Question

How many different A.P’s with integral terms can be formed with first term = 1 and last term 100.

• A. 1
• B. 3
• C. 8
• D. none of these

Answer 1

The correct option is D none of these

Since a and T_n are fixed number of different AP’s will depend on the possible number of common differences. The first AP possible will have first term as 1 and last term as 100. Common difference = 99. Therefore, Common difference can only be factors of 100-1 = 99.

Number of factors of 99 (99= 3²* 11¹) = (2+1 )( 1+1) = 6

Ans = 6

2nd method: - As we know, first term (a) = 1 and last term (nth) = 100

T_n = a + (n - 1)d

100 = 1 + (n - 1)d

99 = (n - 1)d

Now, we need to find out number of factor of 99.

99 = 3² x 11

Number of factors = (2 + 1) x (1 + 1) = 6.