if a, b, c are in G.P., prove that 1/(a2 – b2) + 1/b2 = 1/(b2 – c2)

Question

if a, b, c are in G.P., prove that \(\dfrac{ 1}{(a^2 – b^2)} + \dfrac{1}{b^2} = \dfrac{1}{(b^2 – c^2)}\)

Asked by Aaryan | 1 year ago | 70

0 0

Class 11 Maths Sequences and Series Add Answer

Class 11 Maths Sequences and Series View Answer

Class 11 Maths Sequences and Series View Answer

If a is the G.M. of 2 and \( \dfrac{1}{4}\) find a.

Class 11 Maths Sequences and Series View Answer

Find the geometric means of the following pairs of numbers:

(i) 2 and 8

(ii) a³b and ab³

(iii) –8 and –2

Class 11 Maths Sequences and Series View Answer

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).

Class 11 Maths Sequences and Series View Answer

Accepted Answer

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

Let us consider LHS: \(\dfrac{ 1}{(a^2 – b^2) }+ \dfrac{1}{b^2}\)

Let us take LCM

= \(\dfrac{ (b^2 + a^2 – b^2)}{(a^2 – b^2)b^2}\)

= \(\dfrac{ a^2 }{ (a^2b^2 – b^4)}\)

= \( \dfrac{a^2 }{ a^2(b^2 – c^2)}\)

= \( \dfrac{1} {(b^2 – c^2)}\)

= RHS

LHS = RHS

Hence proved.

Answered by Aaryan | 1 year ago