Logo
Jeeify
MCQMediumJEE 2024Geometric Progression (GP)

JEE Mathematics 2024 Question with Solution

Let a,ar,ar2,…a, ar, ar^2, \ldots be an infinite G.P. If ∑n=0∞arn=57\sum_{n=0}^{\infty} ar^n = 57 and ∑n=0∞a3r3n=9747\sum_{n=0}^{\infty} a^3r^{3n} = 9747, then a+18ra + 18r is equal to:

  • A

    2727

  • B

    4646

  • C

    3838

  • D

    3131

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: ∑n=0∞arn=57\sum_{n=0}^{\infty} ar^n = 57 and ∑n=0∞a3r3n=9747\sum_{n=0}^{\infty} a^3r^{3n} = 9747.

Find: a+18ra + 18r.

For an infinite G.P.,

S=a1−rS = \frac{a}{1-r}

with ∣r∣<1|r| < 1.

From the first series,

a1−r=57\frac{a}{1-r} = 57

so

a=57(1−r)a = 57(1-r)

For the second series, the first term is a3a^3 and the common ratio is r3r^3. Hence,

a31−r3=9747\frac{a^3}{1-r^3} = 9747

Substitute a=57(1−r)a = 57(1-r) into the second equation:

(57(1−r))31−r3=9747\frac{(57(1-r))^3}{1-r^3} = 9747

Using 1−r3=(1−r)(1+r+r2)1-r^3 = (1-r)(1+r+r^2),

573(1−r)3(1−r)(1+r+r2)=9747\frac{57^3(1-r)^3}{(1-r)(1+r+r^2)} = 9747

so

573(1−r)21+r+r2=9747\frac{57^3(1-r)^2}{1+r+r^2} = 9747

Since 573=18519357^3 = 185193 and 9747=185193199747 = \frac{185193}{19},

185193(1−r)21+r+r2=9747\frac{185193(1-r)^2}{1+r+r^2} = 9747

which gives

19(1−r)2=1+r+r219(1-r)^2 = 1+r+r^2

Expanding,

19(1−2r+r2)=1+r+r219(1-2r+r^2) = 1+r+r^2 19−38r+19r2=1+r+r219 - 38r + 19r^2 = 1 + r + r^2 18r2−39r+18=018r^2 - 39r + 18 = 0 6r2−13r+6=06r^2 - 13r + 6 = 0 (3r−2)(2r−3)=0(3r-2)(2r-3) = 0

Thus r=23r = \frac{2}{3} or r=32r = \frac{3}{2}. For an infinite G.P., ∣r∣<1|r| < 1, so

r=23r = \frac{2}{3}

Now,

a=57(1−23)=57⋅13=19a = 57\left(1-\frac{2}{3}\right) = 57\cdot \frac{1}{3} = 19

Therefore,

a+18r=19+18⋅23=19+12=31a + 18r = 19 + 18\cdot \frac{2}{3} = 19 + 12 = 31

The correct option is D.

Using ratio of the two sums

Given: a1−r=57\frac{a}{1-r} = 57 and a31−r3=9747\frac{a^3}{1-r^3} = 9747.

Find: a+18ra + 18r.

Cube the first equation:

(a1−r)3=573\left(\frac{a}{1-r}\right)^3 = 57^3 a3(1−r)3=573\frac{a^3}{(1-r)^3} = 57^3

Now divide this by the second relation:

a3(1−r)3a31−r3=5739747\frac{\dfrac{a^3}{(1-r)^3}}{\dfrac{a^3}{1-r^3}} = \frac{57^3}{9747}

So,

1−r3(1−r)3=19\frac{1-r^3}{(1-r)^3} = 19

Using 1−r3=(1−r)(1+r+r2)1-r^3 = (1-r)(1+r+r^2),

1+r+r2(1−r)2=19\frac{1+r+r^2}{(1-r)^2} = 19

which is equivalent to

1+r+r2=19(1−r)21+r+r^2 = 19(1-r)^2

Solving this gives

r=23r = \frac{2}{3}

as the valid value because ∣r∣<1|r| < 1. Then

a=57(1−r)=57⋅13=19a = 57(1-r) = 57\cdot \frac{1}{3} = 19

Hence,

a+18r=19+12=31a + 18r = 19 + 12 = 31

Therefore, the correct option is D.

Common mistakes

  • Using the second series as ∑a3rn\sum a^3r^n instead of ∑a3r3n\sum a^3r^{3n} is incorrect because the common ratio becomes r3r^3, not rr. First identify the exact first term and common ratio before applying the infinite G.P. sum formula.

  • Ignoring the condition ∣r∣<1|r| < 1 leads to accepting r=32r = \frac{3}{2}, which is invalid for an infinite geometric series. After solving the quadratic, always check which root satisfies convergence.

  • Cancelling factors incorrectly in 1−r3=(1−r)(1+r+r2)1-r^3 = (1-r)(1+r+r^2) can spoil the algebra. Factor 1−r31-r^3 completely and then cancel only the common factor 1−r1-r.

Practice more Geometric Progression (GP) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Geometric Progression (GP)

Related questions