Answer:
Option B
Step-by-step explanation:
Expression representing the series is,
[tex]\sum_{k=6}^{11}(2k+5)[/tex]
Here, k = Number of terms
By placing k = 6, 7, 8, 9, 10, 11 in the expression we can get the terms of the series.
For k = 6,
(2k + 5) = 2(6) + 5
= 17
For k = 7,
(2k + 5) = 2(7) + 5
= 19
For k = 8,
(2k + 5) = 2(8) + 5
= 21
For k = 9,
(2k + 5) = 2(9) + 5
= 23
For k = 10,
0(2k + 5) = 2(10) + 5
= 25
For k = 11,
(2k + 5) = 2(11) + 5
= 27
Therefore, expanded form will be,
17, 19, 21, 23, 25, 27
There, is a common difference in each successive to previous term,
d = 19 - 17
d = 2
Number of terms 'n' = 6
First term = 17
Sum of an arithmetic sequence is given by,
[tex]S=\frac{n}{2}[2a+(n-1)d][/tex]
Therefore, sum of the given sequence will be,
[tex]S=\frac{6}{2}[2(17)+(6-1)2][/tex]
= 3(34 + 10)
S = 132
Option B will be the correct option.
