**Category : **JEE Main & Advanced

If the differences of the successive terms of a series are in A.P. or G.P., we can find \[{{n}^{th}}\] term of the series by the following steps :

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**Step I: **Denote the \[{{n}^{th}}\] term by \[{{T}_{n}}\] and the sum of the series upto \[n\] terms by \[{{S}_{n}}\].

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**Step II: **Rewrite the given series with each term shifted by one place to the right.

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**Step III: **By subtracting the later series from the former, find** \[{{T}_{n}}\]**.

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**Step IV: **From \[{{T}_{n}}\], \[{{S}_{n}}\] can be found by appropriate summation.

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*Example***:** Consider the series 1+ 3 + 6 + 10 + 15 +…..to \[n\] terms. Here differences between the successive terms are \[63,\text{ }106,\text{ }1510,\text{ }\ldots \ldots .\] *i.e., * 2, 3, 4, 5,…… which are in A.P. This difference could be in G.P. also. Now let us find its sum

\[S=1+3+6+10+15+.....+{{T}_{n-1}}+{{T}_{n}}\]

\[S=\,\,\,\,\,\,\,\,\,1+3+6+10+..........+{{T}_{n-1}}+{{T}_{n}}\]

Subtracting, we get

\[0=1+2+3+4+5+.........+({{T}_{n}}-{{T}_{n-1}})-{{T}_{n}}\]

\[\Rightarrow \] \[{{T}_{n}}=1+2+3+4+.........\]to \[n\] terms.

\[\Rightarrow \] \[{{T}_{n}}=\frac{1}{2}n(n+1)\] \[\therefore \] \[{{S}_{n}}=\Sigma {{T}_{n}}=\frac{1}{2}[\Sigma {{n}^{2}}+\Sigma n]\]

= \[\frac{1}{2}\left[ \frac{n(n+1)\,(2n+1)}{6}+\frac{n\,(n+1)}{2} \right]\] = \[\frac{n\,(n+1)\,(n+2)}{6}\].

*play_arrow*Definition*play_arrow*General term of a G.P.*play_arrow*Selection of Terms in a G.P.*play_arrow*Sum of first 'n' terms of a G.P.*play_arrow*Sum of infinite terms of a G.P.*play_arrow*Geometric Mean*play_arrow*Properties of G.P.*play_arrow*Definition*play_arrow*\[{{n}^{th}}\]term of A.G.P.*play_arrow*Sum of A.G.P.*play_arrow*Method for Finding Sum*play_arrow*Method of Difference*play_arrow*Special Series*play_arrow*Properties of Arithmetic, Geometric, Harmonic Means Between Two Given Numbers*play_arrow*Relation Between A.P., G.P. and H.P.

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