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Question 33 : An infinite geometric progression a1, a2, a3,... has the property that an = 3(an+1 + an+2 +....) for every n ≥ 1. If the sum a1 + a2 + a3 +...... = 32, then a5 is
The sum up to infinity, a1 + a2 + a3 +...... = 32.
It can also be written as \\frac{𝑎}{1 − 𝑟}) = 32.
We have been given that, an = 3(an+1 + an+2 +....) if n ≥ 1.
When n = 1, we get,
a = 3(a2 + a3 + .....)
⟹ a = \\frac{3𝑎𝑟}{1 − 𝑟}) [Since, a2 + a3 +...... = \\frac{𝑎r}{1 − 𝑟})]
⟹ 1 – r = 3r
⟹ 1 = 4r
⟹ r = \\frac{1}{4})
Now finding the value of a is easy.
a = 32 × (1 - \\frac{1}{4}))
⟹ a = 32 × \\frac{3}{4})
⟹ a = 24
The value of a5 = ar4 , where a = 24 and r4 = (\\frac{1}{4}))4
a5 = 24 × \\frac{1}{4}) × \\frac{1}{4}) × \\frac{1}{4}) × \\frac{1}{4})
Hence a5 = \\frac{3}{32})
The question is "An infinite geometric progression a1, a2, a3,... has the property that an = 3(an+1 + an+2 +....) for every n ≥ 1. If the sum a1 + a2 + a3 +...... = 32, then a5 is"
Choice C is the correct answer
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