Question

Let\[{{a}_{n}}\]the nth term of the G.P of positive numbers. Let \[\sum\limits_{n-1}^{100}{{{a}_{2n}}}=\alpha \]and \[\sum\limits_{n=1}^{100}{{{a}_{2n-1}}}=\beta \],such that \[\alpha \ne \beta \]. Prove that the common ratio of the G.P is \[\dfrac{\alpha }{\beta }\].

Answer 1

Hint: Assume a GP series with first term as ‘a’ and common ratio as ‘r’. Then, expand the given summations accordingly using the formula at sum of ‘n ‘terms in G.P.

Complete step-by-step answer:

Let us assume a Geometric Progression series, with the first term as ‘a’ and common ratio as ‘r’

So, the nth term of the G.P can be written as \[{{a}_{n}}=a.{{r}^{n-1}}\].

Now let us consider,

\[\sum\limits_{n-1}^{100}{{{a}_{2n}}}=\alpha \].

Expanding the summation, we will have

\[{{a}_{2}}+{{a}_{4}}+.........+{{a}_{200}}=\alpha \]

Since we already know that \[{{a}_{n}}=a.{{r}^{n-1}}\], the above equation can be written as,

 \[a.r+a.{{r}^{3}}+.........+a.{{r}^{199}}=\alpha \].

So, it is a G.P series with the first term as ‘\[ar\]’ and the common ratio is ‘\[{{r}^{2}}\]’.

Therefore, let us apply \[{{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}=\]sum of “n” terms in

 G.P.

Applying the sum of n terms formula, we get:


\[\alpha =ar\left( \dfrac{{{\left( {{r}^{2}} \right)}^{100}}-1}{{{r}^{2}}-1} \right)...........(1)\]

Now, let us consider

\[\sum\limits_{n=1}^{100}{{{a}_{2n-1}}}=\beta \].

Expanding the summation, we will have

\[{{a}_{1}}+{{a}_{3}}+.........+{{a}_{199}}=\beta \]
.
Since, \[{{a}_{n}}=a.{{r}^{n-1}}\] we can write the above equation as:

\[a.r+a.{{r}^{2}}+.........+a.{{r}^{199}}=\beta \]

It is a G.P series with the first term as “a” and the common ratio is ‘\[{{r}^{2}}\]’.

Therefore applying the sum of n terms in G.P, that is \[{{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1

\right)}{r-1}\] we have: \[\beta =\dfrac{a\left( {{\left( {{r}^{2}} \right)}^{100}}-1

\right)}{{{r}^{2}}-1}.............(2)\]

Now dividing the equations (1) and (2), we shall have:

\[\dfrac{\alpha }{\beta }=\dfrac{\dfrac{ar\left( {{r}^{200}}-1 \right)}{{{r}^{2}}-1}}{\dfrac{a\left(

{{r}^{200}}-1 \right)}{{{r}^{2}}-1}}\]

\[\dfrac{\alpha }{\beta }=\dfrac{a.r}{a}=r\]

Hence, we have proven that common ratio of G.P is \[\dfrac{\alpha }{\beta }\], when

\[\sum\limits_{n-1}^{100}{{{a}_{2n}}}=\alpha \]and

\[\sum\limits_{n=1}^{100}{{{a}_{2n-1}}}=\beta \] are given.

Note: Sum of ‘n ‘terms of a G.P series is \[{{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}\],where a= first term, r=common ratio, n= no of terms. Properly substitute the first term and the common ratio in the sum of n terms formula and evaluate accordingly to avoid mistakes.