Key Concepts and Formulas

• Geometric Progression (GP): A sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). The $n$-th term is given by $a_n = a_1 r^{n-1}$.
• Sum of a GP: The sum of the first $n$ terms of a GP is $S_n = \frac{a_1(r^n - 1)}{r - 1}$, provided $r \neq 1$. If $r=1$, $S_n = n \cdot a_1$.

Step-by-Step Solution

Step 1: Express the given information in terms of $a_1$ and $r$. We are given a GP with first term $a_1$ and common ratio $r$. The terms are $a_1, a_1r, a_1r^2, a_1r^3, \dots$. We are given:

1. $a_1 < 0$
2. $a_1 + a_2 = 4$
3. $a_3 + a_4 = 16$

Using the formula $a_n = a_1 r^{n-1}$, we can rewrite the given equations: Equation (1): $a_1 + a_1 r = 4 \implies a_1(1+r) = 4$ (Equation A) Equation (2): $a_1 r^2 + a_1 r^3 = 16 \implies a_1 r^2(1+r) = 16$ (Equation B)

Step 2: Solve for the common ratio ($r$). To find $r$, we can divide Equation B by Equation A. Before dividing, we must ensure that the terms we are dividing by are non-zero. If $a_1 = 0$, then $a_1 + a_2 = 0 \neq 4$, so $a_1 \neq 0$. If $1+r = 0$, i.e., $r = -1$, then Equation A becomes $a_1(1-1) = 0 = 4$, which is a contradiction. So, $1+r \neq 0$. Since $a_1 \neq 0$ and $1+r \neq 0$, we can divide Equation B by Equation A: $\frac{a_1 r^2(1+r)}{a_1(1+r)} = \frac{16}{4}$ $r^2 = 4$ Taking the square root of both sides, we get $r = 2$ or $r = -2$.

Step 3: Determine the values of $a_1$ and $r$ that satisfy all conditions. We have two possible values for $r$. We will find the corresponding $a_1$ for each case using Equation A ($a_1(1+r)=4$) and then apply the condition $a_1 < 0$.

Case 1: $r = 2$ Substitute $r=2$ into Equation A: $a_1(1+2) = 4$ $a_1(3) = 4$ $a_1 = \frac{4}{3}$ In this case, $a_1 = \frac{4}{3}$, which is not less than 0. So, this case is rejected.

Case 2: $r = -2$ Substitute $r=-2$ into Equation A: $a_1(1+(-2)) = 4$ $a_1(-1) = 4$ $a_1 = -4$ In this case, $a_1 = -4$, which is less than 0. This case satisfies all conditions. Thus, the first term is $a_1 = -4$ and the common ratio is $r = -2$.

Step 4: Calculate the sum of the first 9 terms ($\sum_{i=1}^9 a_i$). We need to find $S_9$ using the formula $S_n = \frac{a_1(r^n - 1)}{r - 1}$. Substitute $a_1 = -4$, $r = -2$, and $n = 9$: $S_9 = \frac{-4((-2)^9 - 1)}{-2 - 1}$ First, calculate $(-2)^9$: $(-2)^9 = -(2^9) = -512$ Now, substitute this value back into the sum formula: $S_9 = \frac{-4(-512 - 1)}{-3}$ $S_9 = \frac{-4(-513)}{-3}$ $S_9 = \frac{2052}{-3}$ $S_9 = -684$

Step 5: Solve for $\lambda$. We are given that $\sum_{i=1}^9 a_i = 4\lambda$. We found that $\sum_{i=1}^9 a_i = -684$. Therefore, we have the equation: $-684 = 4\lambda$ Divide by 4 to solve for $\lambda$: $\lambda = \frac{-684}{4}$ $\lambda = -171$

Common Mistakes & Tips

• Sign Errors with Powers: Be extremely careful when calculating powers of negative numbers. An odd exponent on a negative base yields a negative result.
• Utilize All Conditions: The condition $a_1 < 0$ is crucial for selecting the correct values of $a_1$ and $r$. Always ensure all given constraints are met.
• Division by Zero: Before dividing equations to solve for variables, always check if any term could be zero, which would invalidate the division.

Summary The problem involves a Geometric Progression where we are given relationships between terms and a condition on the first term. By expressing the given information algebraically, we formed a system of equations that allowed us to solve for the first term ($a_1$) and the common ratio ($r$). The condition $a_1 < 0$ was essential in selecting the correct pair of values for $a_1$ and $r$. Finally, we used the formula for the sum of a GP to calculate the sum of the first 9 terms and then solved for $\lambda$.

The final answer is $\boxed{-171}$.