Question

In a G.P., $1, \frac{1}{2}, \frac{1}{4}, \ldots$, when the first $n$ number of terms are added, the sum is $\frac{1023}{512}$. Then the value of $n$ is

• A. $10$
• B. $12$
• C. $14$
• D. $16$
• E. $18$

Hint:

For G.P. sums, always simplify the denominator early. Converting the result into powers of 2 or 3 often makes solving much faster.

Answer 1

The Correct Option is A

Concept:
A Geometric Progression (G.P.) is a sequence in which each term is obtained by multiplying the previous term by a constant ratio $r$. Key formulas:
• $n^{\text{th}}$ term: $a_n = a \cdot r^{n-1}$
• Sum of first $n$ terms (for $|r|<1$): \[ S_n = \frac{a(1 - r^n)}{1 - r} \]

Step 1: Identify $a$ and $r$.
Given G.P.: \[ 1, \; \frac{1}{2}, \; \frac{1}{4}, \ldots \] Thus, \[ a = 1, r = \frac{1}{2} \]

Step 2: Write the sum formula.
\[ S_n = \frac{1\left(1 - \left(\frac{1}{2}\right)^n\right)}{1 - \frac{1}{2}} \] \[ = \frac{1 - \left(\frac{1}{2}\right)^n}{\frac{1}{2}} \]

Step 3: Simplify the expression.
Dividing by $\frac{1}{2}$ is equivalent to multiplying by 2: \[ S_n = 2\left(1 - \left(\frac{1}{2}\right)^n\right) \] \[ = 2 - 2\left(\frac{1}{2}\right)^n \] \[ = 2 - \frac{2}{2^n} \] \[ = 2 - \frac{1}{2^{n-1}} \]

Step 4: Use the given sum.
\[ 2 - \frac{1}{2^{n-1}} = \frac{1023}{512} \]

Step 5: Express 2 with denominator 512.
\[ 2 = \frac{1024}{512} \] So, \[ \frac{1024}{512} - \frac{1}{2^{n-1}} = \frac{1023}{512} \]

Step 6: Solve for $n$.
\[ \frac{1}{2^{n-1}} = \frac{1024 - 1023}{512} \] \[ = \frac{1}{512} \] \[ 2^{n-1} = 512 \]

Step 7: Convert 512 into power of 2.
\[ 512 = 2^9 \] Thus, \[ 2^{n-1} = 2^9 \] \[ n - 1 = 9 \] \[ n = 10 \]
Final Answer: \[ \boxed{10} \]