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Find the third term of a G.P. whose common ratio is

    \[3\]

and the sum of whose first seven terms is

    \[2186\]

.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
Find the third term of a G.P. whose common ratio is

    \[3\]

and the sum of whose first seven terms is

    \[2186\]

.

From the question it is given that,

Common ratio r =

    \[3\]

    \[{{S}_{7}}~=\text{ }2186\]

We know that,

    \[{{S}_{n}}~=\text{ }a({{r}^{n}}~\text{ }1)/\left( r\text{ }\text{ }1 \right)\]

    \[\begin{array}{*{35}{l}} {{S}_{7}}~=\text{ }a({{3}^{n}}~\text{ }1)/\left( 3\text{ }\text{ }1 \right) \\ 2186\text{ }=\text{ }a({{3}^{n}}~\text{ }1)/2 \\ \end{array}\]

By cross multiplication,

    \[\begin{array}{*{35}{l}} \left( 2186\text{ }\times \text{ }2 \right)\text{ }=\text{ }a({{3}^{7}}~\text{ }1) \\ \left( 4372 \right)\text{ }=\text{ }a\left( 2187\text{ }\text{ }1 \right) \\ 4372\text{ }=\text{ }a2186 \\ a\text{ }=\text{ }4372/2186 \\ a\text{ }=\text{ }2 \\ \end{array}\]

Then,

    \[{{a}_{3}}~=\text{ }a{{r}^{3\text{ }\text{ }1}}\]

    \[\begin{array}{*{35}{l}}</strong> <strong> =\text{ }a{{r}^{2}} \\</strong> <strong> =\text{ }2\text{ }\times \text{ }{{3}^{2}} \\</strong> <strong> =\text{ }2\text{ }\times \text{ }9 \\</strong> <strong> {{a}_{3}}~=\text{ }18 \\</strong> <strong>\end{array}\]

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