Question

The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8. If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to common difference of the A.P., then the sum of all possible values of the first term of the G.P. is:

• A. $$\frac{34}{9}$$
• B. $$\frac{34}{13}$$
• C. $$\frac{32}{9}$$
• D. $$\frac{32}{13}$$

Hint:

Form two equations using the given sums and substitute the shared variables to create a quadratic equation for the first term of the G.P. (which equals the common difference of the A.P.).

Answer 1

The Correct Option is A

We are given the sum of an A.P. and the sum of two terms of a G.P. with intertwined parameters.

Let the AP be $a, a+d, ...$ and the GP be $A, Ar, ...$
Given:
$S_{10}(\text{AP}) = \frac{10}{2}(2a+9d) = 160 \Rightarrow 2a+9d = 32$
$S_{2}(\text{GP}) = A + Ar = 8 \Rightarrow A(1+r) = 8$

Applying constraints:
$a = r$ and $d = A$.
Placing these into our sum equations yields:
1) $2r + 9A = 32$
2) $A + Ar = 8$

From equation (1), we can isolate $A$ in terms of $r$:
$9A = 32 - 2r \Rightarrow A = \frac{32 - 2r}{9}$

Now substitute this into equation (2):
$$\frac{32 - 2r}{9} (1 + r) = 8$$
$$(32 - 2r)(1 + r) = 72$$
$$32 + 32r - 2r - 2r^2 = 72$$
$$-2r^2 + 30r + 32 - 72 = 0$$
$$-2r^2 + 30r - 40 = 0$$
Dividing by $-2$:
$$r^2 - 15r + 20 = 0$$

Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$r = \frac{15 \pm \sqrt{225 - 80}}{2} = \frac{15 \pm \sqrt{145}}{2}$$

We need the sum of possible values of $A$. We know $A = \frac{32 - 2r}{9}$.
Let $r_1, r_2$ be the roots for $r$. Then the possible values for $A$ are:
$A_1 = \frac{32 - 2r_1}{9}$ and $A_2 = \frac{32 - 2r_2}{9}$
Sum $= A_1 + A_2 = \frac{32 - 2r_1 + 32 - 2r_2}{9} = \frac{64 - 2(r_1 + r_2)}{9}$

From the quadratic $r^2 - 15r + 20 = 0$, the sum of roots $(r_1 + r_2) = 15$.
Sum of values of $A = \frac{64 - 2(15)}{9} = \frac{64 - 30}{9} = \frac{34}{9}$.