Find the 19th term of the A.P. \(\frac{5}{6}\), \(\frac{8}{6}\), \(\frac{11}{6}\).................
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\(\frac{5}{6}\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}