Question 24

Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from

$open parentheses n plus 1 close parentheses to the power of t h end exponent space t o space open parentheses 2 n close parentheses to the power of t h end exponent space t e r m space i s space 1 over r to the power of n$

Answer

Let a be the first term and r be the common ratio of the G.P.

$S u m space o f space f i r s t space n space t e r m s space equals fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator open parentheses 1 minus r close parentheses end fraction$

Since there are n terms from (n +1)th to (2n)th term,

Sum of terms from(n + 1)th to (2n)th term = $fraction numerator a subscript n plus 1 end subscript open parentheses 1 minus r to the power of n close parentheses over denominator open parentheses 1 minus r close parentheses end fraction$

a _n +1 = ar ^{n + 1 – 1} = arⁿ

Thus, required ratio = $fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator open parentheses 1 minus r close parentheses end fraction cross times fraction numerator open parentheses 1 minus r close parentheses over denominator a r to the power of n open parentheses 1 minus r to the power of n close parentheses end fraction equals 1 over r to the power of n$