Question

If \[{S_1},{S_2},{S_3}\] are respectively the sum of \[n,2n\] and \[3n\] terms of a G.P., then prove that \[{S_1}\left( {{S_3} - {S_2}} \right) = {\left( {{S_2} - {S_1}} \right)^2}\] .

Answer 1

Hint: Here, we will prove the given equation by using the geometric series formula. A geometric progression is a sequence or series in which any term after the first term is obtained by multiplying the preceding term by a constant called the common ratio.

Formula used: We will use the following formulas:
1.The sum to \[n\] terms is given by the formula \[{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\] where \[a\] be the first term, \[n\] be the number of terms and \[r\] be the common ratio.
2.The square of difference of two numbers is given by the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]

Complete step by step solution:
Let us consider \[a\] to be the first term and \[r\] to be the common ratio.
Now using the formula of the sum to \[n\] terms, we can write
\[{S_1} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\]
\[{S_2} = \dfrac{{a\left( {{r^{2n}} - 1} \right)}}{{r - 1}}\]
and
\[{S_3} = \dfrac{{a\left( {{r^{3n}} - 1} \right)}}{{r - 1}}\]
We have to prove that \[{S_1}\left( {{S_3} - {S_2}} \right) = {\left( {{S_2} - {S_1}} \right)^2}\] . This can be proved by proving L.H.S and R.H.S separately.
Now, first we will first take the expression on the left hand side, \[{S_1}\left( {{S_3} - {S_2}} \right)\].
Substituting the values of \[{S_1},{S_2}\] and \[{S_3}\] in \[{S_1}\left( {{S_3} - {S_2}} \right)\], we get
\[{S_1}\left( {{S_3} - {S_2}} \right) = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\left[ {\dfrac{{a\left( {{r^{3n}} - 1} \right)}}{{r - 1}} - \dfrac{{a\left( {{r^{2n}} - 1} \right)}}{{r - 1}}} \right]\]

Now, taking \[\dfrac{a}{{r - 1}}\] as common from the parenthesis, we get
\[ \Rightarrow {S_1}\left( {{S_3} - {S_2}} \right) = \dfrac{a}{{r - 1}} \times \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\left( {{r^{3n}} - 1 - {r^{2n}} + 1} \right)\]
Multiplying the common terms, we get
\[ \Rightarrow {S_1}\left( {{S_3} - {S_2}} \right) = \left( {\dfrac{{{a^2}}}{{{{(r - 1)}^2}}}} \right) \times \left( {{r^n} - 1} \right)\left( {{r^{3n}} - {r^{2n}}} \right)\]
Multiplying the terms, we get
\[ \Rightarrow {S_1}\left( {{S_3} - {S_2}} \right) = \dfrac{{{a^2}}}{{{{(r - 1)}^2}}}\left( {{r^{4n}} - {r^{3n}} - {r^{3n}} + {r^{2n}}} \right)\]
Adding the like terms, we get
\[ \Rightarrow {S_1}\left( {{S_3} - {S_2}} \right) = \dfrac{{{a^2}}}{{{{(r - 1)}^2}}}\left( {{r^{4n}} - 2{r^{3n}} + {r^{2n}}} \right)\]
\[ \Rightarrow {S_1}\left( {{S_3} - {S_2}} \right) = \dfrac{{{a^2}}}{{{{(r - 1)}^2}}}{r^{2n}}\left( {{r^{2n}} + 1 - 2{r^n}} \right)\]
By using the algebraic identity, \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], we get
\[ \Rightarrow {S_1}\left( {{S_3} - {S_2}} \right) = \dfrac{{{a^2}}}{{{{(r - 1)}^2}}}{r^{2n}}{\left( {{r^n} - 1} \right)^2}\] ………………..\[\left( 1 \right)\]
\[ \Rightarrow {\left( {{S_2} - {S_1}} \right)^2} = {\left( {\dfrac{{a\left( {{r^{2n}} - 1} \right)}}{{r - 1}} - \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}} \right)^2}\]
Taking \[\dfrac{a}{{r - 1}}\] as common outside from the parenthesis, we get
\[ \Rightarrow {\left( {{S_2} - {S_1}} \right)^2} = \dfrac{{{a^2}}}{{{{\left( {r - 1} \right)}^2}}}{\left( {{r^{2n}} - 1 - {r^n} + 1} \right)^2}\]

Subtracting the like terms, we get
\[ \Rightarrow {\left( {{S_2} - {S_1}} \right)^2} = \dfrac{{{a^2}}}{{{{\left( {r - 1} \right)}^2}}}{\left( {{r^{2n}} - {r^n}} \right)^2}\]
By using the algebraic identity, \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], we get
\[ \Rightarrow {\left( {{S_2} - {S_1}} \right)^2} = \dfrac{{{a^2}}}{{{{\left( {r - 1} \right)}^2}}}\left( {{r^{4n}} + {r^{2n}} - 2{r^{3n}}} \right)\]
Taking \[{r^{2n}}\] as common from the parenthesis, we get
\[ \Rightarrow {\left( {{S_2} - {S_1}} \right)^2} = \dfrac{{{a^2}}}{{{{\left( {r - 1} \right)}^2}}}{r^{2n}}\left( {{r^{2n}} + 1 - 2{r^n}} \right)\]
Again, by using the same algebraic identity, we get
\[ \Rightarrow {\left( {{S_2} - {S_1}} \right)^2} = \dfrac{{{a^2}}}{{{{\left( {r - 1} \right)}^2}}}{r^{2n}}{\left( {{r^n} - 1} \right)^2}\] ……………………………\[(2)\]
From \[(1)\] and \[(2)\] , we have \[{S_1}\left( {{S_3} - {S_2}} \right) = {\left( {{S_2} - {S_1}} \right)^2}\]
Therefore, \[{S_1}\left( {{S_3} - {S_2}} \right) = {\left( {{S_2} - {S_1}} \right)^2}\] is proved.

Note: To solve the question we must be clear about using the formula. The terms \[n\], \[2n\] and \[3n\] are in G.P. and \[{S_1},{S_2},{S_3}\] are the respective sum of \[n\], \[2n\] and \[3n\] terms of a G.P. so the sum to \[n\] terms is used. Also, here we are considering \[r\] to be greater than 1, so we used the formula \[{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\]. If
 \[r\] is smaller than 1, we will use the formula \[{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\]. So we need to check the value of \[r\] before applying the formula.