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Home » The first term of G.P. is     and     term is     . Find the sum of its first     terms.
The first term of G.P. is

    \[27\]

and

    \[{{8}^{th}}\]

term is

    \[1/81\]

. Find the sum of its first

    \[10\]

terms.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
The first term of G.P. is

    \[27\]

and

    \[{{8}^{th}}\]

term is

    \[1/81\]

. Find the sum of its first

    \[10\]

terms.

From the question it is given that,

First term a =

    \[27\]

    \[{{8}^{th}}~term\text{ }{{a}_{8}}~=\text{ }1/81\]

Then,

    \[\begin{array}{*{35}{l}} {{a}_{n}}~=\text{ }a{{r}^{n\text{ }\text{ }1}} \\ {{a}_{8}}~=\text{ }a{{r}^{8\text{ }\text{ }1}}~=\text{ }1/81 \\ {{a}_{8}}~=\text{ }a{{r}^{7}}~=\text{ }1/81 \\ a{{r}^{7}}~=\text{ }1/81 \\ 27{{r}^{7}}~=\text{ }1/81 \\ {{r}^{7}}~=\text{ }1/\left( 81\text{ }\times \text{ }27 \right) \\ {{r}^{7}}~=\text{ }1/2187 \\ {{r}^{7}}~=\text{ }1/({{3}^{7}}) \\ r\text{ }=\text{ }1/3 \\ \end{array}\]

So,

    \[\begin{array}{*{35}{l}}</strong> <strong> {{S}_{10}}~=\text{ }a(1\text{ }\text{ }{{r}^{n}})/\left( 1\text{ }\text{ }r \right) \\</strong> <strong> =\text{ }27[1\text{ }\text{ }{{\left( 1/3 \right)}^{10}}]/\left( 1\text{ }\text{ }1/3 \right) \\</strong> <strong> =\text{ }27[1\text{ }\text{ }(1/{{3}^{10}})]/\left( \left( 3\text{ }\text{ }1 \right)/3 \right) \\</strong> <strong> =\text{ }\left( \left( 27\text{ }\times \text{ }3 \right)/2 \right)\text{ }[1\text{ }\text{ }1/{{3}^{10}}] \\</strong> <strong> =\text{ }\left( 81/2 \right)\text{ }[1\text{ }\text{ }1/{{3}^{10}}] \\</strong> <strong>\end{array}\]

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