Question

In a G.P. ${T_2} + {T_5} = 216$ and ${T_4}:{T_6} = 1:4$ and all the terms are integers, then its first term is:-
A) 16
B) 14
C) 12
D) 15

Answer 1

Here we will use the formula of ${n^{th}}$ terms of G.P that is ${T_n} = a{r^{n - 1}}$ , where $a$ represents first term of the series and $r$ is the common ratio of the series, use this formula in the given expression and then simplify it.

Complete step by step answer :
Given: The sum of second and fifth term of the geometric progression series is given as 216 and ratio of fourth term to the sixth term is given as $1:4$ and all the terms are integers.
As the sum of the second and fifth term of the G.P series is given, that is ${T_2} + {T_5} = 216$.
The formula we apply here of ${n^{th}}$ term is G.P is ${T_n} = a{r^{n - 1}}$
Where $a$ represents the first term of the series and $r$ represents the common ratio.
Now, we will apply the formula.
$a{r^{2 - 1}} + a{r^{5 - 1}} = 216 \\\ ar + a{r^4} = 216 \\\ ar\left( {1 + {r^3}} \right) = 216{\text{ }} \to \left( 1 \right) \\\$
And also we have given the ratio which is ${T_4}:{T_6} = 1:4$.
Again we will apply the formula here.

$$\dfrac{{a{r^{4 - 1}}}}{{a{r^{6 - 1}}}} = \dfrac{1}{4} \\\ \dfrac{{a{r^3}}}{{a{r^5}}} = \dfrac{1}{4} \\\ \dfrac{1}{{{r^2}}} = \dfrac{1}{4} \\\ \Rightarrow {r^2} = 4 \\\ {\text{ }}r = 2 \\\$$

As we have formed the common ratio here therefore we will substitute its value in the equation (1).
$ar\left( {1 + {r^3}} \right) = 216 \\\ a\left( 2 \right)\left( {1 + {2^3}} \right) = 216 \\\ 2a\left( 9 \right) = 216 \\\ 18a = 216 \\\ a = \dfrac{{216}}{{18}} \\\ a = 12 \\\$
Hence the first term of the geometric progression series is 12.
Therefore option (c) is correct.
Note: We have found the values accordingly to the hint given in the question, first we have found the sum and then the ratio and applied the formula of ${n^{th}}$ term of G.P and in a step that is $\dfrac{1}{{{r^2}}} = \dfrac{1}{4}$ we have replaced numerator by denominator in both left hand side and right hand side.