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In an A.P., the sum of its first n terms is

    \[\mathbf{6n}\text{ }\text{ }{{\mathbf{n}}^{\mathbf{2}}}\]

. Find its

    \[\mathbf{2}{{\mathbf{5}}^{\mathbf{th}}}\]

term.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
In an A.P., the sum of its first n terms is

    \[\mathbf{6n}\text{ }\text{ }{{\mathbf{n}}^{\mathbf{2}}}\]

. Find its

    \[\mathbf{2}{{\mathbf{5}}^{\mathbf{th}}}\]

term.

Given,

    \[{{S}_{n}}~=\text{ }6n\text{ }\text{ }{{n}^{2}}\]

Now,

    \[{{S}_{1}}~=\text{ }6\left( 1 \right)\text{ }\text{ }{{\left( 1 \right)}^{2}}~=\text{ }6\text{ }\text{ }1\text{ }=\text{ }5\]

So, the first term a =

    \[5\]

And,

    \[\begin{array}{*{35}{l}} {{S}_{2}}~=\text{ }6\left( 2 \right)\text{ }\text{ }{{\left( 2 \right)}^{2}}~=\text{ }12\text{ }\text{ }4\text{ }=\text{ }8 \\ So,\text{ }{{T}_{1}}~+\text{ }{{T}_{2}}~=\text{ }8 \\ 5\text{ }+\text{ }{{T}_{2}}~=\text{ }8 \\ {{T}_{2}}~=\text{ }8\text{ }\text{ }5\text{ }=\text{ }3 \\ \end{array}\]

So, the common difference

    \[\begin{array}{*{35}{l}} d\text{ }=\text{ }{{T}_{2}}~\text{ }{{T}_{1}} \\ d\text{ }=\text{ }3\text{ }\text{ }5 \\ =\text{ }-2 \\ \end{array}\]

Thus, 25th term in the A.P is given by

    \[\begin{array}{*{35}{l}}</strong> <strong> {{T}_{25}}~=\text{ }5\text{ }+\text{ }\left( 25\text{ }\text{ }1 \right)\left( -2 \right) \\</strong> <strong> =\text{ }5\text{ }+\text{ }24\left( -2 \right) \\</strong> <strong> =\text{ }5\text{ }\text{ }48 \\</strong> <strong> =\text{ }-43 \\</strong> <strong>\end{array}\]

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