Question

Let $a_1, a_2, a_3,...$ be a G.P. such that $a_1 < 0, a_1 + a_2 = 4$ and $a_3 + a_4 = 16$. If $\sum_{i=1}^{9} a_i = 4\lambda$, then $\lambda$ is equal to:

• A. -171
• B. 171
• C. $\frac{511}{3}$
• D. -513

Answer 1

Answer: (A)

-171

Answer 2

The problem provides information about a Geometric Progression (G.P.) and asks us to find the value of $\lambda$ based on the sum of its first nine terms.

Let the first term of the G.P. be $a$ and the common ratio be $r$. The $n$-th term of a G.P. is given by $a_n = ar^{n-1}$.

We are given the following conditions:

1. $a_1 < 0 \implies a < 0$.
2. $a_1 + a_2 = 4$. Substituting the terms in terms of $a$ and $r$: $a + ar = 4$ $a(1+r) = 4$ (Equation 1)
3. $a_3 + a_4 = 16$. Substituting the terms in terms of $a$ and $r$: $ar^2 + ar^3 = 16$ $ar^2(1+r) = 16$ (Equation 2)

Now, we can solve these two equations to find the values of $a$ and $r$. Divide Equation 2 by Equation 1: $\frac{ar^2(1+r)}{a(1+r)} = \frac{16}{4}$ $r^2 = 4$ This gives two possible values for $r$: $r = 2$ or $r = -2$.

Case 1: $r = 2$ Substitute $r=2$ into Equation 1: $a(1+2) = 4$ $3a = 4$ $a = \frac{4}{3}$ However, this value of $a$ contradicts the given condition $a_1 < 0$ (i.e., $a < 0$). Therefore, $r=2$ is not the correct common ratio.

Case 2: $r = -2$ Substitute $r=-2$ into Equation 1: $a(1+(-2)) = 4$ $a(1-2) = 4$ $a(-1) = 4$ $a = -4$ This value of $a$ satisfies the condition $a_1 < 0$. Thus, the first term is $a = -4$ and the common ratio is $r = -2$.

Next, we need to find the sum of the first nine terms of this G.P., denoted as $\sum_{i=1}^{9} a_i = S_9$. The sum of the first $n$ terms of a G.P. is given by the formula $S_n = \frac{a(1-r^n)}{1-r}$, provided $r \neq 1$. Here, $a = -4$, $r = -2$, and $n = 9$.

$S_9 = \frac{-4(1-(-2)^9)}{1-(-2)}$ Calculate $(-2)^9$: $(-2)^9 = -(2^9) = -512$

Substitute this value back into the sum formula: $S_9 = \frac{-4(1-(-512))}{1+2}$ $S_9 = \frac{-4(1+512)}{3}$ $S_9 = \frac{-4(513)}{3}$

Now, simplify the expression: $513 \div 3 = 171$ $S_9 = -4 \times 171$ $S_9 = -684$

The problem states that $\sum_{i=1}^{9} a_i = 4\lambda$. So, we have: $-684 = 4\lambda$

To find $\lambda$, divide both sides by 4: $\lambda = \frac{-684}{4}$ $\lambda = -171$