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

Hint: 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
\[{{a}_{n}}=a{{r}^{n-1}}\]
\[\begin{align}
  & \Rightarrow {{a}_{11}}=a{{r}^{11-1}} \\
 & \Rightarrow {{a}_{11}}=a{{r}^{10}} \\
\end{align}\]
Similarly, we can get
\[{{a}_{13}}=a{{r}^{12}}\]
\[{{a}_{15}}=a{{r}^{14}}\]
Now multiply \[{{a}_{11}}\] and \[{{a}_{15}}\]terms, we get
\[\begin{align}
  & \Rightarrow {{a}_{11}}{{a}_{15}}=a{{r}^{10}}\times a{{r}^{14}} \\
 & \Rightarrow {{a}_{11}}{{a}_{15}}=a{{r}^{24}} \\
\end{align}\]
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.