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Home » The     term of a G.P. is     and the sum of its n terms is     . If its common ratio is     , then find its first term.
The

    \[{{n}^{th}}\]

term of a G.P. is

    \[128\]

and the sum of its n terms is

    \[255\]

. If its common ratio is

    \[2\]

, then find its first term.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
The

    \[{{n}^{th}}\]

term of a G.P. is

    \[128\]

and the sum of its n terms is

    \[255\]

. If its common ratio is

    \[2\]

, then find its first term.

From the question it is given that,

The nth term of a G.P.

    \[{{T}_{n}}~=\text{ }128\]

The sum of its n terms

    \[{{S}_{n}}~=\text{ }255\]

Common ratio r =

    \[2\]

We know that,

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

    \[128\text{ }=\text{ }a{{2}^{n\text{ }\text{ }1}}\]

    \[a\text{ }=\text{ }128/{{2}^{n\text{ }\text{ }1}}~\]

… [equation (i)]

Also we know that,

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

    \[255\text{ }=\text{ }a({{2}^{n}}~\text{ }1)/\left( 2\text{ }\text{ }1 \right)\]

By cross multiplication we get,

    \[255\text{ }=\text{ }a({{2}^{n}}~\text{ }1)\]

    \[a\text{ }=\text{ }255/({{2}^{n}}~\text{ }1)\]

… [equation (ii)]

Now, consider both the equation(i) and equation (ii)

    \[255/({{2}^{n}}~\text{ }1)\text{ }=\text{ }128/({{2}^{n\text{ }\text{ }1}})\]

By cross multiplication we get,

    \[\begin{array}{*{35}{l}} 255\text{ }\times \text{ }{{2}^{n\text{ }\text{ }1}}~=\text{ }128({{2}^{n}}~\text{ }1) \\ 255\text{ }\times \text{ }{{2}^{n\text{ }\text{ }1~}}=\text{ }128\text{ }\times \text{ }{{2}^{n}}~\text{ }128 \\ (255\text{ }\times \text{ }{{2}^{n}})/2\text{ }=\text{ }128\text{ }\times \text{ }{{2}^{n}}~\text{ }128 \\ 255\text{ }\times \text{ }{{2}^{n}}~=\text{ }256\text{ }\times \text{ }{{2}^{n}}~\text{ }256 \\ 256\text{ }\times \text{ }{{2}^{n}}~\text{ }255\text{ }\times \text{ }{{2}^{n}}~=\text{ }256 \\ \end{array}\]

By simplification,

    \[\begin{array}{*{35}{l}} {{2}^{n}}~=\text{ }256 \\ {{2}^{n}}~=\text{ }{{2}^{8}} \\ \end{array}\]

By comparing both LHS and RHS, we get,

Then,

    \[\begin{array}{*{35}{l}} 128\text{ }=\text{ }a{{2}^{7}} \\ 128\text{ }=\text{ }a\text{ }\times \text{ }128 \\ a\text{ }=\text{ }128/128 \\ a\text{ }=\text{ }1 \\ \end{array}\]

Therefore, the value of a is

    \[1\]

.

i

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