




A NAND gate produces an output that is the negation of the AND gate output. The output (\( Y \)) is given by: \[ Y = \overline{A \cdot B}, \] where \( A \) and \( B \) are the inputs to the NAND gate. The truth table for a NAND gate is as follows:

Step-by-Step Analysis of the Inputs and Outputs: - When both \( A = 0 \) and \( B = 0 \), the output \( Y = 1 \).
- When \( A = 0 \) and \( B = 1 \), the output \( Y = 1 \). - When \( A = 1 \) and \( B = 0 \), the output \( Y = 1 \).
- When both \( A = 1 \) and \( B = 1 \), the output \( Y = 0 \).
Now analyze the given input waveforms for \( A \) and \( B \):
1. For each interval where \( A \) and \( B \) are given, calculate \( A \cdot B \).
2. Take the negation (\( \overline{A \cdot B} \)) to find the output \( Y \).
From the given inputs and truth table, the output waveform matches Option (2).



Which of the following circuits has the same output as that of the given circuit?

Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to
