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Home » The nth term of an A.P. is     . Find the common difference.
The nth term of an A.P. is

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

. Find the common difference.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
The nth term of an A.P. is

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

. Find the common difference.

From the question it is given that,

nth term is

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

So,

    \[{{T}_{n}}~=\text{ }6n\text{ }+\text{ }2\]

Now, we start giving values,

    \[1,\text{ }2,\text{ }3,\text{ }\ldots \]

in the place of n, we get,

    \[\begin{array}{*{35}{l}} {{T}_{1}}~=\text{ }\left( 6\text{ }\times \text{ }1 \right)\text{ }+\text{ }2\text{ }=\text{ }6\text{ }+\text{ }2\text{ }=\text{ }8 \\ {{T}_{2}}~=\text{ }\left( 6\text{ }\times \text{ }2 \right)\text{ }+\text{ }2\text{ }=\text{ }12\text{ }+\text{ }2\text{ }=\text{ }14 \\ {{T}_{3}}~=\text{ }\left( 6\text{ }\times \text{ }3 \right)\text{ }+\text{ }2\text{ }=\text{ }18\text{ }+\text{ }2\text{ }=\text{ }20 \\ {{T}_{4}}~=\text{ }\left( 6\text{ }\times \text{ }4 \right)\text{ }+\text{ }2\text{ }=\text{ }24\text{ }+\text{ }2\text{ }=\text{ }26 \\ \end{array}\]

Therefore, A.P. is

    \[8,\text{ }14,\text{ }20,\text{ }26,\text{ }\ldots \]

So, common difference d =

    \[14\text{ }\text{ }8\text{ }=\text{ }6\]

i

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