Given the arithmetic progression 2, 6, 10, 14,... find the sum from the fifth term to the sixteenth term
The first term of the arithmetic progression is 2, and the common difference is 6 - 2 = 4.
The nth term of an arithmetic progression can be calculated using the formula:
\[a_n = a_1 + (n-1)d\]
where \(a_n\) is the nth term, \(a_1\) is the first term, n is the position of the term in the sequence, and d is the common difference.
For the fifth term, n = 5, so we have:
\[a_5 = 2 + (5-1)4\]
\[a_5 = 2 + (4)4\]
\[a_5 = 2 + 16 = 18\]
For the sixteenth term, n = 16, so we have:
\[a_{16} = 2 + (16-1)4\]
\[a_{16} = 2 + (15)4\]
\[a_{16} = 2 + 60 = 62\]
To find the sum of the terms from the fifth term to the sixteenth term, we can use the formula for the sum of an arithmetic series. The sum of an arithmetic series can be calculated using the formula:
\[S_n = \frac{n}{2}(a_1 + a_n)\]
where \(S_n\) is the sum of the first n terms, \(a_1\) is the first term, \(a_n\) is the nth term, and n is the number of terms.
For the sum from the fifth term to the sixteenth term, we have:
\[S = \frac{16}{2}(18 + 62)\]
\[S = 8(80)\]
\[S = 640\]
Therefore, the sum from the fifth term to the sixteenth term is 640.