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Solved Subjective Question on Progression and Series Set 1

Posted on - 10-02-2017

JEE Math P And S

IIT JEE

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
&⇒
.

 
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