# The sum of 18 terms of an arithmetic series is 189 and the first term is 2. What is the 16th term.

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An arithmetic series is one where the difference between subsequent terms of the series is a constant. If the first term of a series is a, and the common difference between subsequent terms is d, the nth term of the series is equal to a + (n-1)*d

The sum of n terms of an arithmetic series is equal to (n/2)*(2*a + (n-1)*d)

The sum of 18 terms of an arithmetic series is 189 and the first term is 2. Let the common difference be d.

Using the formula given earlier 189 = (18/2)*(2*2 + 17*d)

189 = 9*(4 + 17d)

17d = 17

d = 1

As d = 1 and a = 2, the 16th term is 2 + 15*1 = 17

**The 16th term of the series is 17**

The sum of an arithmetic progression is given by the equation:-

Sum, S = (n/2)*[2a + (n-1)*d]

where n = number of terms that is to be added; a = 1st term of the progression

d = common difference between two consecutive terms in the progression

Thus, as per question

sum of first 18 terms, S = (18/2)*[2*2 + (18-1)*d]

or, 189 = 9*(4 + 17d)

or, 189 = 36 + 153d

or, d = 1

Now, In an arithmetic progression , the value of nth term is given by the equation:-

a(n) = a + (n-1)*d

or, a(16) = 2 + (16-1)*1

or, a(16) = 2 + 15

or, a(16) = 17