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
SOLUTION: Solution on next page
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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
⇒
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