Posted on - 10-02-2017

IIT JEE

Suppose
a_{1}, a_{2}, a_{3}........is an A.P. and b_{1},
b_{2}, b_{3}.......is a G.P. . Then the Progression a_{1}b_{1},
a_{2}b_{2}, ....... is said to be an arithmetico-geometric
progression (A.G.P). Hence an arithmetico-geometric progression is of the form
ab, (a+d)br, (a+2d)br^{2}, (a+3d)br^{3},.........

The sum S_{n}
of first n terms of an A.G. P. is obtained in the following way :.

S_{n} = ab +
(a + d)br + (a + 2d)br^{2} +.........+(a + (n - 2)d)br^{n-2} +
(a + (n - 1)d)br^{n-1}.

Multiply both sides by r, so that

r S_{n} = abr + (a + d)br^{2}+.........+(a
+ (n - 3)d)br^{n-2} + (a + (n - 2)d)br^{n-1} + (a + (n - 1)d)br^{n}.

Subtracting, we get

(1 - r)S_{n} = ab + dbr + dbr^{2}
+.......+dbr^{n - 2} + dbr^{n - 1} - (a + (n - 1)d)br^{n}.

=

&⇒

If -1 < r < 1, the sum of the
infinite number of terms of the progression is

= .

**Find the sum of the series 1****.2 + 2.22
+ 3.22 + ….. + 100.2100.**.

S = 1.2 + 2.2^{2}
+ 3.2^{3} + …. + 100.2^{100}.

2S = 1.2^{2}
+ 2.2^{3} + …. + 99.2^{100} + 100.2^{101}.

&⇒ –S = 1.2 + 1.2^{2} + 1.2^{3}
+ …. + 1.2^{100} – 100.2^{101}.

&⇒ -S = 1.2- 100.2^{101}

&⇒
S = -2^{101} + 2 + 100.2^{101} = 99.2^{101} + 2.