IIT JEE

The p th term of an
A.P. is a and q th term is b. Prove that sum of it's

(p + q) terms is

Let x be the first term and y be the c . d of A.P. .

a = x + (p – 1 ) d

b = x + ( q – 1)d

&⇒ d = .
. . . . (1)

so, x = a -

=

Hence, S_{p+q}
= .

If the sum of m terms of an
arithmetical progression is equal to the sum of either the next n terms or
the next p terms, prove that

(m + n)= (m +p) .

Let first term =
a c. d = d, S_{m} = sum of first m terms.

Then given S_{m
}= S_{m + n} –S_{m} = S_{m + p} –S_{m}

S_{m} = S_{m
+ n} –S_{m}

&⇒ 2S_{m} = S_{m
+ n}

&⇒ 2

2a [2m –m –n] = d [m^{2
}+ n^{2} + 2mn –m –n –2m^{2} + 2m]

2a [m –n] = d [ n^{2}
–m^{2} + 2mn + m –n ] ….(1).

also 2 S_{m}
= S_{m + p }

&⇒ 2a [m –p] = d [p^{2}
–m^{2} + 2mp +m –p] ….(2).

from (1) and (2)

&⇒

&⇒

&⇒

&⇒

&⇒

Sum of the series to n terms

Here
t_{r} =

=

=

&⇒ Sum S_{n}
= = .

For positive real numbers x, y, z prove that

Let x, y, z be three numbers with weights x, y, z respectively. Then.

(weighted A. M. ≥ weighted G. M.)

. . . (A)

Again let be three numbers with weights x, y, z respectively then

(weighted A.M ≥ weighted G.M.)

&⇒

&⇒ x^{x}y^{y}z^{z}
≥ . . .
(B)

Using (A) and (B) , we get the result .

If a, b, x, y are positive natural numbers such that then prove that .

Consider the positive numbers

a^{x},
a^{x}, ….ky times and b^{y}, b^{y}, ….kx times .

For all these numbers,

AM =

= .

GM =

= = ….(1)

As , , i.e. x + y = xy

\ (1) becomes or .