Correct Answer: Option B

Explanation

Given the 3rd and 5th terms of an A.P. are 6 and 10 respectively, we have:

The formula for the \( n \)-th term of an A.P. is given by:
\(T_n = a + (n-1)d\)
where \( a \) is the first term and \( d \) is the common difference.

 For the 3rd term:
\(T_3 = a + 2d = 6 \quad \text{(Equation 1)}\)

 For the 5th term:
\(T_5 = a + 4d = 10 \quad \text{(Equation 2)}\)
From these equations, we have:
\(a + 2d = 6 \tag{1}\)
\(a + 4d = 10 \tag{2}\)
Subtract Equation 1 from Equation 2:
\((a + 4d) - (a + 2d) = 10 - 6\)
This simplifies to:
\(2d = 4\)
Thus,
\(d = 2\)
Substituting \( d = 2 \) into Equation 1:
\(a + 2(2) = 6\)
This simplifies to:
\(a + 4 = 6\)
Thus,
\(a = 2\)
The first term \( a \) is \( 2 \) and the common difference \( d \) is \( 2 \).

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