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.

\(\frac{1}{9}\)

Correct answer is A

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

= \(\frac{2 - 3}{3}\)

= \(\frac{-1}{3}\)