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CAT 2017 Slot 2 — QA Question 33

Infinite Geometric ProgressionEasy

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

Answer & solution

Correct answer: 3/32

  • A

    1/32

  • B

    2/32

  • 3/32

  • D

    4/32

Solution

Easy

The defining relation an=3(an+1+an+2+⋯ )a_n=3(a_{n+1}+a_{n+2}+\cdots) ties any term to the tail sum after it. Use the infinite-GP tail-sum formula to find the common ratio rr, then use the total sum =32=32 to find the first term, and finally compute a5a_5.

1

Find the common ratio. Let an=arn−1a_n=ar^{n-1}. The tail after ana_n is itself a GP with first term arnar^{n} and ratio rr, summing to arn1−r\dfrac{ar^{n}}{1-r}.

arn−1=3⋅arn1−r⇒ 1=3r1−r(divide by arn−1)⇒ 1−r=3r⇒ r=14\begin{aligned} &ar^{n-1}=3\cdot\frac{ar^{n}}{1-r}\\ &\Rightarrow\ 1=\frac{3r}{1-r} \quad\text{(divide by }ar^{n-1}\text{)}\\ &\Rightarrow\ 1-r=3r\\ &\Rightarrow\ r=\tfrac14 \end{aligned}
2

Find the first term. The total sum of the infinite GP is 3232.

a1−r=32⇒ a1−14=32(from step 1)⇒ 4a3=32⇒ a=24\begin{aligned} &\frac{a}{1-r}=32\\ &\Rightarrow\ \frac{a}{1-\frac14}=32 \quad\text{(from step 1)}\\ &\Rightarrow\ \frac{4a}{3}=32\\ &\Rightarrow\ a=24 \end{aligned}
3

Compute a5a_5. a5=ar4a_5=ar^{4}.

a5=24⋅(14)4=24⋅1256⇒ a5=24256=332\begin{aligned} &a_5=24\cdot\left(\tfrac14\right)^{4}=24\cdot\frac{1}{256}\\ &\Rightarrow\ a_5=\frac{24}{256}=\frac{3}{32} \end{aligned}
a5=332a_5=\dfrac{3}{32}

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CAT 2017 Slot 2 QA Q33: An infinite geometric progression a 1 , a 2 , a 3 , … has the property that a n = 3( a n +1 + a n +2 + — Solution | TheCATExam