The sum of the first 15 terms of the sequence a_n = 10n + 21 is

Solution:

The series described by the sequence a_n = 10n + 21 is

a₁ = 10(1) + 21 = 31

a₂ = 10(2) + 21 = 41

a₃ = 10(3) + 21 = 51

Hence say we can series is an arithmetic progression with first term a = 31 and the common difference d is 10.

The sum of an arithmetic progression is given by the expression:

\(S_{n} =\frac{n}{2}[2a + (n-1)d ]\)

\(S_{15} =\frac{15}{2}[2(31) + (15-1)10 ]\)

\(S_{15} =\frac{15}{2}[62 + (14)10 ]\)

\(S_{15} =\frac{15}{2}[62 + 140]\)

\(S_{n} =\frac{15}{2}[202]\)

\(S_{n} =15(101)\)

\(S_{n} = 1515\)

---

The sum of the first 15 terms of the sequence a_n = 10n + 21 is

Summary:

The sum of the first 15 terms of the sequence a_n = 10n + 21 is \(S_{n} = 1515\)