 # How many terms are there in the AP

Question:

How many terms are there in the AP $1 \frac{5}{6}, 1 \frac{1}{6}, 1 \frac{1}{6}, \frac{-1}{6}, \frac{-5}{6}, \ldots,-16 \frac{1}{6} ?$

Solution:

To Find: we need to find number of terms in the given AP.

Given: The series is $1 \frac{5}{6}, 1 \frac{1}{6}, 1 \frac{1}{6}, \frac{-1}{6}, \frac{-5}{6}, \ldots,-16 \frac{1}{6}$.

$a_{1}=1 \frac{5}{6}=\frac{11}{6}, a_{2}=1 \frac{1}{6}=\frac{7}{6}, d=\left(\frac{7}{6}\right)-\left(\frac{11}{6}\right)=\frac{-4}{6}$ and $a_{n}=-16 \frac{1}{6}=\frac{-95}{6}$

(Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $A P$ )

Formula Used: $a_{n}=a+(n-1) d$

$a_{n}=\frac{-97}{6}=a_{1}+(n-1)\left(\frac{-2}{3}\right)$

$\frac{-97}{6}-\frac{11}{6}=(n-1)\left(\frac{-2}{3}\right)$ [subtract $\frac{11}{6}$ on both sides]

$\frac{-108}{6}=(n-1)\left(\frac{-2}{3}\right)$ [Multiply both side by $\frac{-3}{2}$ ] or [Divide both side by $\frac{-2}{3}$ ]

$27=(n-1)[$ add 1 on both sides $]$

$n=28$

So there are 28 terms in this AP.