Courses
Courses for Kids
Free study material
Free LIVE classes
More

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

Last updated date: 23rd Mar 2023
Total views: 306.9k
Views today: 6.85k
Verified
306.9k+ views
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.

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.