IIT JEE

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.

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 = c^{2} .
. . (3).

Substituting for a and b in (1), (using relations (2) and (3)), we get

c^{2} + 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.

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.

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 = AR^{m -
1}, 12 = AR^{n - 1}, 27 = AR^{t- 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).

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 .

Given b^{2}
= ac, x = , y =

Consider =

= 2= 2.

Again, a = 2x – b, c = 2y –b

Hence
b^{2} =( 2x –b)(2y–b) = 4xy - 2b(x + y) + b^{2}

&⇒ .

Let x = 1 + 3a + 6a^{2} + 10a^{3}
+ ....., |a| < 1,.

y = 1 + 4b + 10b^{2} + 20b^{3}
+ ......, |b| < 1.

Find S = 1 + 3(ab) + 5(ab)^{2}
+ ........ in terms of x and y.

We have x= 1 + 3a +
6a^{2} + 10a^{3} + ..........

&⇒ ax = a + 3a^{2}
+ 6a^{3} + .............

&⇒ x (1 - a) = 1 + 2a + 3a^{2}
+ 4a^{3} + ......... . . . (1).

&⇒ x(1 - a)a = a + 2a^{2}
+ 3a^{3} + ................. . . . (2).

Subtracting, (2) from (1), we get

x(1 - a)^{2}
= 1 + a + a^{2} + ..... =

&⇒ x = . .
. . (3)

Further, we have y =
1 + 4b + 10b^{2} + 20b^{3} + .... . . . (4).

&⇒ yb = b + 4b^{2} + 10b^{3}
+ ...... . . . (5).

Subtracting, (5)
from (4), we get y(1 - b) = 1 + 3b + 6b^{2} + 10b^{3} + .....

= 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 =

= .

A sequence of real
numbers a_{1}, a_{2}, a_{3}, . . .. , a_{n} is
such that a_{1} = 0,

|a_{2}| = |a_{1}+1|, |a_{3}| = |a_{2}+1|, . . .
.,|a_{n}| = |a_{n-1}+1| . Prove that .

|a_{i}| = |a_{i-1}+1|
for i = 2, . . . , n .

Squaring
we have a_{i}^{2} = a_{i-1}^{2} + 2a_{i-1 }+
1

&⇒ a_{i}^{2}
– a_{i-1}^{2} = 2a_{i-1} +1

&⇒

&⇒

&⇒

&⇒ (as )

&⇒ 2

&⇒ .