Question

Let \(a_{1},a_{2},a_{3},\dots\) are in G.P. such that \(n>m\), \(a_{n}>a_{m}\) and \(a_{1}+a_{n}=66\), \(a_{2}\cdot a_{n-1}=128\). If \(\sum_{r=1}^{n}a_{r}=126\), then \(n\) is:

• A. 11
• B. 8
• C. 6
• D. 64

Hint:

Once you find that $a_1 = 2$ and $r = 2$, you can quickly double-check your work by writing out the terms of the series manually: $2 + 4 + 8 + 16 + 32 + 64$. Summing these 6 numbers gives exactly $126$, confirming that $n=6$ is correct without having to do any complex algebra.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In a Geometric Progression, the product of terms equidistant from the beginning and the end is constant. That is, \( a_1 \cdot a_n = a_2 \cdot a_{n-1} = a_3 \cdot a_{n-2} = \dots \). This property allows us to find the values of \( a_1 \) and \( a_n \) using the given sum and product.
Step 2: Key Formula or Approach:
1. Use \( a_1 \cdot a_n = a_2 \cdot a_{n-1} = 128 \).
2. Use \( a_1 + a_n = 66 \) to solve for \( a_1 \) and \( a_n \) using a quadratic equation.
3. Use the sum of G.P. formula: \( S_n = \frac{a_n r - a_1}{r - 1} \) or \( \frac{a_1(r^n - 1)}{r - 1} \).
Step 3: Detailed Explanation:
Let \( a_1 \) and \( a_n \) be roots of \( x^2 - (a_1 + a_n)x + (a_1 a_n) = 0 \).
\[ x^2 - 66x + 128 = 0 \] \[ (x - 64)(x - 2) = 0 \] The roots are 64 and 2. Since \( a_n>a_m \) and \( n>m \), the sequence is increasing. Thus: \[ a_1 = 2 \quad \text{and} \quad a_n = 64 \] Now, for a G.P., \( a_n = a_1 r^{n-1} \): \[ 64 = 2 \cdot r^{n-1} \implies r^{n-1} = 32 \] The sum is given as \( S_n = 126 \). Using the formula \( S_n = \frac{a_n r - a_1}{r - 1} \): \[ 126 = \frac{64r - 2}{r - 1} \] \[ 126r - 126 = 64r - 2 \] \[ 62r = 124 \implies r = 2 \] Now substitute \( r = 2 \) into \( r^{n-1} = 32 \): \[ 2^{n-1} = 2^5 \] \[ n - 1 = 5 \implies n = 6 \] Step 4: Final Answer:
The number of terms \( n \) is 6.