Sir, please help me to solve the following problems relating arithmetic & geometric progressions.

Asked by  | 19th Sep, 2008, 02:44: PM

Expert Answer:

Q1. for A.P.  a=1, S9 =9/2 (2*1+(9-1)d) = 369 ==> d=10

a9 = a+(9-1)d = 81

so for G.P. a1 =a=1 and a9 = 81 = ar8 =r8 ==> r8=81 ==> r = 3

so seventh term of G.P. = ar6 = (3)6 = 27

please post the next question separately.

Answered by  | 19th Sep, 2008, 11:29: PM

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