Class 11 Mathematics - Chapter Sequence and Series NCERT Solutions | Which term of the following sequences:

Welcome to the NCERT Solutions for Class 11th Mathematics - Chapter Sequence and Series. This page offers a step-by-step solution to the specific question from Exercise 3, Question 5: which term of the following sequences....
Question 5

Which term of the following sequences:

2 comma space 2 square root of 2 comma 4 comma space...

Answer

(a) The given sequence is 2 comma space 2 square root of 2 comma 4 comma space...

Here, a = 2 and r = fraction numerator 2 square root of 2 over denominator 2 end fraction equals square root of 2

Let the nth term of the given sequence be 128.

a subscript n space end subscript equals space a r to the power of n minus 1 end exponent
rightwards double arrow open parentheses 2 close parentheses open parentheses square root of 2 close parentheses to the power of n minus 1 end exponent space equals space 128

rightwards double arrow open parentheses 2 close parentheses open parentheses 2 close parentheses to the power of fraction numerator n minus 1 over denominator 2 end fraction end exponent equals open parentheses 2 close parentheses to the power of 7

rightwards double arrow open parentheses 2 close parentheses to the power of fraction numerator n minus 1 over denominator 2 end fraction plus 1 end exponent equals open parentheses 2 close parentheses to the power of 7

s o space fraction numerator n minus 1 over denominator 2 end fraction plus 1 equals 7
rightwards double arrow fraction numerator n minus 1 over denominator 2 end fraction space equals space 6
rightwards double arrow n space minus 1 space equals space 12
rightwards double arrow n equals 13

Thus, the 13th term of the given sequence is 128.

(b) The given sequence is square root of 3 space comma space 3 comma space 3 square root of 3 space comma space... space i s space 729 ?

Here, a= square root of 3 space a n d space r space equals space fraction numerator 3 over denominator square root of 3 end fraction space equals space square root of 3

Let the nth term of the given sequence be 729.

a subscript n equals a r to the power of n minus 1 end exponent
s o
a r to the power of n minus 1 end exponent space equals space 729
rightwards double arrow open parentheses square root of 3 close parentheses space open parentheses square root of 3 close parentheses to the power of n minus 1 end exponent space equals space 729

rightwards double arrow open parentheses 3 close parentheses to the power of 1 half end exponent open parentheses 3 close parentheses to the power of fraction numerator n minus 1 over denominator 2 end fraction end exponent equals open parentheses 3 close parentheses to the power of 6

rightwards double arrow open parentheses 3 close parentheses to the power of 1 half plus fraction numerator n minus 1 over denominator 2 end fraction end exponent equals open parentheses 3 close parentheses to the power of 6
s o
1 half plus fraction numerator n minus 1 over denominator 2 end fraction equals 6
rightwards double arrow fraction numerator 1 plus n minus 1 over denominator 2 end fraction equals 6
rightwards double arrow n equals 12

Thus, the 12th term of the given sequence is 729.

(c) The given sequence is 1 third comma space 1 over 9 comma space 1 over 27 comma space... space

Here, 

a=1 third space space a n d space r space equals space 1 over 9 obelus divided by 1 third equals 1 third

Let the nth term of the given sequence be 1 over 19683.

a subscript n space equals space a r to the power of n minus 1 end exponent
s o
a r to the power of n minus 1 end exponent equals 1 over 19683
rightwards double arrow open parentheses 1 third close parentheses open parentheses 1 third close parentheses to the power of n minus 1 end exponent equals 1 over 19683
rightwards double arrow open parentheses 1 third close parentheses to the power of n equals open parentheses 1 third close parentheses to the power of 9
rightwards double arrow n equals 9

Thus, the 9th term of the given sequence is 1 over 19683.

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