Question

13. Find the $2^{nd}$ term and $r^{th}$term of an AP whose $6^{th}$ term is $12$ and ${18}^{th}$ term is $22$.

Answer 1

Step 1. Note the given data:

Given: $6^{th}$ term of an A.P. $12$ and ${18}^{th}$ term of an A.P. $22$.

Let $a$be the first term of the AP and $d$ be the common difference.

Step 2. Finding the $6^{th}$and ${18}^{th}$terms of the AP in terms of $a$ and $d$:

The formula of the $n^{th}$term of an AP is $t_n=a+\left(n-1\right)d$

The $6^{th}$term of AP is $t_6=a+\left(6-1\right)d$

$=a+5d$

The ${18}^{th}$term of AP is $t_{18}=a+\left(18-1\right)d$

$=a+17d$

Step 3. Equating $6^{th}$ term and ${18}^{th}$term with $12$ and $22$ respectively:

$a+5d=12$ …..(i)

$a+17d=22$ ….(ii)

Step 4. Subtracting equation (i) from (ii):

$$\begin{gathered} \underline{22=a+17d \\ 12=a+5d \\ ---} \\ 10=0+12d \end{gathered}$$

$10=12d$

Divide by $12$ on both sides

$$\begin{gathered} \frac{12d}{12}=\frac{10}{12} \\ d=\frac{5}{6} \end{gathered}$$

Step 5. Putting the value of $d$ in equation (i):

$$\begin{gathered} 12=a+5\times \frac{5}{6} \\ 12=a+\frac{25}{6} \end{gathered}$$

Subtracting $\frac{25}{6}$in both sides of the equation

$$\begin{gathered} a+\frac{25}{6}-\frac{25}{6}=12-\frac{25}{6} \\ a=\frac{72-25}{6} \\ a=\frac{47}{6} \end{gathered}$$

Step 6. Find the $2^{nd}$ term for AP:

The formula of the $n^{th}$term an AP is $t_n=a+\left(n-1\right)d$

Here, $n=2,a=\frac{47}{6},d=\frac{5}{6}$

The second term is

$$\begin{gathered} t_2=\frac{47}{6}+\left(2-1\right)\frac{5}{6} \\ {t}_2=\frac{47}{6}+\frac{5}{6} \\ {t}_2=\frac{52}{6} \end{gathered}$$

Step 7. Find the $r^{th}$ term for AP:

$$\begin{gathered} a_r=\frac{47}{6}+\left(r-1\right)\frac{5}{6} \\ {t}_r=\frac{47}{6}+\frac{5}{6}r-\frac{5}{6} \\ {t}_r=\frac{42}{6}+\frac{5}{6}r \\ {t}_r=7+\frac{5}{6}r \end{gathered}$$

Hence, the second term of the AP is $t_2=\frac{52}{6}$ and the $r^{th}$term of the AP is $t_r=7+\frac{5}{6}r$.