Question

which sequence follows the rule (a_n=a_{n - 1}+7)?
a) 11, 18, 25, 32, ...
b) -15, -7, 1, 8, ...
c) -9, -2, 4, 9, ...
d) 17, 24, 30, 38, ...
question 2 (5 points)
evaluate the series: (sum_{a = 1}^{7}(3a^{2}-4))
a) 420
b) 406
c) 249

Explanation:

Step1: Check sequence rule for first question

For a sequence \(a_n=a_{n - 1}+7\), the common - difference between consecutive terms should be 7.
For option A: \(18−11 = 7\), \(25−18 = 7\), \(32−25 = 7\).
For option B: \(-7-(-15)=8
eq7\), \(1 - (-7)=8
eq7\), \(8 - 1 = 7\) (not consistent).
For option C: \(-2-(-9)=7\), \(4-(-2)=6
eq7\), \(9 - 4 = 5
eq7\).
For option D: \(24−17 = 7\), \(30−24 = 6
eq7\), \(38−30 = 8
eq7\).

Step2: Evaluate series for second question

We know that \(\sum_{a = 1}^{7}(3a^{2}-4)=3\sum_{a = 1}^{7}a^{2}-\sum_{a = 1}^{7}4\).
The sum of the squares of the first \(n\) positive integers is given by \(\sum_{k = 1}^{n}k^{2}=\frac{n(n + 1)(2n + 1)}{6}\). When \(n = 7\), \(\sum_{a=1}^{7}a^{2}=\frac{7\times(7 + 1)\times(2\times7+1)}{6}=\frac{7\times8\times15}{6}=140\).
And \(\sum_{a = 1}^{7}4=4\times7 = 28\).
Then \(3\sum_{a = 1}^{7}a^{2}-\sum_{a = 1}^{7}4=3\times140-28=420 - 28=392\) (There seems to be an error in the options for the second - part. But following the correct procedure).

Answer:

A. 11, 18, 25, 32, ... (for the first question)
(Note: The options for the second question may be incorrect based on the correct calculation of the series)