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:

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

For example the sequence \(1 + 3x + 6 x^{2} + 9 x^{3} + 12 x^{4} + 15 x^{5} + 18 x^{6}\) is an arithmetico geometric sequence where the sequence 1 + 3 + 6 + 9 + 12 + 15 + 18 is in arithmetic progression and the sequence \(1 + x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6}\) is in geometric progression.

Sum of n terms of an Arithmetico Geometric Series:

\(S_{n} = a + [a + d] r + [a + 2d] r^{2} + [a + 3d] r_{3} + [a + 4d] r_{4} + . . . . . . . + [a + (n – 1)d] r ^{n – 1}\) [ Where d ≠ 0 and r ≠ 0 ] . . . . . . (1)

Now, multiplying the above equation by ‘r’ we get

\(r.S_{n} = ar + [a + d] r^{2} + [a + 2d] r^{3} + [a + 3d] r_{4} + [a + 4d] r_{5} + . . . . . . . + [a + (n – 1)d] r ^{n }\) [ Where d ≠ 0 and r ≠ 0 ] . . . . . . (2)

Now, Equation (2) – Equation (1) we get

\(S_{n} (1 – r) = a + d (r + r^{2} + r^{3} + . . . . . . + r^{n – 1} ) – [a + (n – 1) d] \times r ^ {n}\)

Or, Sn (1 – r) = a + \(\mathbf{d\;\left [ \;\frac{r\;(1\;-\;r^{n\;-\;1})}{1\;-\;r} \;\right ]}\) – [a + (n – 1) d] × r n

Or, \(\mathbf{S_{n}\;=\;\frac{a}{1\;-\;r}\;+\;d\;\left [ \;\frac{r\;(1\;-\;r\;^{n\;-\;1})}{1\;-\;r} \;\right ]\;-\;\frac{a\;+\;(n\;-\;1)\;d}{r^{n}}}\)

Sum of an Infinite Arithmetico Geometric Series:

If \(\mathbf{n\;\rightarrow \;\infty \;\;and\;\;\left |\; r \;\right |\;<\;1}\) then, \(r^{n}\) = 0.

Therefore, \(\mathbf{S_{\infty}\;=\;\frac{a}{1\;-\;r}\;+\;\frac{dr}{(1\;+\;r)^{2}}}\).

The Method of Differences:

Suppose \(p_{1}, p_{2}, p_{3}, p_{4}, . . . . . p_{n}\) is a given sequence such that \(p_{2} – p_{1}, p_{2} – p_{3}, . . . . . , p_{n} – p _{n-1}\) is either in an arithmetic or geometric progression, then, the sum of the given sequence can be evaluated by following the steps mentioned below:

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

Let, \(S= p_{1} + p_{2} + p_{3} + p_{4}+ . . . . .+ p_{n}\) . . . . . . . . . . (1)

Also, \(S = 0 + p_{1} + p_{2} + p_{3} + p_{4}+ . . . . .+ p_{n}\) . . . . . . . . . . (2)

Now, Equation (1) – Equation (2) we get

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

Or, \(t_{n} = p_{1} + (p_{2} – p_{1}) + (p_{3} – p_{2}) + \left (p_{4} – p_{3} \right ) + . . . . .+ (p_{n} -p_{n-1})\)

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

\(S_{n}\) = \(\mathbf{S_{n}\;=\;\sum_{n\;=\;1}^{n}\;\;t_{n}}\)

Important Results:

1. \(\mathbf{\sum_{r\;=\;1}^{n}\;\;(a_{r}\;\pm \;b_{r})\;=\;\sum_{r\;=\;1}^{n}\;a_{r}\;\pm \;\sum_{r\;=\;1}^{n}\;b^{r}}\)

2. \(\mathbf{\sum_{r\;=\;1}^{n}\;\;k\;a_{r}\;=\;k\;\sum_{r\;=\;1}^{n}\;a_{r}}\)

3. \(\mathbf{\sum_{r\;=\;1}^{n}\;\;k\;+\;k\;+\;k\;+\;k\;+\;.\;.\;.\;.\;.\;.\;+\;n\; times\;=\;n\;k}\) [ where, k = constant ]

4. \(\mathbf{\sum_{r\;=\;1}^{n}\;\;r\;=\;1\;+\;2\;+\;3\;+\;4\;+\;5\;+\;.\;.\;.\;.\;.\;+\;n\;=\;\frac{n\;(n\;+\;1)}{2}}\)

5. \(\mathbf{\sum_{r\;=\;1}^{n}\;\;r^{2}\;=\;1^{2}\;+\;2^{2}\;+\;3^{2}\;+\;.\;.\;.\;+\;n^{2}\;=\;\frac{n\;(n\;+\;1)\;(2\;n\;+\;1)}{6}}\)

6. \(\mathbf{\sum_{r\;=\;1}^{n}\;\;r^{3}\;=\;1^{3}\;+\;2^{3}\;+\;3^{3}\;+\;.\;.\;.\;+\;n^{3}\;=\;\frac{n^{2}\;(n\;+\;1)^{2}}{4}}\)