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If ${{S}_{1}},{{S}_{2}}$ and ${{S}_{3}}$ are respectively the sum of n, 2n and 3n terms of a GP, then prove that ${{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)={{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}$.

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Last updated date: 13th Jul 2024
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Answer
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Hint: Assume a geometric progression having its first term as a and the common ratio as r. Use the formula for sum of geometric progression i.e. $S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ and find the sum of n, 2n and 3n terms of this GP.

Complete step-by-step answer:

Before proceeding with the question, we must know the formula that will be required to solve this question.
For a geometric progression with its first term as a and the common ratio as r, the sum of the first n terms of this GP is given by the formula,
$S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ . . . . . . . . . . . . . . . (1)
In this question, we have to prove ${{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)={{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}$ where ${{S}_{1}},{{S}_{2}}$ and ${{S}_{3}}$ are respectively the sum of n, 2n and 3n terms of a GP.
Let us assume a geometric progression having its first term as a and the common ratio as r.
Using formula (1), the sum of n terms is equal to,
${{S}_{1}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$
Using formula (1), the sum of 2n terms is equal to,
\[{{S}_{2}}=\dfrac{a\left( {{r}^{2n}}-1 \right)}{r-1}\]
Using formula (1), the sum of 3n terms is equal to,
${{S}_{3}}=\dfrac{a\left( {{r}^{3n}}-1 \right)}{r-1}$
Since we have to prove ${{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)={{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}$, let us first find ${{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)$. Substituting ${{S}_{1}},{{S}_{2}}$ and ${{S}_{3}}$, we get,
$\begin{align}
  & {{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}\left( \dfrac{a\left( {{r}^{3n}}-1 \right)}{r-1}-\dfrac{a\left( {{r}^{2n}}-1 \right)}{r-1} \right) \\
 & \Rightarrow {{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}\left( \dfrac{a}{r-1} \right)\left( \left( {{r}^{3n}}-1 \right)-\left( {{r}^{2n}}-1 \right) \right) \\
 & \Rightarrow {{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)={{\left( \dfrac{a}{r-1} \right)}^{2}}\left( {{r}^{n}}-1 \right)\left( {{r}^{3n}}-1-{{r}^{2n}}+1 \right) \\
 & \Rightarrow {{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)={{\left( \dfrac{a}{r-1} \right)}^{2}}\left( {{r}^{n}}-1 \right)\left( {{r}^{3n}}-{{r}^{2n}} \right) \\
 & \Rightarrow {{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)={{\left( \dfrac{a}{r-1} \right)}^{2}}\left( {{r}^{n}}-1 \right)\left( \left( {{r}^{2n}} \right)\left( {{r}^{n}}-1 \right) \right) \\
 & \Rightarrow {{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)={{\left( \dfrac{a}{r-1} \right)}^{2}}\left( {{r}^{2n}} \right){{\left( {{r}^{n}}-1 \right)}^{2}}.............\left( 2 \right) \\
\end{align}$
Now, we will find ${{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}$. Substituting ${{S}_{1}}$ and ${{S}_{2}}$, we get,
\[\begin{align}
  & {{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}={{\left( \dfrac{a\left( {{r}^{2n}}-1 \right)}{r-1}-\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1} \right)}^{2}} \\
 & \Rightarrow {{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}={{\left( \dfrac{a}{r-1} \right)}^{2}}{{\left( \left( {{r}^{2n}}-1 \right)-\left( {{r}^{n}}-1 \right) \right)}^{2}} \\
 & \Rightarrow {{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}={{\left( \dfrac{a}{r-1} \right)}^{2}}{{\left( {{r}^{2n}}-1-{{r}^{n}}+1 \right)}^{2}} \\
 & \Rightarrow {{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}={{\left( \dfrac{a}{r-1} \right)}^{2}}{{\left( {{r}^{2n}}-{{r}^{n}} \right)}^{2}} \\
 & \Rightarrow {{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}={{\left( \dfrac{a}{r-1} \right)}^{2}}{{\left( {{r}^{n}}\left( {{r}^{n}}-1 \right) \right)}^{2}} \\
 & \Rightarrow {{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}={{\left( \dfrac{a}{r-1} \right)}^{2}}{{\left( {{r}^{n}} \right)}^{2}}{{\left( {{r}^{n}}-1 \right)}^{2}} \\
 & \Rightarrow {{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}={{\left( \dfrac{a}{r-1} \right)}^{2}}{{r}^{2n}}{{\left( {{r}^{n}}-1 \right)}^{2}}.....................\left( 3 \right) \\
\end{align}\]
Comparing equation (2) and equation (3), we can say,
 ${{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)={{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}$
Hence proved.

Note: This question can also be solved by assuming the first term of the GP as 1 instead of a variable a. If we assume the first term of the GP as 1, our calculations become much simpler than in the case we have assumed the first term as a.