For a, b 0, let 5a – b, 2a + b, a + 2b be in A.P. and (b + 1)2 , ab + 1, (a – 1)2 are in G.P. then compute (a–1 + b–1).

Question

For a, b > 0, let 5a – b, 2a + b, a + 2b be in A.P. and (b + 1)2 , ab + 1, (a – 1)2 are in G.P. then compute (a–1 + b–1).

Yash Baheti · Answer

Hi,

From the ap condition find one relation between a & b, which will come out as b=2a.
Now use the gp condition and put the result from ap condition, you will get a equation in single variable, either in a or b.
Solve this equation to get the values of one variable and from ap condition u will get second variable as well.

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For a, b > 0, let 5a – b, 2a + b, a + 2b be in A.P. and (b + 1)2 , ab + 1, (a – 1)2 are in G.P. then compute (a–1 + b–1).

For a, b > 0, let 5a – b, 2a + b, a + 2b be in A.P. and (b + 1)2
, ab + 1, (a – 1)2
are in G.P. then
compute (a–1 + b–1).

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For a, b > 0, let 5a – b, 2a + b, a + 2b be in A.P. and (b + 1)2 , ab + 1, (a – 1)2 are in G.P. then compute (a–1 + b–1).

For a, b > 0, let 5a – b, 2a + b, a + 2b be in A.P. and (b + 1)2, ab + 1, (a – 1)2 are in G.P. thencompute (a–1 + b–1).

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For a, b > 0, let 5a – b, 2a + b, a + 2b be in A.P. and (b + 1)2
, ab + 1, (a – 1)2
are in G.P. then
compute (a–1 + b–1).