# What is the 17th term in the arithmetic sequence in which a_6 is 101 and a_9 is 83?

Jun 14, 2016

${17}^{t h}$ term is $35$

#### Explanation:

${n}^{t h}$ term of an arithmetic sequence $\left\{a , a + d , a + 2 d , \ldots \ldots . .\right\}$, whose first term is $a$ and difference between a term and its preceeding term is $d$, is $a + \left(n - 1\right) d$.

As ${6}^{t h}$ term is $101$, we have $a + 5 d = 101$ and .......(1)

as ${9}^{t h}$ term is $83$, we have $a + 8 d = 83$. ......(2)

Subtracting (1) from (2), we get $3 d = - 18$ i.e, $d = - 6$

Putting this in (1), we get $a - 30 = 101$ or $a = 131$

Hence, ${17}^{t h}$ term is $131 + 16 \times \left(- 6\right) = 131 - 96 = 35$