The sum of the first 11th term of the sequence is 638
The sequence in question is an arithmetic progression. The formula for finding the sum of an A.P is expressed as:
[tex]S_n=\frac{n}{2}(2a+(n-1)d)[/tex]
a is the first term = 28
d is the common difference = 34 - 28 = 40 - 34 = 6
n is the number of terms = 11
Substitute the given parameters into the expression
[tex]S_{11}=\frac{11}{2}(2(28)+(11-1)(6))\\S_{11}=\frac{11}{2}(56+(10((6))\\S_{11}=\frac{11}{2}(116)\\S_{11}=11\times 58\\S_{11}=638[/tex]
Hence the sum of the first 11th term of the sequence is 638
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