NCERT/CBSE MATHEMATICS CLASS 11 textbook - TutorBreeze.com

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Answers to NCERT/CBSE MATH (Class XI)textbook. 9. SEQUENCES AND SERIES. SOLUTION: SOLUTION: Solution on next page. Solution on next page ...
NCERT/CBSE MATHEMATICS CLASS 11 textbook http://www.TutorBreeze.com MISCELLANEOUS EXERCISES Answers to NCERT/CBSE MATH (Class XI)textbook 9. SEQUENCES AND SERIES

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NCERT/CBSE MATHEMATICS CLASS 11 textbook http://www.TutorBreeze.com

Let the GP be given by : a , ar , ar 2 , ar3 , ar n 1 1 1 1 1 Then the reciprocal GP is given by : , , 2 , 3, a ar ar ar n ar We know that the sumof first n terms of a GP whose First term = a and Common ratio = r is a ( r n −1) Sn = r −1 a ( r n −1) ⇒S = r −1  1   1   1 −  n a   r   = 1− r ⇒R=  1 ar n-1 (1 − r ) −1 r a ( r n −1) S r − 1 = a 2 r n-1 ⇒ = 1− rn R ar n-1 (1 − r ) S ⇒ = a 2 r n-1 R n n S ⇒   =  a 2 r n-1  R n

 S n   2n n( n-1)  n  = a r  R  

⇒

2

3

n

n 1+ 2+3+.+ n

Now, P = a × ar × ar × ar × ar = a r ⇒P=a

n

n( n-1) r 2 2

n( n-1)   n n-1 2 n ⇒ P =  a r 2  =  a 2n r ( )       

 Sn  2 n =P R 

⇒

Please do not copy the answer given here Write to us for help

=a

n

n( n-1) r 2