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.

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Solution

## 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 $a+\left(n-1\right)d$.So, the $3rd$ term is given as$\begin{array}{c}12=a+\left(3-1\right)d\\ 12=a+2d\\ a=12-2d.......\left(1\right)\end{array}$Similarly, the last term is given as$\begin{array}{c}106=a+\left(50-1\right)d\\ 106=a+49d\\ a=106-49d.......\left(2\right)\end{array}$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,$\begin{array}{c}{a}_{29}=8+\left(29-1\right)2\\ {a}_{29}=8+28×2\\ {a}_{29}=8+56\\ {a}_{29}=64\end{array}$Hence, the ${29}^{th}$term is $64$.

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