5th Term of Infinite Geometric Progression

CAT 2017 Slot 2 · QA · Medium · Progressions

An infinite geometric progression a1,a2,a3,a_1, a_2, a_3, \dots has the property that an=3(an+1+an+2+)a_n = 3(a_{n+1} + a_{n+2} + \dots) for every n1n \ge 1. If the sum a1+a2+a3+=32a_1 + a_2 + a_3 + \dots = 32, then a5a_5 is

  1. A.

    1/32

  2. B.

    2/32

  3. C.

    3/32

  4. D.

    4/32

Answer

C

Explanation

Let the first term be aa and common ratio be rr.

Given an=3×(sum of terms after an)a_n = 3 \times (\text{sum of terms after } a_n): a1=3(a2+a3+)a_1 = 3(a_2 + a_3 + \dots) a1=3(a1r1r)a_1 = 3 \left(\frac{a_1 r}{1 - r}\right) 1=3r1r    1r=3r    4r=1    r=141 = \frac{3r}{1 - r} \implies 1 - r = 3r \implies 4r = 1 \implies r = \frac{1}{4}

Total sum = a11r=32    a13/4=32    a1=24\frac{a_1}{1 - r} = 32 \implies \frac{a_1}{3/4} = 32 \implies a_1 = 24.

5th term a5=a1r4=24×(14)4=24×1256=332a_5 = a_1 r^4 = 24 \times \left(\frac{1}{4}\right)^4 = 24 \times \frac{1}{256} = \frac{3}{32}.

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