Example 1
In an increasing G.P.,
the sum of the first and the last term is 66, the product of the second and the
last but one term is 128, and the sum of all the terms is 126. How many terms
are there in the progression.
Solution:
Let a
be the first term, r be the common difference and n be the number of terms
Since, a+ a rn-1
= 66 . . . . (1) .
Also, ar. arn-2
= 128.
&⇒ a2rn-1 = 128 .
. . . (2).
and , 
.
. . . . (3)
From (1) and (2),
a
&⇒ a2 – 66 a + 128 = 0
&⇒ (a – 64 ) (a – 2) = 0
so, a = 2 or 64
when a = 2. Then
4.rn-1 = 128 ( from (2).
&⇒ rn-1 = 32
&⇒ rn = 32r
so, from (3), 2
&⇒ 32r –1 = 63r – 63
&⇒ 31r = 62
&⇒ r = 2
so, rn =
64 = 26
&⇒ n = 6.
Similarly if a =
64, therefore r = 2 and then n = 6
Hence
number of terms = 6.
Example 2
The sum of two numbers
is 
and an even number of AMs are
inserted between the them such that their sum exceeds their number by 1.
Find the number of means inserted .
Solution:
Let n
(being even) AMs inserted between the a and b,
&⇒ a, A1,
A2, . . . . ., An, b are in AP and (n+2) terms are
there altogether. .
Now
since a +b = A1+ An = A2 + An-1
= . . . . constant. .
Also A1
+ A2 + . . . . . + An = n +1 (given).
&⇒ n
= n + 1
&⇒ n
= n + 1
&⇒ 13n = 12n +12
&⇒ n = 12. .
Example 3
f(x) = x – 2x2
+ 3x3 – …. + (–1)n–1 n.xn .
where x ∈
then find f¢(1/2). .
Solution:
Here
f¢(x) = 1 – 22x
+ 32x2 – 42x3+ ….+ (–1)n–1
n2xn–1.
\ f¢
= 
\ 
Subtracting 
Let Sn
= 
– 
Sn
= –
Subtracting
+ 
= 
terms + (–1)n
= 
= 
Thus 
\ f¢= 
.
Example 4
If a, b, c
are real positive quantities, show that 
.
Solution:
We
know 
….(1)
and
&⇒ 
….(2)
Now (a
–b) 2 + (b –c)2 + (c –a)2 ≥ 0
&⇒ a2
+ b2 + c2≥
ab + bc + ac
&⇒ (a
+ b + c)2 ≥ 3 (ab + bc + ac)
….(2)
From
(1) and (2)
….(3)
Form
(1) and (3)
equility
hold when a = b = c.
Example 5
In a centre test, there
are p questions, in this test 2p–r students give wrong answers to at
least r question ( 1 £ r £ p). If total
number of wrong answers given is 2047, find the value of p. .
Solution:
Number
of students giving wrong answers to at least r questions
= 2p
–r
Number
of students giving wrong answers to at least (r +1) questions = 2p–r–1
\ Number of students
giving wrong answers to exactly r questions
= 2p–r
– 2p–r–1. Also number of students giving wrong answers to exactly
p questions .
= 2p–p
= 2 0 = 1.
\ Total number of wrong
answers
1(2p–1
– 2p–2) + 2(2p –z – 2 p–3) + …. + (p–1) (21
– 20) + n(20).
= 2p–1
+ (–2p–2 + 2.2p–2) + (–2.2p–3 + 3.2p–3)
+ … + {–(n–1)20 + n.20} .
= 2p–1
+ 2p–2 + 2p–3 + …. + 20 = 2p –1 .
&⇒ 2p – 1 =
2047
&⇒ 2p =
2048 = 211
&⇒ p = 11
