SELECT * FROM question_mgmt as q WHERE id=1039 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=6 AND chapterId=80 and ex_no='3' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 11th mathematics Sequence and Series show that the ratio of the sum of first nbsp n nbs

Question:

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

Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is 1 over r to the power of n.

 


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