The sum of the first 30 terms of the sequence a_n = 6n + 5 is

Solution:

The sequence is given by the explicit expression 

a_n = 6n + 5

Therefore

The sum of the first 30 terms of the sequence an = 6n + 5 is

a₁ = 6(1) + 5 = 11

a₂ = 6(2) + 5 = 17

a₃ = 6(3) + 5 = 23 and so on…

The sequence is an arithmetic progression with first term a = 11 and difference d = 6

The sum of first n terms is given by:

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

The sum of the first 30 terms is given by:

\(S_{30}=\frac{30}{2}[2(11) + (30-1)(6)]\)

\(S_{30}=15[22 + (29)(6)]\)

\(S_{30}=15[22 + 174]\)

\(S_{30}=15[196]\)

\(S_{30}=2940\)

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The sum of the first 30 terms of the sequence a_n = 6n + 5 is

Summary:

The sum of the first 30 terms of the sequence a_n = 6n + 5 is 2940