Find the 19th term of the A.P. \(\frac{5}{6}\), \(\frac{8}{6}\), \(\frac{11}{6}\).................

7\(\frac{1}{2}\)

9\(\frac{1}{2}\)

9\(\frac{5}{6}\)

Correct answer is D

first term (a) = \(\frac{5}{6}\)

common difference = \(\frac{8}{6}\) - \(\frac{5}{6}\) → \(\frac{3}{6}\) or \(\frac{1}{2}\)

A.P formula → T\(_n\) = a + (n - 1)d
T\(_n\) = \(\frac{5}{6}\) + (19 - 1)\(\frac{1}{2}\)
T\(_n\) = \(\frac{5}{6}\) + 9

→ 9\(\frac{5}{6}\)

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