Find the 6th term of the sequence \(\frac{2}{3} \frac{7}{15} \frac{4}{15}\),...
A
-\(\frac{1}{3}\)
B
-\(\frac{1}{5}\)
C
-\(\frac{1}{15}\)
D
\(\frac{1}{9}\)
Verified Correct Answer: A.
Step-by-Step Mathematical Solution
a = \(\frac{2}{3}\), d = \(\frac{7}{15}\) - \(\frac{2}{3}\)
= 7 - 10
= \(\frac{-3}{15}\)
d = - \(\frac{-1}{5}\)
T6 = a + 5d
= \(\frac{2}{3}\) + 5(\(\frac{-1}{5}\)
= \(\frac{2}{3}\) - 1
= \(\frac{2 - 3}{3}\)
= \(\frac{-1}{3}\)
Source: Myschool Past Questions Archive (Record #50121)