Question

Find four numbers in G.P in which the sum of the extreme terms is 112 and the sum of middle terms is 48.

Answer 1

Hint – In this question let the 4 terms be $a,ar,a{r^2},a{r^3}$. Use the constraints to get the equation $a + a{r^3} = 112$, and $ar + a{r^2} = 48$. Solve for the value of a and r that is the first term and the common ratio. This will help getting the 4 numbers of G.P.
Complete step-by-step answer:
Let the four term G.P be
$a,ar,a{r^2},a{r^3}$
Where, (a) is the first term and (r) is the common ratio.
Now it is given that the sum of extreme ends is 112.
$ \Rightarrow a + a{r^3} = 112$
$ \Rightarrow a\left( {1 + {r^3}} \right) = 112$.................. (1)
And it is also given that the sum of middle terms is 48.
$ \Rightarrow ar + a{r^2} = 48$
$ \Rightarrow ar\left( {1 + r} \right) = 48$................... (2)
Now from equation (1) we have
$ \Rightarrow a = \dfrac{{112}}{{1 + {r^3}}}$
Substitute this value in equation (2) we have,
$ \Rightarrow \left( {\dfrac{{112}}{{1 + {r^3}}}} \right)r\left( {1 + r} \right) = 48$
Now as we know that $\left( {{a^3} + {b^3}} \right) = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$ so use this property in above equation we have,
$ \Rightarrow \left( {\dfrac{{112}}{{\left( {1 + r} \right)\left( {1 + {r^2} - r} \right)}}} \right)r\left( {1 + r} \right) = 48$
Now simplify this we have,
$ \Rightarrow 112r = 48\left( {1 + {r^2} - r} \right)$
$ \Rightarrow 48{r^2} - 160r + 48 = 0$
Now divide by 16 throughout we have,
$ \Rightarrow 3{r^2} - 10r + 3 = 0$
Now factorize this equation we have,
$ \Rightarrow 3{r^2} - 9r - r + 3 = 0$
$ \Rightarrow 3r\left( {r - 3} \right) - 1\left( {r - 3} \right) = 0$
$ \Rightarrow \left( {r - 3} \right)\left( {3r - 1} \right) = 0$
\[ \Rightarrow r = 3,\dfrac{1}{3}\]
Now from equation (1) we have,
If r = 3,
$ \Rightarrow a\left( {1 + {3^3}} \right) = 112$
$ \Rightarrow a = \dfrac{{112}}{{28}} = 4$
So we check whether this is right or wrong
$a + a{r^3} = 4 + 4\left( {{3^3}} \right) = 4 + 108 = 112$
So this is same value as in equation (1)
If $r = \dfrac{1}{3}$
$ \Rightarrow a\left( {1 + {{\left( {\dfrac{1}{3}} \right)}^3}} \right) = 112$
$ \Rightarrow a\left( {\dfrac{{28}}{{27}}} \right) = 112$
$ \Rightarrow a = 108$
So we check whether this is right or wrong
$a + a{r^3} = 108 + 108{\left( {\dfrac{1}{3}} \right)^3} = 108 + \dfrac{{108}}{{27}} = 108 + 4 = 112$
So this is also same value as in equation (1)
So these values are also the correct one,
So the G.P series is
$a,ar,a{r^2},a{r^3} = 4,4\left( 3 \right),4\left( {{3^2}} \right),4\left( {{3^3}} \right) = 4,12,36,108$
Or,
$a,ar,a{r^2},a{r^3} = 108,108\left( {\dfrac{1}{3}} \right),108\left( {\dfrac{1}{{{3^2}}}} \right),108\left( {\dfrac{1}{{{3^3}}}} \right) = 108,36,12,4$
So this is the required four term series of a G.P.

Note – A series is said to be in G.P if and only if the common ratio that is the ratio of consecutive terms of the series remains constant throughout the series, if we have a look at the general series of G.P than it can be taken as \[{a_1},{\text{ }}{a_1}r,{\text{ }}{a_1}{r^2}{\text{, }}{a_1}{r^3}{\text{, }}........................\infty \], then the common ratio is $\dfrac{{{a_1}r}}{{{a_1}}} = r{\text{ or }}\dfrac{{{a_1}{r^2}}}{{{a_1}r}} = r$, clearly it remains constant throughout, thus our prediction for four numbers in G.P is absolutely right.