Let a, ar, ar², … Be an Infinite G.P. If the Sum of arⁿ From n = 0 to ∞ Is 57 and the Sum of a³r³ⁿ Is 9747, Then a + 18r Is Equal To

JEE Main 2025 — 22 January, Morning Shift. Previous Year Question.

Problem

Let $a, ar, ar^2, \ldots$ be an infinite G.P. If $\displaystyle\sum_{n=0}^{\infty}ar^n=57$ and $\displaystyle\sum_{n=0}^{\infty}a^3r^{3n}=9747$, then $a+18r$ is equal to:

Key insight. The second sum $\sum a^3r^{3n}$ is itself an infinite GP — with first term $a^3$ and common ratio $r^3$ — so it sums to $\dfrac{a^3}{1-r^3}$. Since $1-r^3$ factors as $(1-r)(1+r+r^2)$, dividing the cube of the first sum by the second leaves a clean equation in $r$ alone.

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Approach

Write both infinite sums using the standard formula for the sum of an infinite GP. Cube the first equation and divide it by the second — this eliminates $a$ entirely, using the factorisation $1-r^3=(1-r)(1+r+r^2)$, and leaves a quadratic purely in $r$. Solve it, discard the root that makes $|r|\geq1$ (since the GP must converge), then find $a$ and compute $a+18r$.

Solution

Step 1 — Write both sums using the infinite GP formula

$$\sum_{n=0}^{\infty}ar^n = \frac{a}{1-r} = 57 \quad \text{…(i)}$$

$$\sum_{n=0}^{\infty}a^3r^{3n} = \frac{a^3}{1-r^3} = 9747 \quad \text{…(ii)}$$

Step 2 — Cube equation (i) and divide by equation (ii)

$$\frac{a^3}{(1-r)^3} = 57^3, \qquad \frac{a^3}{1-r^3} = 9747$$

Dividing the first by the second:

$$\frac{1-r^3}{(1-r)^3} = \frac{57^3}{9747}$$

Step 3 — Simplify the right-hand side

$$57 = 3\times19, \qquad 57^3 = 3^3\times19^3 = 185193$$

$$9747 = 27\times361 = 3^3\times19^2$$

$$\frac{57^3}{9747} = \frac{3^3\times19^3}{3^3\times19^2} = 19$$

Step 4 — Simplify the left-hand side using 1 − r³ = (1−r)(1+r+r²)

$$\frac{(1-r)(1+r+r^2)}{(1-r)^3} = \frac{1+r+r^2}{(1-r)^2} = 19$$

Step 5 — Cross-multiply and solve for r

$$1+r+r^2 = 19(1-r)^2 = 19(1-2r+r^2)$$

$$1+r+r^2 = 19-38r+19r^2$$

$$18r^2-39r+18=0$$

Dividing by 3:

$$6r^2-13r+6=0$$

Step 6 — Solve the quadratic

$$r = \frac{13\pm\sqrt{169-144}}{12} = \frac{13\pm5}{12}$$

$$r = \frac{18}{12}=\frac32 \quad \text{or} \quad r=\frac{8}{12}=\frac23$$

Step 7 — Reject the non-convergent root

An infinite GP only converges for $|r|<1$, so $r=\dfrac32$ is rejected. Hence $r=\dfrac23$.

Step 8 — Find a using equation (i)

$$a = 57(1-r) = 57\left(1-\frac23\right) = 57\times\frac13 = 19$$

Step 9 — Compute a + 18r

$$a+18r = 19+18\times\frac23 = 19+12 = 31$$

Answer

Common mistakes

• Dividing the two equations in the wrong direction. Whether you compute $\dfrac{\text{(i)}^3}{\text{(ii)}}$ or its reciprocal determines which side of the resulting quadratic carries the $19$ — getting this backwards leads to a quadratic with no real roots, a clear signal to double back and check the division direction.
• Keeping the root $r=\dfrac32$. Since the problem explicitly describes an infinite GP, convergence requires $|r|<1$ — a root outside this range must always be discarded, no matter how clean it looks.

Practise next

• Let $a,ar,ar^2,\ldots$ be an infinite GP with $\sum_{n=0}^\infty ar^n=20$ and $\sum_{n=0}^\infty a^3r^{3n}=1520$. Find $a+10r$, using the same cube-and-divide approach.