# How many terms of G.P \[3,{{3}^{2}},{{3}^{3}}.....\] are needed to give the sum 120?

Answer

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Hint: We know that the given terms are in G.P. So we will use the formula of the sum of n terms of G.P that is \[{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\] and equate it to 120 to get the value of n that is the number of terms.

Here, we are given a series \[3,{{3}^{2}},{{3}^{3}}.....\] such that the sum of the terms is 120. We have to find the number of terms in the given series.

First of all, let us take the total number of terms as n.

We know that in geometric progression (G.P), the terms are as follows:

\[a,ar,a{{r}^{2}},a{{r}^{3}}.....\]

where ‘a’ is the first term and ‘r’ is the common ratio.

Now by substituting a = r = 3 in the above terms, we get the new G.P as,

\[3,{{3}^{2}},{{3}^{3}},{{3}^{4}}......\text{n terms}\]

This is the series given in the question. So now, we know that the series which is given in the question is in G.P.

We know that the sum of n terms of G.P is,

\[{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]

For series, \[3,{{3}^{2}},{{3}^{3}},{{3}^{4}}......\text{n terms}\], we know that a = 3 and r = 3. So, by substituting the value of a and r in the above equation, we get,

\[{{S}_{n}}=\dfrac{3\left( 1-{{3}^{n}} \right)}{\left( 1-3 \right)}\]

We are given the sum of terms of this series is 120. So by substituting \[{{S}_{n}}=120\] in the above equation, we get,

\[120=\dfrac{3\left( 1-{{3}^{n}} \right)}{\left( 1-3 \right)}\]

Or, \[\dfrac{3\left( 1-{{3}^{n}} \right)}{\left( -2 \right)}=120\]

By multiplying \[\left( \dfrac{-2}{3} \right)\] on both the sides of the above equation, we get,

\[\left( \dfrac{-2}{3} \right)\left( \dfrac{3}{-2} \right)\left( 1-{{3}^{n}} \right)=\left( \dfrac{-2}{3} \right)\left( 120 \right)\]

Or, \[\left( 1-{{3}^{n}} \right)=-80\]

Or, \[{{3}^{n}}=80+1\]

\[\Rightarrow {{3}^{n}}=81\]

We can write \[81={{3}^{4}}\]. So, we get,

\[\Rightarrow {{3}^{n}}={{3}^{4}}\]

We know that when \[{{a}^{p}}={{a}^{q}}\] then p = q for all the values of ‘a’ except -1, 0 and 1. By using this in the above equation, we get n = 4.

So our total terms are 4 and that are \[3,{{3}^{2}},{{3}^{3}},{{3}^{4}}\] or 3, 9, 27, 81.

Note: Here students can cross-check their answer as follows:

We are given that the sum of the terms is 120. We have got the terms as 3, 9, 27, 81. So their sum would be 3 + 9 + 27 + 81 = 120. So, our answer is correct.

Here, we are given a series \[3,{{3}^{2}},{{3}^{3}}.....\] such that the sum of the terms is 120. We have to find the number of terms in the given series.

First of all, let us take the total number of terms as n.

We know that in geometric progression (G.P), the terms are as follows:

\[a,ar,a{{r}^{2}},a{{r}^{3}}.....\]

where ‘a’ is the first term and ‘r’ is the common ratio.

Now by substituting a = r = 3 in the above terms, we get the new G.P as,

\[3,{{3}^{2}},{{3}^{3}},{{3}^{4}}......\text{n terms}\]

This is the series given in the question. So now, we know that the series which is given in the question is in G.P.

We know that the sum of n terms of G.P is,

\[{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]

For series, \[3,{{3}^{2}},{{3}^{3}},{{3}^{4}}......\text{n terms}\], we know that a = 3 and r = 3. So, by substituting the value of a and r in the above equation, we get,

\[{{S}_{n}}=\dfrac{3\left( 1-{{3}^{n}} \right)}{\left( 1-3 \right)}\]

We are given the sum of terms of this series is 120. So by substituting \[{{S}_{n}}=120\] in the above equation, we get,

\[120=\dfrac{3\left( 1-{{3}^{n}} \right)}{\left( 1-3 \right)}\]

Or, \[\dfrac{3\left( 1-{{3}^{n}} \right)}{\left( -2 \right)}=120\]

By multiplying \[\left( \dfrac{-2}{3} \right)\] on both the sides of the above equation, we get,

\[\left( \dfrac{-2}{3} \right)\left( \dfrac{3}{-2} \right)\left( 1-{{3}^{n}} \right)=\left( \dfrac{-2}{3} \right)\left( 120 \right)\]

Or, \[\left( 1-{{3}^{n}} \right)=-80\]

Or, \[{{3}^{n}}=80+1\]

\[\Rightarrow {{3}^{n}}=81\]

We can write \[81={{3}^{4}}\]. So, we get,

\[\Rightarrow {{3}^{n}}={{3}^{4}}\]

We know that when \[{{a}^{p}}={{a}^{q}}\] then p = q for all the values of ‘a’ except -1, 0 and 1. By using this in the above equation, we get n = 4.

So our total terms are 4 and that are \[3,{{3}^{2}},{{3}^{3}},{{3}^{4}}\] or 3, 9, 27, 81.

Note: Here students can cross-check their answer as follows:

We are given that the sum of the terms is 120. We have got the terms as 3, 9, 27, 81. So their sum would be 3 + 9 + 27 + 81 = 120. So, our answer is correct.

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