a + ar + ar^2 + .....∞ = 15 a^2 + (ar)^2 + (ar^2)^2 + ....∞ = 150. Find ar^3 + ar^4 + ar^6 + ...∞
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26/12/2021 1:52 pm
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a + ar + ar2 + .....∞ = 15
a2 + (ar)2 + (ar2)2 + ....∞ = 150. Find ar3 + ar4 + ar6 + ...∞
(a) 1/2
(b) 2/3
(c) 3/4
(d) 5/4
Answer
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26/12/2021 2:09 pm
Correct answer: (a) 1/2
Explanation:
a + ar + ar2 + .....∞ = 15
⟹ \(\frac{a}{1 - r}\) = 15 ....(1)
a2 + (ar)2 + (ar2)2 + ....∞ = 150
⟹ \(\frac{a^2}{1 - r^2}\) = 150 ....(2)
Square (1) and divide by 2
\(\frac{\frac{a^2}{1-r^2}}{\frac{a^2}{(1-r)^2}}\) = \(\frac{150}{15.15}\)
⟹ \(\frac{1 - r}{1 + r}\) = \(\frac{2}{3}\)
⟹ 3 - 3r = 2 + 2r
⟹ 1 = 5r
⟹ r = \(\frac{1}{5}\)
Substitute in (1)
\(\frac{a}{1 - \frac{1}{5}}\) = 15
\(\frac{5a}{4}\) = 15
a = 12
ar3 + ar4 + ar6 + ...∞
= \(\frac{ar^2}{1 + r^2}\)
= \(\frac{12 \times (\frac{1}{5})^2}{1 - (\frac{1}{5})^2}\)
= \(\frac{\frac{12}{25}}{\frac{24}{25}}\) = \(\frac{1}{2}\)
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