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

Concept: In any finite Geometric Progression ($G.P.$), the product of any two terms that are equidistant from the beginning and the end remains perfectly constant, and is always equal to the product of the first and last terms: $$a_1 \cdot a_n = a_2 \cdot a_{n-1} = a_3 \cdot a_{n-2} = \dots$$ Step 1: Solve the system for the first and last terms.
Using the constant product rule of geometric sequences, we can establish a system of equations for $a_1$ and $a_n$:
• $a_1 + a_n = 66$
• $a_1 \cdot a_n = a_2 \cdot a_{n-1} = 128$ We can find these roots by setting up a quadratic tracking equation $t^2 - (\text{sum})t + (\text{product}) = 0$: $$t^2 - 66t + 128 = 0 \quad \Rightarrow \quad (t - 64)(t - 2) = 0$$ This yields the roots $t = 64$ and $t = 2$. We are given that $a_n > a_m$ for $n > m$, meaning the sequence is strictly increasing. Therefore, the final term must be larger than the first term: $$a_1 = 2 \quad \text{and} \quad a_n = 64$$

Step 2: Relate the common ratio $r$ using the $n$-th term formula.
The standard formula for the $n$-th term of a geometric sequence is $a_n = a_1 \cdot r^{n-1}$. Substituting our values: $$64 = 2 \cdot r^{n-1} \quad \Rightarrow \quad r^{n-1} = 32 \quad \cdots (1)$$

Step 3: Use the series summation formula to find $r$.
The sum of a finite geometric progression is given by $S_n = \frac{a_1(r^n - 1)}{r - 1}$. We are given that $S_n = 126$: $$\frac{2(r^n - 1)}{r - 1} = 126 \quad \Rightarrow \quad \frac{r^n - 1}{r - 1} = 63 \quad \cdots (2)$$ We can split the term $r^n$ as $r \cdot r^{n-1}$. Substitute our expression from equation (1) ($r^{n-1} = 32$) into this term: $$r^n = r \cdot 32 = 32r$$ Now substitute $r^n = 32r$ back into equation (2): $$\frac{32r - 1}{r - 1} = 63 \quad \Rightarrow \quad 32r - 1 = 63(r - 1)$$ $$32r - 1 = 63r - 63 \quad \Rightarrow \quad 63 - 1 = 63r - 32r$$ $$62 = 31r \quad \Rightarrow \quad r = 2$$

Step 4: Calculate the total number of terms $n$.
Substitute the common ratio $r = 2$ back into equation (1) to solve for $n$: $$2^{n-1} = 32 \quad \Rightarrow \quad 2^{n-1} = 2^5$$ Equating the exponents: $$n - 1 = 5 \quad \Rightarrow \quad n = 6$$ This matches the corrected selection value for option (C).