Question

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

• A. \( 100 \)
• B. \( 101 \)
• C. \( 99 \)
• D. \( 102 \)

Hint:

\textbf{Arithmetic Progression.} An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant. The \(n^{th}\) term of an A.P. is given by \(a_n = a + (n - 1)d\), where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms.

Answer 1

The Correct Option is D

The given sequence is an arithmetic progression (A.P.) with the first term \(a = 3\) and the common difference \(d = 7 - 3 = 4\). The last term of the A.P. is given as \(l = 407\). We need to find the number of terms \(n\) in this A.P. The formula for the \(n^{th}\) term of an A.P. is: $$ a_n = a + (n - 1)d $$ In this case, \(a_n\) is the last term \(l\). So, we have: $$ l = a + (n - 1)d $$ Substitute the given values: $$ 407 = 3 + (n - 1)4 $$ Now, we need to solve for \(n\): $$ 407 - 3 = (n - 1)4 $$ $$ 404 = 4(n - 1) $$ Divide both sides by 4: $$ \frac{404}{4} = n - 1 $$ $$ 101 = n - 1 $$ Add 1 to both sides: $$ n = 101 + 1 $$ $$ n = 102 $$ Therefore, there are 102 terms in the given arithmetic progression.