Question

An A.P. consists of $50$ terms of which $3rd$ term is $12$ and the last term is $106$. Find the ${29}^{th}$ term.

Answer 1

Solving A.P with the given terms:

Step 1:Calculate the equation of $3rd$ term and last term.

Given that an A.P. consists of $50$ terms of which $3rd$ term is $12$ and the last term is $106$.

Assume that the first term is $a$and the common difference is $d$.

Know that the $nth$term is given as $a+(n-1)d$.

So, the $3rd$ term is given as

$\begin{array}{c}12=a+(3-1)d\\ 12=a+2d\\ a=12-2d\mathrm{.......}\left(1\right)\end{array}$

Similarly, the last term is given as

$\begin{array}{c}106=a+(50-1)d\\ 106=a+49d\\ a=106-49d\mathrm{.......}\left(2\right)\end{array}$

Step 2: Calculate the common difference and the first term.

Equate equations $\left(1\right)\text{\hspace{0.17em}and\hspace{0.17em}}\left(2\right)$.

$\begin{array}{c}12-2d=106-49d\\ 49d-2d=106-12\\ 47d=94\\ d=2\end{array}$

Apply $d=2$ in equation $\left(2\right)$.

So,

$\begin{array}{c}a=106-49\left(2\right)\\ a=106-98\\ a=8\end{array}$

Step 3: Calculate the ${29}^{th}$term.

Use $a=8$ and $d=2$ to find the ${29}^{th}$ term.

Therefore,

$\begin{array}{c}{a}_{29}=8+(29-1)2\\ a_{29}=8+28\times 2\\ a_{29}=8+56\\ a_{29}=64\end{array}$

Hence, the ${29}^{th}$term is $64$.