# What is the 13th term in an arithmetic sequence whose first term is 10 and whose 4th term is 1?

Jun 22, 2016

$- 26.$

#### Explanation:

Let the first term of an A.P. be a , common diff. be d , $d \ne 0 ,$ & the ${n}^{t h}$ term be ${t}_{n} .$

Then, we have the formula : ${t}_{n} = a + \left(n - 1\right) d \ldots \ldots \ldots \ldots \left(1\right)$

In our case, $a = 10 , n = 4 , {t}_{4} = 1.$Using these values in $\left(1\right) ,$

$1 = 10 + 3 d ,$ giving, $d = - 3.$

Hence, in $\left(1\right)$, $a = 10 , d = - 3 ,$ so, ${t}_{n} = 10 - 3 \left(n - 1\right) = 10 - 3 n + 3 = 13 - 3 n .$

$\therefore$ Reqd. Term$= {t}_{13} = 13 - 39 = - 26.$