Question 39

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Jambmaths question: 

Find the sum to infinity $\tfrac{1}{2},\tfrac{1}{6},\tfrac{1}{18}---$

Option A: 

$\tfrac{2}{3}$

Option B: 

$\tfrac{1}{3}$

Option C: 

$\tfrac{3}{4}$

Option D: 

1

Jamb Maths Solution: 

$\begin{align}  & {{S}_{\infty }}=\frac{a}{1-r} \\ & a=\frac{1}{2},\text{ }r=\frac{\tfrac{1}{6}}{\tfrac{1}{2}}=\frac{1}{3} \\ & S\infty =\frac{\tfrac{1}{2}}{1-\tfrac{1}{3}}=\frac{\tfrac{1}{2}}{\tfrac{2}{3}}=\frac{1}{2}\times \frac{3}{2}=\frac{3}{4} \\\end{align}$

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