Example 1
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 
Solution:
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, Sp+q
=
.
Example 2
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)
.
Solution:
Let first term =
a c. d = d, Sm = sum of first m terms.
Then given Sm
= Sm + n –Sm = Sm + p –Sm
Sm = Sm
+ n –Sm
&⇒ 2Sm = Sm
+ n
&⇒ 2
2a [2m –m –n] = d [m2
+ n2 + 2mn –m –n –2m2 + 2m]
2a [m –n] = d [ n2
–m2 + 2mn + m –n ] ….(1).
also 2 Sm
= Sm + p
&⇒ 2a [m –p] = d [p2
–m2 + 2mp +m –p] ….(2).
from (1) and (2)
&⇒ 
&⇒ 
&⇒ 
&⇒ 
&⇒ 

Example 3
Sum of the series to n
terms 
Solution:
Here
tr = 
= 
= 
&⇒ Sum Sn
=
=
.
Example 4
For positive real numbers x, y,
z prove that 
Solution:
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.)
&⇒ 
&⇒ xxyyzz
≥
. . .
(B)
Using (A) and (B)
, we get the result .
Example 5
If a, b, x, y are
positive natural numbers such that
then
prove that
.
Solution:
Consider
the positive numbers
ax,
ax, ….ky times and by, by, ….kx times .
For
all these numbers,
AM = 
=
.
GM = 
=
=
….(1)
As
,
, i.e. x + y
= xy
\ (1) becomes
or
.