Question

The ${{11}^{th}}$, ${{13}^{th}}$ and ${{15}^{th}}$terms of any G.P. are in
A. G.P.
B. A.P.
C. H.P.
D. A.G.P

Answer 1

This question is based on the series i.e. geometric progression. Find the all terms given in the question i.e. ${{11}^{th}}$, ${{13}^{th}}$ and ${{15}^{th}}$terms using the general formula of the ${{n}^{th}}$term of GP series. After finding the terms observe the relation between the given terms. By the relation we can find the final answer.

Complete step by step answer:
Geometric progression or in short we can say G.P., it is a kind of series or sequence in which the common ratio between the two consecutive terms is the same. Or in simple terms we can say that GP is a series in which a new term is obtained by multiplying the preceding term with some constant value $r$known as the common ratio. Let us assume that the given series is ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},.....$ . We can say that the given series is in GP if and only if it follows the condition of GP i.e. $\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}$ or by further simplification we can say that
${{a}_{2}}^{2}={{a}_{1}}{{a}_{3}}$ $.......(1)$
If we are required to find the ${{n}^{th}}$ term of the GP series, the formula used to find the ${{n}^{th}}$term is
${{a}_{n}}=a{{r}^{n-1}}$
Where,
{{a}_{n}}=$$$${{n}^{th}}term of the GP series
$a=$ first term of the GP series
$r=$ common difference
If we consider $a$ as the first term and $r$ as the common difference of the series then according to the formula the GP series will become $a,a{{r}^{1}},a{{r}^{2}},a{{r}^{3}},a{{r}^{4}},.....$.
To solve the question let us first find ${{11}^{th}}$, ${{13}^{th}}$ and ${{15}^{th}}$terms of the GP series
Use the formula of ${{n}^{th}}$ term of the GP series, we get
${{a}_{n}}=a{{r}^{n-1}}$

& \Rightarrow {{a}_{11}}=a{{r}^{11-1}} \\\ & \Rightarrow {{a}_{11}}=a{{r}^{10}} \\\ \end{aligned}$$ Similarly, we can get $${{a}_{13}}=a{{r}^{12}}$$ $${{a}_{15}}=a{{r}^{14}}$$ Now multiply $${{a}_{11}}$$ and $${{a}_{15}}$$terms, we get $$\begin{aligned} & \Rightarrow {{a}_{11}}{{a}_{15}}=a{{r}^{10}}\times a{{r}^{14}} \\\ & \Rightarrow {{a}_{11}}{{a}_{15}}=a{{r}^{24}} \\\ \end{aligned}$$ Or we can express the following expression as, $$\Rightarrow {{a}_{11}}{{a}_{15}}={{(a{{r}^{12}})}^{2}}$$ From the above expression we can say that, $${{a}_{13}}^{2}={{a}_{11}}{{a}_{15}}$$ And, from equation $$(1)$$ we also know that this is the condition of a GP series. Hence we can conclude that the $${{11}^{th}}$$, $${{13}^{th}}$$ and $${{15}^{th}}$$terms of any G.P. series are in GP i.e. geometric progression. **So, the correct answer is “Option A”.** **Note:** If we want to calculate the sum of GP series then we can determine it by the direct use of the formula and the formula used is $$Sum=\dfrac{a({{r}^{n}}-1)}{r-1}$$, where $$a$$is the first term of the series, $$r$$ is the common ratio between the consecutive terms and $$n$$is the number of terms present in the given GP series.