Question

Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from ${{\left( n+1 \right)}^{th}}$ to ${{\left( 2n \right)}^{th}}$ is $\dfrac{1}{{{r}^{n}}}$.

Answer 1

Hint: From the given series of geometric sequences, we find the general term of the series. We find the formula for ${{t}_{n}}$, the \[{{n}^{th}},2{{n}^{th}},\] term of the series. From the given sequence we find the common ratio which is the ratio between two consecutive terms. We put the values to get the ratio of the sum of first n terms of a G.P. to the sum of terms from ${{\left( n+1 \right)}^{th}}$ to ${{\left( 2n \right)}^{th}}$.

Complete step-by-step solution:
We have been given a series of geometric sequences.
We express the geometric sequence in its general form.
We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
The first term be ${{t}_{1}}$ and the common ratio be $r$ where $r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}}$.
We can express the general term ${{t}_{n}}$ based on the first term and the common ratio.
The formula being ${{t}_{n}}={{t}_{1}}{{r}^{n-1}}$.
The sum of the first n terms is ${{S}_{n}}={{t}_{1}}\dfrac{{{r}^{n}}-1}{r-1}$.
Putting the value of 2n in place of n we get the sum of the first 2n terms is ${{S}_{2n}}={{t}_{1}}\dfrac{{{r}^{2n}}-1}{r-1}$.
The sum of terms from ${{\left( n+1 \right)}^{th}}$ to ${{\left( 2n \right)}^{th}}$ is ${{S}_{2n}}-{{S}_{n}}={{t}_{1}}\dfrac{{{r}^{2n}}-1}{r-1}-{{t}_{1}}\dfrac{{{r}^{n}}-1}{r-1}={{t}_{1}}\left( \dfrac{{{r}^{2n}}-1}{r-1}-\dfrac{{{r}^{n}}-1}{r-1} \right)$
The simplified form is ${{S}_{2n}}-{{S}_{n}}={{t}_{1}}\left( \dfrac{{{r}^{2n}}-1}{r-1}-\dfrac{{{r}^{n}}-1}{r-1} \right)={{t}_{1}}\left( \dfrac{{{r}^{2n}}-{{r}^{n}}}{r-1} \right)$.
Now the ratio of the sum of first n terms of a G.P. to the sum of terms from ${{\left( n+1 \right)}^{th}}$ to ${{\left( 2n \right)}^{th}}$ is $\dfrac{{{S}_{n}}}{{{S}_{2n}}-{{S}_{n}}}=\dfrac{{{t}_{1}}\dfrac{{{r}^{n}}-1}{r-1}}{{{t}_{1}}\left( \dfrac{{{r}^{2n}}-{{r}^{n}}}{r-1} \right)}=\dfrac{\left( {{r}^{n}}-1 \right)}{{{r}^{n}}\left( {{r}^{n}}-1 \right)}=\dfrac{1}{{{r}^{n}}}$.

Note: The sequence is an increasing sequence where the common ratio is a positive number. The common difference will never be calculated according to the difference of greater number from the lesser number. The ratio formula should always be according $r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}}$.