Question

A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms corresponding to odd places, then find its common ratio.

Answer 1

Hint: At first take sequence as $a,ar,a{{r}^{2}},a{{r}^{3}}..........a{{r}^{2n-1}}$ then according to the question write equation as\[a+ar+a{{r}^{2}}+.....a{{r}^{2n-1}}=\ 5\left( a+a{{r}^{2}}+.....a{{r}^{2n}} \right)\], then use the formula of the sum which is \[\dfrac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\], where a is first term, r is common ratio and n is number of terms.

Complete step-by-step solution -
In the question, a G.P. consists of an even term. If the sum of all term is 5 times the sum of terms occupying odd spaces, we have to find the common ratio.
Before proceeding we will learn briefly about geometric progression.
In mathematics, a geometric progression, also known as a geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio. For, for example, the sequence, 2, 6, 18, 54……… is a geometric progression with common ratio 3. Similarly, 10, 5, 2.5, 1.25 ………. Is a geometric sequence with common ratio $\dfrac{1}{2}$.
There is also a general form of geometric sequence which is,
$a,ar,a{{r}^{2}},a{{r}^{3}}..........a{{r}^{2n-1}}$
Where, $r\ne 0$ and a is first term or start value.
The nth term of the sequence is generally represented as $a{{r}^{n-1}}$.
The sum of sequence is also represented by formula: \[\dfrac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\]
Where r is a common ratio not equal to ‘0’ and n is a number of terms and a is the first term.
So, in the sequence of G.P. let there be 2n terms with a as first term and r as common ratio according to the question,
\[\text{sum of all terms}=\ 5\left( \text{sum of term occupying odd places} \right)\]
\[{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+.....{{a}_{2n}}=\ 5\left( {{a}_{1}}+{{a}_{3}}+{{a}_{5}}+.....{{a}_{2n-1}} \right)\]
Here sequence is $a,ar,a{{r}^{2}},a{{r}^{3}}..........a{{r}^{2n-1}}$
So,
\[{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+.....a{{r}^{2n-1}}=\ 5\left( {{a}_{1}}+a{{r}^{2}}+{{a}_{5}}+.....a{{r}^{2n-2}} \right)\]
Hence on simplifying we get,
\[a\left\{ \dfrac{{{r}^{2n}}-1}{r-1} \right\}\ =\ 5a\left\{ \dfrac{{{\left( {{r}^{2}} \right)}^{n}}-1}{{{r}^{2}}-1} \right\}\]
Hence, on simplifying we get.
\[1\ =\ \dfrac{5}{r+1}\]
So, \[r+1\ =\ 5\]
\[r\ =\ 4\]
Hence, the common ratio is 4.

Note: Instead of using \[\dfrac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\] students can use \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\], it will fetch same results. Also, if $r<1$ and we have to find the total sum up to infinity we will use the formula \[\dfrac{a}{\left( 1-r \right)}\].