If \(a_{n}= 3 - 4n \), then what is \(a_{1} + a_{2} + a_{3} + ...... + a_{n}\) equal to \( [ 1 + 2 + 3 + ..... + n = \frac{n(n + 1)}{2}]\)
Correct Answer: Description for Correct answer:
\( \Large a_{1} = 3 - 4 = -1 \)
\( \Large a_{2} = 3 - 8 = 5\)
\( \Large a_{31} = 3 - 12 = -9\)
\( \Large a_{n} = 3 - 4n\)
Now, \( \Large a_{1} + a_{2} +a_{3}+ ... + a_{n} = -1 + -5 + -9,....(3 - 4n)\)
\( \Large s_{n} = \frac{n}{2} (-1 + (3 - 4n)) \)
\( \Large s_{n} = \frac{n}{2} (-1 + 3 - 4n) \)
\( \Large s_{n} = \frac{n}{2} (2 - 4n) = n(1 - 2n) \)
=-n(2n - 1)
Option (B)is correct.
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Exams >> CDSE >> CDS Maths(1)
