An infinite geometric progression a 1- a 2- a 3- - has the property that a n -3 a n -1- a n -2- for every n - 1- If the suma 1- a 2- a 3- -32- then a 5 isA- 1 - 32B- 2 - 32C- 3 - 32D- 4 - 32

The correct option is C 3/32 According to question, a_n = 3(a_n+1 + a_n+2 + …) From the above equation we can conclude that every nth term is equal to three ti ...

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Question

An infinite geometric progression $a_{1}, a_{2}, a_{3}, . . . . .$ has the property that $a_{n} = 3 (a_{n + 1} + a_{n + 2} + . . . . . .)$ for every $n \geq 1.$ If the sum $a_{1}, a_{2}, a_{3}, . . . . . = 32,$ then $a_{5}$ is

1/32

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2/32

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3/32

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4/32

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Solution

The correct option is C
3/32

According to question, $a_{n} = 3 (a_{n + 1} + a_{n + 2} + \dots)$
From the above equation we can conclude that every nth term is equal to three times the sum of all the terms after the nth term i.e. 1st term i.e. 1st term is equal to 3 times the sum of all terms starting from 2nd term and 2nd term is equal to three times the sum of all the terms starting from 3rd term.
So, $a_{1} + a_{2} + a_{3} + a_{4} + \dots = 32 \Rightarrow a_{1} + (a_{1} / 3) = 32 \Rightarrow a_{1} = 24$
Then, $a_{2} + a_{3} + a_{4} + a + 5 + \dots = 32 - 24 = 8 \Rightarrow a_{2} + (a_{2} / 3) = 8 \Rightarrow a_{2} = 6$
So, common ratio = $a_{2} / a_{1} = 6 / 24 = \frac{1}{4}$
Therefore, $a_{5} = a r^{5 - 1} = a r^{4}$ = 24 x $(\frac{1}{4})^{4} = \frac{3}{32}$

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