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

To solve this problem, we need to establish the relationships between the Arithmetic Progression (A.P.) and the Geometric Progression (G.P.) based on the given information.

Let the first term of the A.P. be $a$ and its common difference be $d$.
Let the first term of the G.P. be $A$ and its common ratio be $r$.

1. Information about the A.P.: The sum of the first $n$ terms of an A.P. is given by the formula $S_n = \frac{n}{2} [2a + (n-1)d]$.
For $n=10$, we have $S_{10} = 160$:
$$\frac{10}{2} [2a + (10-1)d] = 160$$
$$5[2a + 9d] = 160$$
$$2a + 9d = 32$$ ... (Equation 1)

2. Information about the G.P.: The sum of the first two terms of a G.P. is $A + Ar = 8$.
$$A(1 + r) = 8$$ ... (Equation 2)

3. Connecting conditions: We are told that:
- First term of A.P. ($a$) = common ratio of G.P. ($r$) $\Rightarrow a = r$.
- First term of G.P. ($A$) = common difference of A.P. ($d$) $\Rightarrow A = d$.

Substituting these into Equation 1 and Equation 2:
From Eq 1: $2r + 9A = 32 \Rightarrow r = \frac{32 - 9A}{2}$
From Eq 2: $A(1 + r) = 8$

Substitute the expression for $r$ into Equation 2:
$$A\left(1 + \frac{32 - 9A}{2}\right) = 8$$
$$A\left(\frac{2 + 32 - 9A}{2}\right) = 8$$
$$A(34 - 9A) = 16$$
$$34A - 9A^2 = 16$$
$$9A^2 - 34A + 16 = 0$$

The possible values of the first term of the G.P. ($A$) are the roots of this quadratic equation. The sum of all possible values of $A$ is the sum of the roots of the equation $ax^2 + bx + c = 0$, which is $-\frac{b}{a}$.

Sum of values of $A = -\frac{-34}{9} = \frac{34}{9}$.