, we get:
${\text{S = }}\dfrac{{{\text{a}}\left( {{{\text{r}}^{\text{n}}} - 1} \right)}}{{{\text{r - 1}}}} \\ \Rightarrow 60 = \dfrac{{{\text{a}}\left( {{{\text{r}}^4} - 1} \right)}}{{{\text{r - 1}}}} = \dfrac{{{\text{a}}\left( {({{\text{r}}^2} + 1)({\text{r - 1)(r + 1)}}} \right)}}{{{\text{r - 1}}}} = {\text{a}}\left( {({{\text{r}}^2} + 1){\text{(r + 1)}}} \right) \\$
$\Rightarrow {\text{a}}\left( {({{\text{r}}^2} + 1){\text{(r + 1)}}} \right) = 60$ (2)

Now, we arithmetic mean of two numbers a and b is given as:
A.M. = $\dfrac{{{\text{a + b}}}}{2}$ (3)
Putting the values in equation 3, we get:
$18 = \dfrac{{{\text{a + a}}{{\text{r}}^3}}}{2} = \dfrac{{{\text{a(}}{{\text{r}}^3} + 1)}}{2} \\ \Rightarrow {\text{a(}}{{\text{r}}^3} + 1) = 36 \\ \Rightarrow {\text{a = }}\dfrac{{36}}{{{\text{(}}{{\text{r}}^3} + 1)}}. \\$ (4)
Putting the value of ‘a’ in equation 2. We get:
$\dfrac{{36}}{{{{\text{r}}^3} + 1}}\left( {({{\text{r}}^2} + 1){\text{(r + 1)}}} \right) = 60 \\ \Rightarrow 36\left( {({{\text{r}}^2} + 1){\text{(r + 1)}}} \right) = 60\left( {{{\text{r}}^3} + 1} \right) \\$
On rearranging the terms on both sides, we get:
$60{{\text{r}}^3} - 36{{\text{r}}^3} - 36{{\text{r}}^2} - 36{\text{r + 60 - 36 = 0}} \\ \Rightarrow {\text{24}}{{\text{r}}^3} - 36{{\text{r}}^2} - 36{\text{r + 24 = 0}} \\$
On solving the above equation, we get:
r=2, -1 and 0.5
But we have assumed and used the formula for G.P. with r>1.
Therefore r=2.
Putting the value of ‘r’ in equation 4, we get:
${\text{a = }}\dfrac{{36}}{{{\text{(}}{{\text{r}}^3} + 1)}} = \dfrac{{36}}{{{2^3} + 1}} = \dfrac{{36}}{9} = 4$
$\therefore$ First term = a= 4
Second term = ar = 8
Third term = ${\text{a}}{{\text{r}}^2}$ =16
Fourth term = ${\text{a}}{{\text{r}}^3}$ = 32.

Note: In this type of question, we should remember the formula for finding the sum of terms of a G.P. which is given by ,${\text{S = }}\dfrac{{{\text{a}}\left( {{{\text{r}}^{\text{n}}} - 1} \right)}}{{{\text{r - 1}}}}$ , where r>1. We should also know how to find the A.M. of given numbers. Arithmetic mean is calculated as the ratio of sum of given numbers and total their total number.