there are how many terms in the sequence 31​, 32​, 33​, 34​, ​..., 131.

Answer 1

The given sequence is an arithmetic sequence.

The first term is a = 31.

The common difference is d = 1.

The nth term is given by the formula

a_n = a + (n-1)*d

The last term (n-th term) is 131.

The number of terms, n, is given by

131 = 31 + (n-1)*1

131 = 31 + n-1

131 = 30 + n

131 - 30 = 30 + n - 30

101 = n

Answer: 101