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

Here is the step-by-step solution, breaking down the algebraic manipulation clearly for your students:

1. Set up the equations from the given infinite G.P. sums:

For an infinite Geometric Progression with first term $$a$$ and common ratio $$r$$ (where $$|r| < 1$$), the sum is $$\frac{a}{1-r}$$.

From the first given sum:

$$\frac{a}{1-r} = 57 \implies a = 57(1-r)$$  --- (Equation 1)

The second series is $$\sum_{n=0}^{\infty} a^3 r^{3n} = a^3 + a^3r^3 + a^3r^6 + \dots$$

This is another infinite G.P. where the first term is $$a^3$$ and the common ratio is $$r^3$$.

$$\frac{a^3}{1-r^3} = 9747$$  --- (Equation 2)

2. Substitute and simplify:

Recall the algebraic identity $$1 - r^3 = (1-r)(1 + r + r^2)$$.

Substitute this and Equation 1 into Equation 2:

$$\frac{[57(1-r)]^3}{(1-r)(1+r+r^2)} = 9747$$

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

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

To avoid massive calculations, let's factor the right side. Notice that $$9747 \div 57 = 171$$, and $$171 \div 57 = 3$$. So, $$9747 = 57^2 \times 3$$.

$$\frac{57^3(1-r)^2}{1+r+r^2} = 57^2 \times 3$$

$$\frac{57(1-r)^2}{1+r+r^2} = 3$$

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

3. Solve the resulting quadratic equation:

Cross-multiply to get:

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

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

Divide the entire equation by 3 to simplify:

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

Factorize by splitting the middle term (sum = -13, product = 36):

$$6r^2 - 9r - 4r + 6 = 0$$

$$3r(2r - 3) - 2(2r - 3) = 0$$

$$(3r - 2)(2r - 3) = 0$$

This gives two possible values for the common ratio: $$r = \frac{2}{3}$$ or $$r = \frac{3}{2}$$.

Since an infinite G.P. only has a finite sum when $$|r| < 1$$, we must reject $$\frac{3}{2}$$. Therefore, $$r = \frac{2}{3}$$.

4. Find $$a$$ and calculate the final expression:

Substitute $$r = \frac{2}{3}$$ back into Equation 1:

$$a = 57\left(1 - \frac{2}{3}\right) = 57\left(\frac{1}{3}\right) = 19$$

Finally, calculate the required value for $$a + 18r$$:

$$a + 18r = 19 + 18\left(\frac{2}{3}\right)$$

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