Solved Subjective Question on Progression and Series Set 2

Posted on - 19-02-2017

JEE Math P And S

IIT JEE

Example 1

If |x| < 1 and |y| < 1 then prove that

(x +y) +(x² + xy + y²) + (x³ +x²y+xy² + y³) + . .. .¥ = .

Solution:

The given sum S = (x + y) +(x² + xy + y²) + . . . . .

=

=

= == .

Example 2

Let S = and show that assuming not all a_i ‘ s are equal .

Solution:

AM₁ = GM₁ = (P denotes product sign)

and AM₁ > GM₁ (as a_i ‘s are not all equal) . . . . (1) .

Again AM₂ = and GM₂ =

and AM₂ > GM₂ .. . . . (2).

By multiplying (1) and (2)

> 1

&⇒

&⇒ .

Example 3

If n is a root of the equation x²(1 – ac) – x(a² + c²) – (1 + ac) = 0 and if n harmonic means are inserted between a and c show that the difference between the first and last mean is equal to ac(a –c). .

Solution:

Given: n² (1 –ac) – n(a² +c²) –(1 + ac) = 0 . . . (1).

also, a, H₁ H₂ H₃ ……..H_n c are in H.P.

To prove H₁ –H_n = ac(a –c)

Now are in A.P.

&⇒ or d =

Example 4

Find the sum to n terms of the series

Solution:

The rth terms of the series is given by

t_r =

&⇒ t_{r + 1} =

&⇒ rt_r = (r + 4)t_{r +1}
&⇒ rt_r - (r + 1)t_{r + 1} = 3t_{r + 1}

Putting r = 1, 2, ...., n - 1 we get.

1t₁ - 2t₂ = 3t₂

2t₂ - 3t₃ = 3t₃

. . .

. . .

. . .

(n - 1)t_{n - 1} - nt_n = 3t_n

Adding the above equations, we get

t₁ - nt_n = 3[t₂ + t₃ + .... + t_n] .

&⇒ 4t₁ - nt_n = 3[t₁ + t₂ + .... + t_n].

&⇒

&⇒ t₁ + t₂ + ..... + t_n = - .

Example 5

Let a, b, c be three distinct positive real numbers in G.P., then
prove that a² + 2bc – 3ac > 0 . .

Solution:

Since a, b, c ∈ R⁺ and distinct

&⇒ AM > GM > HM

since b = and consider AM and HM of a and c,

&⇒

From first inequality (a +c) > 2b

&⇒ a² +ac – 2ab > 0

From second inequality b(a +c) > 2ac

&⇒ 2ab + 2bc –4ac > 0

Adding the two inequalities a² +2bc –3ac > 0 . .

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