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

Question

Accepted Answer

The given sequence is an arithmetic progression (A.P.) with:  
- First term $$ a = 3 $$  
- Common difference $$ d = 4 $$  
- Last term $$ T_n = 407 $$.

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

Substitute the values:  
$$
407 = 3 + (n - 1)4
$$

Simplify:  
$$
407 - 3 = (n - 1)4
$$  
$$
404 = (n - 1)4
$$  
$$
n - 1 = \frac{404}{4} = 101
$$  
$$
n = 101 + 1 = 102
$$

Thus, the total number of terms in the A.P. is **102**.

Answer

Answer

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