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Question: If \({{S}_{1}},{{S}_{2}}\) and \({{S}_{3}}\) are respectively the sum of n, 2n and 3n terms of a GP,...

If S1,S2{{S}_{1}},{{S}_{2}} and S3{{S}_{3}} are respectively the sum of n, 2n and 3n terms of a GP, then prove that S1(S3−S2)=(S2−S1)2{{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)={{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}.

Explanation

Solution

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=a(rn−1)r−1S=\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=a(rn−1)r−1S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1} . . . . . . . . . . . . . . . (1)
In this question, we have to prove S1(S3−S2)=(S2−S1)2{{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)={{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}} where S1,S2{{S}_{1}},{{S}_{2}} and S3{{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,
S1=a(rn−1)r−1{{S}_{1}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}
Using formula (1), the sum of 2n terms is equal to,
S2=a(r2n−1)r−1{{S}_{2}}=\dfrac{a\left( {{r}^{2n}}-1 \right)}{r-1}
Using formula (1), the sum of 3n terms is equal to,
S3=a(r3n−1)r−1{{S}_{3}}=\dfrac{a\left( {{r}^{3n}}-1 \right)}{r-1}
Since we have to prove S1(S3−S2)=(S2−S1)2{{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right)={{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}, let us first find S1(S3−S2){{S}_{1}}\left( {{S}_{3}}-{{S}_{2}} \right). Substituting S1,S2{{S}_{1}},{{S}_{2}} and S3{{S}_{3}}, we get,
S1(S3−S2)=a(rn−1)r−1(a(r3n−1)r−1−a(r2n−1)r−1) ⇒S1(S3−S2)=a(rn−1)r−1(ar−1)((r3n−1)−(r2n−1)) ⇒S1(S3−S2)=(ar−1)2(rn−1)(r3n−1−r2n+1) ⇒S1(S3−S2)=(ar−1)2(rn−1)(r3n−r2n) ⇒S1(S3−S2)=(ar−1)2(rn−1)((r2n)(rn−1)) ⇒S1(S3−S2)=(ar−1)2(r2n)(rn−1)2.............(2) \begin{aligned} & {{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{aligned}
Now, we will find (S2−S1)2{{\left( {{S}_{2}}-{{S}_{1}} \right)}^{2}}. Substituting S1{{S}_{1}} and S2{{S}_{2}}, we get,

& {{\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{aligned}$$ 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.