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CBSE Class 11-science Answered

Let a,b,c be respectively the sum of the first terms , the next n terms of a G.P .Show that a, b , c are in G.P.
Asked by Vineeth K | 22 Feb, 2015, 08:43: AM
answered-by-expert Expert Answer
L e t space x space b e space t h e space f i r s t space t e r m space a n d space r space b e space t h e space c o m m o n space r a t i o space o f space t h e space G. P. a equals fraction numerator x left parenthesis 1 minus r to the power of n right parenthesis over denominator 1 minus r end fraction....... left parenthesis i right parenthesis b equals fraction numerator x left parenthesis 1 minus r to the power of 2 n end exponent right parenthesis over denominator 1 minus r end fraction minus a rightwards double arrow a plus b equals fraction numerator x left parenthesis 1 minus r to the power of 2 n end exponent right parenthesis over denominator 1 minus r end fraction....... left parenthesis i i right parenthesis c equals fraction numerator x left parenthesis 1 minus r to the power of 3 n end exponent right parenthesis over denominator 1 minus r end fraction minus left parenthesis a plus b right parenthesis rightwards double arrow a plus b plus c equals fraction numerator x left parenthesis 1 minus r to the power of 3 n end exponent right parenthesis over denominator 1 minus r end fraction....... left parenthesis i i i right parenthesis left parenthesis i i right parenthesis divided by left parenthesis i right parenthesis comma fraction numerator a plus b over denominator a end fraction equals fraction numerator left parenthesis 1 minus r to the power of 2 n end exponent right parenthesis over denominator left parenthesis 1 minus r to the power of n right parenthesis end fraction 1 plus b over a equals 1 plus r to the power of n b equals a r to the power of n left parenthesis i i i right parenthesis divided by left parenthesis i i right parenthesis comma fraction numerator a plus b plus c over denominator a plus b end fraction equals fraction numerator left parenthesis 1 minus r to the power of 3 n end exponent right parenthesis over denominator left parenthesis 1 minus r to the power of 2 n end exponent right parenthesis end fraction 1 plus fraction numerator c over denominator a plus b end fraction equals fraction numerator left parenthesis 1 plus r to the power of n plus r to the power of 2 n end exponent right parenthesis over denominator left parenthesis 1 plus r to the power of n right parenthesis end fraction fraction numerator c over denominator a plus b end fraction equals fraction numerator r to the power of 2 n end exponent over denominator 1 plus r to the power of n end fraction fraction numerator c over denominator a plus a r to the power of n end fraction equals fraction numerator r to the power of 2 n end exponent over denominator 1 plus r to the power of n end fraction fraction numerator c over denominator a left parenthesis 1 plus r to the power of n right parenthesis end fraction equals fraction numerator r to the power of 2 n end exponent over denominator 1 plus r to the power of n end fraction c equals a r to the power of 2 n end exponent W e space c a n space s a y space t h a t space a comma a r to the power of n comma a r to the power of 2 n end exponent space a r e space i n space G P therefore a comma b comma c space a r e space i n space G P.
Answered by Prasenjit Paul | 23 Feb, 2015, 11:48: AM

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