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:

Sn=a+[a+d]r+[a+2d]r2+[a+3d]r3+[a+4d]r4+.......+[a+(n1)d]rn1S_{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+6x2+9x3+12x4+15x5+18x61 + 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+x2+x3+x4+x5+x61 + x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6} is in geometric progression.

Sum of n terms of an Arithmetico Geometric Series:

Sn=a+[a+d]r+[a+2d]r2+[a+3d]r3+[a+4d]r4+.......+[a+(n1)d]rn1S_{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.Sn=ar+[a+d]r2+[a+2d]r3+[a+3d]r4+[a+4d]r5+.......+[a+(n1)d]rnr.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

Sn(1r)=a+d(r+r2+r3+......+rn1)[a+(n1)d]×rnS_{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 + d  [  r  (1    rn    1)1    r  ]\mathbf{d\;\left [ \;\frac{r\;(1\;-\;r^{n\;-\;1})}{1\;-\;r} \;\right ]} – [a + (n – 1) d] × r n

Or, Sn  =  a1    r  +  d  [  r  (1    r  n    1)1    r  ]    a  +  (n    1)  drn\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 n        and      r    <  1\mathbf{n\;\rightarrow \;\infty \;\;and\;\;\left |\; r \;\right |\;<\;1} then, rnr^{n} = 0.

Therefore, S  =  a1    r  +  dr(1  +  r)2\mathbf{S_{\infty}\;=\;\frac{a}{1\;-\;r}\;+\;\frac{dr}{(1\;+\;r)^{2}}}.

The Method of Differences:

Suppose p1,p2,p3,p4,.....pnp_{1}, p_{2}, p_{3}, p_{4}, . . . . . p_{n} is a given sequence such that p2p1,p2p3,.....,pnpn1p_{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=p1+p2+p3+p4+.....+pnS= p_{1} + p_{2} + p_{3} + p_{4}+ . . . . .+ p_{n} . . . . . . . . . . (1)

Also, S=0+p1+p2+p3+p4+.....+pnS = 0 + p_{1} + p_{2} + p_{3} + p_{4}+ . . . . .+ p_{n} . . . . . . . . . . (2)

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

0=p1+(p2p1)+(p3p2)+(p4p3)+.....+(pnpn1)tn0 = p_{1} + (p_{2} – p_{1}) + (p_{3} – p_{2}) + \left (p_{4} – p_{3} \right ) + . . . . .+ (p_{n} -p_{n-1}) – t_{n}

Or, tn=p1+(p2p1)+(p3p2)+(p4p3)+.....+(pnpn1)t_{n} = p_{1} + (p_{2} – p_{1}) + (p_{3} – p_{2}) + \left (p_{4} – p_{3} \right ) + . . . . .+ (p_{n} -p_{n-1})

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

SnS_{n} = Sn  =  n  =  1n    tn\mathbf{S_{n}\;=\;\sum_{n\;=\;1}^{n}\;\;t_{n}}

Important Results:

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

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

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

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

  5. r  =  1n    r2  =  12  +  22  +  32  +  .  .  .  +  n2  =  n  (n  +  1)  (2  n  +  1)6\mathbf{\sum_{r\;=\;1}^{n}\;\;r^{2}\;=\;1^{2}\;+\;2^{2}\;+\;3^{2}\;+\;.\;.\;.\;+\;n^{2}\;=\;\frac{n\;(n\;+\;1)\;(2\;n\;+\;1)}{6}}

  6. r  =  1n    r3  =  13  +  23  +  33  +  .  .  .  +  n3  =  n2  (n  +  1)24\mathbf{\sum_{r\;=\;1}^{n}\;\;r^{3}\;=\;1^{3}\;+\;2^{3}\;+\;3^{3}\;+\;.\;.\;.\;+\;n^{3}\;=\;\frac{n^{2}\;(n\;+\;1)^{2}}{4}}