Question

Find the 13th term of an A.P 11/5, 1, -1/5, -7/5.....

Answer 1

Step-by-step explanation:

$\Huge\underline{\underline{\sf ANSWER}}\\ a = t1 = \frac{11}{5} \\ t2 = 1 \\ d = t2 - t1 = 1 - \frac{11}{5} = \frac{ - 6}{5} \\ \\ tn = a + (n - 1)d \\ t13 = \frac{11}{5} + (12)( - \frac{6}{5} ) \\ t13 = \frac{11}{5} - \frac{72}{5} \\ t13 = \frac{ - 61}{5} \\ \\ So, \: 13th \: term \: of \: A. P \: is \: \frac{ - 61}{5}$