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Find the infinite sum of 10,5,2.5,1.25...

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User Lambros
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2 Answers

1 vote
to the risk of sounding redundant.

notice it goes from 10, to 5, to 2.5 and so forth, the next term is half the previous one, namely the multiplier, or "common ratio" is 1/2.

10 * 1/2 is 5, 5 * 1/2 is 2.5 and so on.

now, the Sum is going to infinity, however, when the common ratio is a fraction less than 1 and more than 0, either negative or positive, it is convergent, namely it reaches a limit, it doesn't go to neverland.

so, in this case, the common ratio is indeed | r | < 1, and therefore, it does reach a limit, keeping in mind the first term's value is 10.


\bf \textit{Sum of an infinite sequence}\\\\ S=\sum\limits_(i=0)^\infty~a_1\cdot r^i,\quad |~r~|~<~1\implies \cfrac{a_1}{1-r}\quad \begin{cases} a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ a_1=10\\ r=(1)/(2) \end{cases} \\\\\\ S=\cfrac{10}{1-(1)/(2)}\implies S=\cfrac{10}{(1)/(2)}\implies S=\cfrac{(10)/(1)}{(1)/(2)}\implies S=\cfrac{10}{1}\cdot \cfrac{2}{1}\implies S=20
answered
User MitoRibo
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8.2k points
1 vote
Notice this is a geometric progression since each number multiplied by some factor equals the next number in the sequence, in this case,


10 * r = 5 \Rightarrow r=0.5

Then by applying the formula for sum to infinity of a geometric progression,

S_\infty = (a)/(1-r) = (10)/(1-0.5) = 20
answered
User Neil Benn
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7.5k points

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