Question

the common ratio of a geometric progression whose first term is 3 last term is 3072 and the sum of the series is 4095

Answer 1

r = 3 , last term = 486 , sum = 728

last term = arn-1 = 486 ..................1

sum = a(rn - 1)/r-1 = 728 ................2

divide both

728/486 = (rn-1)/[(r-1)(rn-1)

r = 3 so

728/486 = 3n-1/2x3n-1

729 - 1 / 243 = 3n - 1 /3n-1

36 - 1 / 35 = 3n - 1 / 3n-1

comparing both sides we get

n = 6 or n-1 = 5

total terms = 6

from eq 1 , arn-1 = 486

a 35 = 486 = 243x2 = 35x2

a = 2

firat term is 2 & total terms are 6 ,

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