SELECT * FROM question_mgmt as q WHERE id=1028 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 how many terms of g p 3 32 33 hellip are nee

Question:

How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?

Answer:

The given G.P. is 3, 32, 33, …

Let n terms of this G.P. be required to obtain the sum as 120.

S subscript n space equals space fraction numerator a open parentheses r to the power of n space minus space 1 close parentheses over denominator r minus 1 end fraction

Here, a = 3 and r = 3

∴ S subscript n space equals space 120 space equals space fraction numerator 3 space open parentheses 3 to the power of n space minus 1 close parentheses over denominator 3 minus 1 end fraction rightwards double arrow 120 space equals space fraction numerator 3 open parentheses 3 to the power of n space minus 1 close parentheses over denominator 2 end fraction rightwards double arrow fraction numerator 120 space cross times 2 over denominator 3 end fraction space equals space 3 to the power of n space minus 1 rightwards double arrow 3 to the power of n space minus 1 space equals space 80 rightwards double arrow 3 to the power of n space equals space 81 rightwards double arrow 3 to the power of n space equals space 3 to the power of 4

∴  n = 4

Thus, four terms of the given G.P. are required to obtain the sum as 120.


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