Show that the ration of sum of first n terms of a GP tto sum of terms from (n+1)th term to 2n th term is 1/r^n

Asked by lovemaan5500 | 13th Jan, 2018, 03:07: PM

Expert Answer:

begin mathsize 16px style Sum space of space straight n space terms space of space GP space is space
straight S subscript straight n equals straight a open parentheses fraction numerator 1 minus straight r to the power of straight n over denominator 1 minus straight r end fraction close parentheses............. left parenthesis straight i right parenthesis
Sum space of space terms space from space left parenthesis straight n plus 1 right parenthesis to the power of th space to space left parenthesis 2 straight n right parenthesis to the power of th space term
equals sum space of space straight a space GP space from space 1 to the power of st space to space open parentheses 2 straight n close parentheses to the power of th space term minus sum space of space straight a space GP space from space 1 to the power of st space to space straight n to the power of th space term
equals straight S subscript 2 straight n end subscript minus straight S subscript straight n
equals straight a open parentheses fraction numerator 1 minus straight r to the power of 2 straight n end exponent over denominator 1 minus straight r end fraction close parentheses minus straight a open parentheses fraction numerator 1 minus straight r to the power of straight n over denominator 1 minus straight r end fraction close parentheses
equals fraction numerator straight a over denominator 1 minus straight r end fraction open square brackets open parentheses 1 minus straight r to the power of 2 straight n end exponent close parentheses minus open parentheses 1 minus straight r to the power of straight n close parentheses close square brackets
equals fraction numerator straight a over denominator 1 minus straight r end fraction open square brackets 1 minus straight r to the power of 2 straight n end exponent minus 1 plus straight r to the power of straight n close square brackets
equals fraction numerator straight a over denominator 1 minus straight r end fraction open parentheses straight r to the power of straight n minus straight r to the power of 2 straight n end exponent close parentheses.............. left parenthesis ii right parenthesis
Required space ratio equals fraction numerator straight a open parentheses fraction numerator 1 minus straight r to the power of straight n over denominator 1 minus straight r end fraction close parentheses over denominator fraction numerator straight a over denominator 1 minus straight r end fraction open parentheses straight r to the power of straight n minus straight r to the power of 2 straight n end exponent close parentheses end fraction
equals fraction numerator 1 minus straight r to the power of straight n over denominator straight r to the power of straight n minus straight r to the power of 2 straight n end exponent end fraction
equals fraction numerator 1 minus straight r to the power of straight n over denominator straight r to the power of straight n open parentheses 1 minus straight r to the power of straight n close parentheses end fraction
equals 1 over straight r to the power of straight n
end style

Answered by Sneha shidid | 15th Jan, 2018, 09:55: AM