How many terms are there in the A. P. 3, 7, 11, ......... 407 ?

The correct option is 102.

To find the number of terms in the given Arithmetic Progression (A.P.):
3, 7, 11, ..., 407
we can use the standard formula for the n-th term of an A.P.

The formula for the n-th term of an A.P. is:
$$a_n=a+(n-1)d$$

From the given A.P., we can identify the following values:
- The first term (a) is 3.
- The common difference (d) is the difference between any two consecutive terms: 7 - 3 = 4.
- The last term ($a_n$) is 407.

Now, we substitute these values into the formula to solve for the number of terms (n):
$$407=3+(n-1)\times 4$$

Subtract 3 from both sides of the equation:
$$407-3=(n-1)\times 4$$

$$404=(n-1)\times 4$$

Next, divide both sides by 4:
$$\frac{404}{4}=n-1$$

$$101=n-1$$

Finally, add 1 to both sides to find n:
$$n=101+1$$

$$n=102$$

Therefore, there are 102 terms in the given Arithmetic Progression.