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Arithmetico Geometric Series IIT JEE Study Material

An arithmetico geometric series is obtained by term-by-term multiplication of a GP with the corresponding terms of an AP. The general term of an arithmetico geometric series is given by:

$$\begin{array}{l}S_{n} = a + [a + d] r + [a + 2d] r^{2} + [a + 3d] r_{3} + [a + 4d] r_{4} + . . . . . . . + [a + (n – 1)d] r ^{n – 1}\end{array}$$

For example, the sequence 1 + 3x + 6 x2 + 9 x3 + 12 x4 + 15 x5 + 18 x6 is an arithmetico geometric sequence where the sequence 1 + 3 + 6 + 9 + 12 + 15 + 18 is in arithmetic progression and the sequence 1 + x + x2 + x3 + x4 + x5 + x6Β is in geometric progression.

The Sum of n Terms of an Arithmetico Geometric Series

$$\begin{array}{l}S_{n} = a + [a + d] r + [a + 2d] r^{2} + [a + 3d] r_{3} + [a + 4d] r_{4} + . . . . . . . + [a + (n – 1)d] r ^{n – 1}\end{array}$$
[ Where d β  0 and r β  0 ] . . . . . . (1)

Now, multiplying the above equation by βrβ, we get

$$\begin{array}{l}r.S_{n} = ar + [a + d] r^{2} + [a + 2d] r^{3} + [a + 3d] r_{4} + [a + 4d] r_{5} + . . . . . . . + [a + (n – 1)d] r ^{n }\end{array}$$
[ Where d β  0 and r β  0 ] . . . . . . (2)

Now, Equation (2) β Equation (1), we get

$$\begin{array}{l}S_{n} (1 – r) = a + d (r + r^{2} + r^{3} + . . . . . . + r^{n β 1} ) β [a + (n – 1) d] \times r ^ {n}\end{array}$$

Or

$$\begin{array}{l}S_n (1 – r) = a + \mathbf{d\;\left [ \;\frac{r\;(1\;-\;r^{n\;-\;1})}{1\;-\;r} \;\right ]} – [a + (n – 1) d]\times r^n\end{array}$$

Or,

$$\begin{array}{l}\mathbf{S_{n}\;=\;\frac{a}{1\;-\;r}\;+\;d\;\left [ \;\frac{r\;(1\;-\;r\;^{n\;-\;1})}{1\;-\;r} \;\right ]\;-\;\frac{a\;+\;(n\;-\;1)\;d}{r^{n}}}\end{array}$$

The Sum of an Infinite Arithmetico Geometric Series

If n β β, and |r| < 1, then rn = 0.

Therefore,

$$\begin{array}{l}\mathbf{S_{\infty}\;=\;\frac{a}{1\;-\;r}\;+\;\frac{dr}{(1\;+\;r)^{2}}}\end{array}$$

The Method of Differences:

Suppose p1, p2, p3, p4,…., pn is a given sequence such that (p2 – p1), (p3 – p2),…., (pn – pn-1) is either in an arithmetic or geometric progression. The sum of the given sequence can be evaluated by the steps mentioned below:

1. Finding the nth term of the given sequence (tn):

Let,

$$\begin{array}{l}S= p_{1} + p_{2} + p_{3} + p_{4}+ . . . . .+ p_{n}. . . . . . . . . . (1)\end{array}$$

Also,

$$\begin{array}{l}S = 0 + p_{1} + p_{2} + p_{3} + p_{4}+ . . . . .+ p_{n}. . . . . . . . . . (2)\end{array}$$

Now, Equation (1) β Equation (2), we get;

$$\begin{array}{l}0 = p_{1} + (p_{2} – p_{1}) + (p_{3} – p_{2}) + \left (p_{4} – p_{3} \right ) + . . . . .+ (p_{n} -p_{n-1}) – t_{n}\end{array}$$

Or,

$$\begin{array}{l}t_{n} = p_{1} + (p_{2} – p_{1}) + (p_{3} – p_{2}) + \left (p_{4} – p_{3} \right ) + . . . . .+ (p_{n} -p_{n-1})\end{array}$$

1. Evaluating the sum of the given sequence using its nth term:

$$\begin{array}{l}\mathbf{S_{n}\;=\;\sum_{n\;=\;1}^{n}\;\;t_{n}}\end{array}$$

Important Results

1. $$\begin{array}{l}\mathbf{\sum_{r\;=\;1}^{n}\;\;(a_{r}\;\pm \;b_{r})\;=\;\sum_{r\;=\;1}^{n}\;a_{r}\;\pm \;\sum_{r\;=\;1}^{n}\;b^{r}}\end{array}$$
2. $$\begin{array}{l}\mathbf{\sum_{r\;=\;1}^{n}\;\;k\;a_{r}\;=\;k\;\sum_{r\;=\;1}^{n}\;a_{r}}\end{array}$$
3. $$\begin{array}{l}\mathbf{\sum_{r\;=\;1}^{n}\;\;k\;+\;k\;+\;k\;+\;k\;+\;.\;.\;.\;.\;.\;.\;+\;n\; times\;=\;n\;k}\end{array}$$
[ where, k = constant ]
4. $$\begin{array}{l}\mathbf{\sum_{r\;=\;1}^{n}\;\;r\;=\;1\;+\;2\;+\;3\;+\;4\;+\;5\;+\;.\;.\;.\;.\;.\;+\;n\;=\;\frac{n\;(n\;+\;1)}{2}}\end{array}$$
5. $$\begin{array}{l}\mathbf{\sum_{r\;=\;1}^{n}\;\;r^{2}\;=\;1^{2}\;+\;2^{2}\;+\;3^{2}\;+\;.\;.\;.\;+\;n^{2}\;=\;\frac{n\;(n\;+\;1)\;(2\;n\;+\;1)}{6}}\end{array}$$
6. $$\begin{array}{l}\mathbf{\sum_{r\;=\;1}^{n}\;\;r^{3}\;=\;1^{3}\;+\;2^{3}\;+\;3^{3}\;+\;.\;.\;.\;+\;n^{3}\;=\;\frac{n^{2}\;(n\;+\;1)^{2}}{4}}\end{array}$$