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Question: If \[{S_1},{S_2},{S_3}\] are respectively the sum of \[n,2n\] and \[3n\] terms of a G.P., then prove...

If S1,S2,S3{S_1},{S_2},{S_3} are respectively the sum of n,2nn,2n and 3n3n terms of a G.P., 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

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 nn terms is given by the formula Sn=a(rn−1)r−1{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}} where aa be the first term, nn be the number of terms and rr be the common ratio.
2.The square of difference of two numbers is given by the algebraic identity (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab

Complete step by step solution:
Let us consider aa to be the first term and rr to be the common ratio.
Now using the formula of the sum to nn terms, we can write
S1=a(rn−1)r−1{S_1} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}
S2=a(r2n−1)r−1{S_2} = \dfrac{{a\left( {{r^{2n}} - 1} \right)}}{{r - 1}}
and
S3=a(r3n−1)r−1{S_3} = \dfrac{{a\left( {{r^{3n}} - 1} \right)}}{{r - 1}}
We have to prove that S1(S3−S2)=(S2−S1)2{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, S1(S3−S2){S_1}\left( {{S_3} - {S_2}} \right).
Substituting the values of S1,S2{S_1},{S_2} and S3{S_3} in S1(S3−S2){S_1}\left( {{S_3} - {S_2}} \right), we get
S1(S3−S2)=a(rn−1)r−1[a(r3n−1)r−1−a(r2n−1)r−1]{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 ar−1\dfrac{a}{{r - 1}} as common from the parenthesis, we get
⇒S1(S3−S2)=ar−1×a(rn−1)r−1(r3n−1−r2n+1)\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
⇒S1(S3−S2)=(a2(r−1)2)×(rn−1)(r3n−r2n)\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
⇒S1(S3−S2)=a2(r−1)2(r4n−r3n−r3n+r2n)\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
⇒S1(S3−S2)=a2(r−1)2(r4n−2r3n+r2n)\Rightarrow {S_1}\left( {{S_3} - {S_2}} \right) = \dfrac{{{a^2}}}{{{{(r - 1)}^2}}}\left( {{r^{4n}} - 2{r^{3n}} + {r^{2n}}} \right)
⇒S1(S3−S2)=a2(r−1)2r2n(r2n+1−2rn)\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, (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab, we get
⇒S1(S3−S2)=a2(r−1)2r2n(rn−1)2\Rightarrow {S_1}\left( {{S_3} - {S_2}} \right) = \dfrac{{{a^2}}}{{{{(r - 1)}^2}}}{r^{2n}}{\left( {{r^n} - 1} \right)^2} ………………..(1)\left( 1 \right)
⇒(S2−S1)2=(a(r2n−1)r−1−a(rn−1)r−1)2\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 ar−1\dfrac{a}{{r - 1}} as common outside from the parenthesis, we get
⇒(S2−S1)2=a2(r−1)2(r2n−1−rn+1)2\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
⇒(S2−S1)2=a2(r−1)2(r2n−rn)2\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, (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab, we get
⇒(S2−S1)2=a2(r−1)2(r4n+r2n−2r3n)\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 r2n{r^{2n}} as common from the parenthesis, we get
⇒(S2−S1)2=a2(r−1)2r2n(r2n+1−2rn)\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
⇒(S2−S1)2=a2(r−1)2r2n(rn−1)2\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)(2)
From (1)(1) and (2)(2) , we have S1(S3−S2)=(S2−S1)2{S_1}\left( {{S_3} - {S_2}} \right) = {\left( {{S_2} - {S_1}} \right)^2}
Therefore, S1(S3−S2)=(S2−S1)2{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 nn, 2n2n and 3n3n are in G.P. and S1,S2,S3{S_1},{S_2},{S_3} are the respective sum of nn, 2n2n and 3n3n terms of a G.P. so the sum to nn terms is used. Also, here we are considering rr to be greater than 1, so we used the formula Sn=a(rn−1)r−1{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}. If
rr is smaller than 1, we will use the formula Sn=a(1−rn)1−r{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}. So we need to check the value of rr before applying the formula.