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 \( \sum\limits_{i = 1}^9 {{a_i}} = 4\lambda \), then \( \lambda \) is equal to:

• A. 171
• B. -171
• C. -513
• D. \( {{511} \over 3} \)

Hint:

Use the two given pair-sums to find the common ratio, apply the condition a₁ \(\lt\) 0 to fix the sign, then use the GP sum formula.

Answer 1

The Correct Option is B

Step 1: \(a_1+a_2=a(1+r)=4\) and \(a_3+a_4=ar^2(1+r)=16\). Dividing: \(r^2=4\Rightarrow r=\pm2\).
Step 2: If \(r=2\): \(a=4/3\gt0\), rejected since \(a_1\lt0\). If \(r=-2\): \(a(1-2)=4\Rightarrow a=-4\).
Step 3: \(S_9=\dfrac{a(1-r^9)}{1-r}=\dfrac{-4(1+512)}{3}=-684\).
Step 4: \(4\lambda=-684\Rightarrow\lambda=-171\).
Answer: (B) \(-171\)