In an infinite g.P. Each term is equal to three times the sum of all the terms that follow it and the sum of the first two terms is 15. Find the sum of the series up to infinity.
Answers
tn=3{sum of infinite -Sn}
ar^n-1=3{a/(1-r) -a(1-rⁿ)/(1-r)}a
ar^n-1=3{a-a(1-rⁿ)/1-r}
ar^n-1=3{a-a+arⁿ/1-r}
ar^n-1=3{arⁿ/1-r}
arⁿ/r=3{arⁿ/1-r}
1/r=3/1-r
1-r=3r
r=1/4
given :
a+ar=15
a(1+r)=15
a=15/(1+r)
a=15/(1+¼)
a=15/5/4
a=12
thus,
sum of ∞=a/(1-r)
=12/1-¼
=12/¾
=16(answer)
sum of the series up to infinity. = 16
Step-by-step explanation:
Let say first term = a
& common Ratio = r
Next term = ar
sum of the first two terms is 15
a + ar = 15
a(1 + r) = 15
Each term is equal to three times the sum of all the terms that follow it
a = 3 (ar + ar² +..............................+ ∞)
a = 3 ar/(1 - r)
=> 1 - r = 3r
=> 4r = 1
=> r = 1/4
a(1 + 1/4) = 15
=> a * 5/4 = 15
=> a = 12
so GP is
12 , 3 , 3/4 ..............................
Sum = 12/(1 - 1/4) = 48/3 = 16
sum of the series up to infinity. = 16
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