# What is AGP – How to Solve AGP Series

Here you will learn what is agp (Arithmetico Geometric Series) and how to solve agp series.

Let’s begin –

## What is AGP (Arithmetico – Geometric series)

A series, each term of which is formed by multiplying the corresponding term of an A.P. & G.P. is called the Arithmetico-Geometric Series, e.g. 1 + 3x + 5$$x^2$$ + 7$$x^3$$ + ……..

Here 1, 3, 5, ……. are in A.P. & 1, x, $$x^2$$, $$x^3$$ …….. are in G.P.

### (a) Sum of N terms of an arithmetico-geometric series :

Let $$S_n$$ = a + (a + d)r + …….. + [a + (n-1)d]$$r^{n-1}$$

then $$S_n$$ = $$a\over {1-r}$$ + $$dr({1}-{r}^{n-1})\over (1-r)^2$$ – $$[a + (n-1)d]r^2\over {1-r}$$, r $$\ne$$ 1

### (b) Sum of Infinity :

If 0 < |r| < 1 & n $$\rightarrow$$ $$\infty$$   $$\displaystyle \lim_{x \to \infty}$$ $$r^n$$ = 0, $$S_{\infty}$$ = $$a\over {1-r}$$ + $$dr\over (1-r)^2$$

Example : Find the sum of series 4 – 9x + 16$$x^2$$ – 25$$x^3$$ + 36$$x^4$$ – 49$$x^5$$ + ……. $$\infty$$

Solution : Let S = 4 – 9x + 16$$x^2$$ – 25$$x^3$$ + 36$$x^4$$ – 49$$x^5$$ + ……. $$\infty$$

-Sx = -4x + 9$$x^2$$ – 16$$x^3$$ + 25$$x^4$$ – 36$$x^5$$ + ……. $$\infty$$

On Subtraction, we get

S(1 + x) = 4 – 5x + 7$$x^2$$ – 9$$x^3$$ + 11$$x^4$$ – 13$$x^5$$ + ……. $$\infty$$

-S(1 + x)x = -4x + 5$$x^2$$ – 7$$x^3$$ + 9$$x^4$$ – 11$$x^5$$ + ……. $$\infty$$

On Subtraction, we get

S$$(1 + x)^2$$ = 4 – x + 2$$x^2$$ – 2$$x^3$$ + 2$$x^4$$ – 2$$x^5$$ + ……. $$\infty$$

= 4 – x + 2$$x^2$$(1 – x + $$x^2$$ – $$x^3$$ + ……. $$\infty$$) = 4 – x + $$2x^2\over {1+x}$$ = $${4 + 3x + x^2}\over {1+x}$$

S = $${4 + 3x + x^2}\over (1+x)^3$$

#### RESULTS

(a)  $${\sum}_{r=1}^{n‎}$$r = $$n(n + 1)\over 2$$    (sum of the first n natural numbers)

(b)  $${\sum}_{r=1}^{n‎} r^2$$ = $$n(n + 1)(2n + 1)\over 6$$    (sum of the squares of the first n natural numbers)

(c)  $${\sum}_{r=1}^{n‎} r^3$$ = $$n^2(n + 1)^2\over 4$$    (sum of the cubes of the first n natural numbers)

(d)  $${\sum}_{r=1}^{n‎} r^4$$ = $$n(n + 1)(2n + 1)(3n^2 + 3n -1)\over 30$$

(e)  $${\sum}_{r=1}^{n‎}$$(2r – 1) = $$n^2$$    (sum of the first n odd natural numbers)

(f)  $${\sum}_{r=1}^{n‎}$$2r = n(n + 1)    (sum of the first n even natural numbers)

Note :

If $$n^{th}$$ terms of a sequence is given by $$T_n$$ = $$an^3$$ + $$bn^2$$ + cn + d where a, b, c, d are constants,

then sum of n terms $$S_n$$ = $$\sum T_n$$ = a$$\sum n^3$$ + b$$\sum n^2$$ + c$$\sum n$$ + $$\sum d$$

This can be evaluated using the above results.

Example : Sum upto 16 terms of the series $$1^3\over 1$$ + $$1^3 + 2^3\over 1 + 3$$ + $$1^3 + 2^3 + 3^3\over 1 + 3 + 5$$ +…….. is

Solution : $$t_n$$ = $$1^3 + 2^3 + 3^3 + …. + n^3\over 1 + 3 + 5 + ….. + (2n – 1)$$

= $${n^2(n + 1)^2\over 4}\over n/2{[2+2(n – 1)]}$$ = $${n^2(n + 1)^2\over 4}\over n^2$$ = $${(n + 1)^2\over 4}$$ = $$n^2\over 4$$ + $$n\over 4$$ + $$1\over 4$$

$$\therefore$$    $$S_n$$ = $$\sum t_n$$ + $$1\over 4$$$$\sum n^2$$ + $$1\over 2$$$$\sum n$$ + $$1\over 4$$$$\sum 1$$ = $$1\over 4$$.$$n(n + 1)(2n + 1)\over 6$$ + $$1\over 2$$.$$n(n + 1)\over 2$$ + $$1\over 4$$.n

$$\therefore$$    $$S_{16}$$ = $$16.17.33\over 24$$ + $$16.17\over 4$$ + $$16\over 4$$ = 446.