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if the sum to the infinity of the series 3+5r+7r²+.......is 44/9 , then find 'r'.
one year ago

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
one year ago

S=44/9
S=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-r
44+44r²-88r=27-9r
44r²-79r+17=0
r=79±√6241-2992/2*44
r=79±57/88
r=1/4 or 1.54
But r=1.54 does not satisfy the equation
So r=1/4
r=79±57/88
r=
one year ago

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}^2
Solving the above equation we will get the answer
That is 1/4
one year ago

As the sum of the above question is 44/9
And by substituting the value
44+44r^2_88r=27_9r

Note _ means subtraction
one year ago
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