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if the sum to the infinity of the series 3+5r+7r²+.......is 44/9 , then find 'r'.
Dear Priyanshu S = 3 + 5r + 7r^2 + 9r^3 + ….................. rS = + 3r + 5r^2 + 7r^3 + …........... S (1 – r) = 3 + 2r + 2r^2 + 2r^3 + …....... S (1 – r) = 3 + 2r / (1 – r) 44/9 * (1 – r)^2 = 3 – 3r + 2r 44 + 44 r^2 – 88r = 27 – 9r 44 r^2 – 79r + 14 = 0 find r by solving quadratic equation
S=44/9S=3+5r+7r2+9r3++........-rS= -3r+5r³+7r4+.....S(1-r)=3+2r+2r²+2r³+............ Up to ∞S(1-r)=3+2r/1-r(Sum of infinite gp=a/1-r)44/9(1-r)=3-r/1-r44+44r²-88r=27-9r44r²-79r+17=0r=79±√6241-2992/2*44r=79±57/88r=1/4 or 1.54But r=1.54 does not satisfy the equation So r=1/4r=79±57/88r=
As s infinity is equal to a/1_r+Dr/{1_r}^2 a=3 ;d=2 ;r=r 3/1_r+2r/{1-r}^2Solving the above equation we will get the answer That is 1/4
As the sum of the above question is 44/9And by substituting the value44+44r^2_88r=27_9r Note _ means subtraction
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