Example 1
Find the numbers a,b,c
between 2 and 18 such that (i) their sum is 25, (ii) the numbers 2, a, b are
consecutive terms of an A.P. and .
(iii) the numbers
b,c,18 are consecutive terms of a G.P.
Solution:
We have a + b + c =
25 . . . (1).
2, a, b are in A.P.
&⇒ 2a = 2 + b . . .
(2).
Also b, c, 18 are in
G.P
&⇒ 18b = c2 .
. . (3).
Substituting for a
and b in (1), (using relations (2) and (3)), we get
c2 + 12c
– 288 = 0
&⇒ (c – 12) (c + 24) =
0
&⇒ c = 12, or – 24
Since the numbers
lie between 2 and 18,
we take c = 12,
&⇒ b = 8, a = 5.
Example 2
Does
there exist a G.P. containing 27, 8, and 12 as three of its terms ? If it
exists, how many such progressions are possible.
Solution:
Let 8 be the m th,
12 the n th and 27 be the t th terms of a G.P. whose first term is A and common
ratio is R.
Then 8 = ARm -
1, 12 = ARn - 1, 27 = ARt- 1
&⇒ 2m - 2n = n - t and 3m - 3n
= m - t
&⇒ 2m + t = 3n and 2m
+ t = 3n
&⇒ 
There are infinity
of sets of values of m,n,t which satisfy this relation. For example, take m =
1, then 
By giving
different values to k we get integral values of n and t. Hence there are
infinite number of G.P.'s whose terms may be 27, 8, 12 (not consecutive).
Example 3
If a, b and c be in
G.P. and x, y be the arithmetic means between a , b and b, c respectively
then prove that 
and 
.
Solution:
Given b2
= ac, x = 
, y = 
Consider
= 
= 2
= 2.
Again,
a = 2x – b, c = 2y –b
Hence
b2 =( 2x –b)(2y–b) = 4xy - 2b(x + y) + b2
&⇒ 
.
Example 4
Let x = 1 + 3a + 6a2 + 10a3
+ ....., |a| < 1,.
y = 1 + 4b + 10b2 + 20b3
+ ......, |b| < 1.
Find S = 1 + 3(ab) + 5(ab)2
+ ........ in terms of x and y.
Solution:
We have x= 1 + 3a +
6a2 + 10a3 + ..........
&⇒ ax = a + 3a2
+ 6a3 + .............
&⇒ x (1 - a) = 1 + 2a + 3a2
+ 4a3 + ......... . . . (1).
&⇒ x(1 - a)a = a + 2a2
+ 3a3 + ................. . . . (2).
Subtracting, (2)
from (1), we get
x(1 - a)2
= 1 + a + a2 + ..... = 
&⇒ x = 
. .
. . (3)
Further, we have y =
1 + 4b + 10b2 + 20b3 + .... . . . (4).
&⇒ yb = b + 4b2 + 10b3
+ ...... . . . (5).
Subtracting, (5)
from (4), we get y(1 - b) = 1 + 3b + 6b2 + 10b3 + .....
= 
using (3)
&⇒ y = 
.
. .. (6)
Now S = 1 + 3(ab) +
5(ab)2+... . . . (7).
&⇒ (ab)S = (ab) + 3(ab)2
+ .... . . .. (8).
Subtracting, (8)
from (7), we get
S(1 - ab) = 1 + 2ab
+ 2(ab)2 + .....
S = 
.
. . . (9)
From (3) and (6), we
get a = 
so that S = 
= 
.
Example 5
A sequence of real
numbers a1, a2, a3, . . .. , an is
such that a1 = 0,
|a2| = |a1+1|, |a3| = |a2+1|, . . .
.,|an| = |an-1+1| . Prove that 
.
Solution:
|ai| = |ai-1+1|
for i = 2, . . . , n .
Squaring
we have ai2 = ai-12 + 2ai-1 +
1
&⇒ ai2
– ai-12 = 2ai-1 +1
&⇒ 
&⇒ 
&⇒ 
&⇒ 
(as 
)
&⇒ 2
&⇒ 
.
