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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 โฅ 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{๐}{1 โ ๐}) = 32.
We have been given that, an = 3(an+1 + an+2 +....) if n โฅ 1.
When n = 1, we get,
a = 3(a2 + a3 + .....)
โน a = \\frac{3๐๐}{1 โ ๐}) [Since, a2 + a3 +...... = \\frac{๐r}{1 โ ๐})]
โน 1 โ r = 3r
โน 1 = 4r
โน r = \\frac{1}{4})
Now finding the value of a is easy.
a = 32 ร (1 - \\frac{1}{4}))
โน a = 32 ร \\frac{3}{4})
โน a = 24
The value of a5 = ar4 , where a = 24 and r4 = (\\frac{1}{4}))4
a5 = 24 ร \\frac{1}{4}) ร \\frac{1}{4}) ร \\frac{1}{4}) ร \\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 โฅ 1. If the sum a1 + a2 + a3 +...... = 32, then a5 is"
Choice C is the correct answer
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