Visual SolutionPYQ 2024 · Apr 9 Shift 2Tricky

Question

Let

a, ar, ar^2, \ldots

be an infinite G.P. If

\sum_{n=0}^{\infty} ar^n = 57

and

\sum_{n=0}^{\infty} a^3 r^{3n} = 9747

, then

a + 18r

is equal to

(A)

46

(B)

38

(C)

31

(D)

27

Solution Path

From the two infinite GP sums, solve

6r^2 - 13r + 6 = 0

to get

r = \frac{2}{3}

a = 19

. Answer:

a + 18r = 31

01Question Setup

1/4

Infinite GP with

\sum ar^n = 57

and

\sum a^3 r^{3n} = 9747

. Find

a + 18r

a + 18r = \;?

02Set Up Equations

2/4

From

\dfrac{a}{1-r} = 57

and

\dfrac{a^3}{1-r^3} = 9747

, derive

\dfrac{(1-r)^2}{1+r+r^2} = \dfrac{1}{19}

\dfrac{(1-r)^2}{1+r+r^2} = \dfrac{1}{19}

03Solve for rKEY INSIGHT

3/4

Quadratic

6r^2 - 13r + 6 = 0

gives

r = \frac{2}{3}

. Then

a = 19

, so

a + 18r = 19 + 12 = 31

a + 18r = 31

04Final Answer

4/4

a + 18r = 31

\boxed{31}

Concepts from this question1 concepts unlocked

★

EASY

When |r| < 1, the infinite GP converges to a/(1-r)

S_\infty = \frac{a}{1-r} \quad (|r| < 1)

Appears in 2-3 JEE questions every year. Often combined with cube/square series.

Convergence problemsRecurring decimalsSum of cubes of GP

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