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in Binomial theorem by (52.7k points)

Let A1, A2, A3, …, Am be the arithmetic means between -2 and 1027 and G1, G2, G3, …, Gn be the geometric means between 1 and 1024. Product of the geometric means is 245 and the sum of arithmetic means is 1025 x 171.

1.  The common difference of the progression A1, A3, A5, …, Am-1 is

(A)  6 

(B)  3

(C)  2

(D)  1

2.  The numbers 2 A171, G25 + 1, 2A172 are in

(A)  AP

(B)  GP

(C)  HP

(D)  AGP

1 Answer

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Best answer

Correct answer 1.  (A) 2. (A)

1.  The common difference of sequence A1, A2,…, Am is 1027 + 2/342 + 1 = 3.

Therefore, the common difference of sequence A1, A3, A5,…, Am-1 is 6.

2.  We have A171 + A172 = -2 + 1027 = 1025. Therefore

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