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Hint: Here we have to find which term consist the value 248 for that we have to first find the value first term followed by common difference and then apply nth term of n AP

“Complete step-by-step answer:”

Given AP series is 3, 8, 13……

Now you have to find out which term is 248

So, let us suppose the last term of an AP, ${a_n} = 248$

First term of an AP, ${a_1} = 3$

Common difference of an AP, $d = \left( {8 - 3} \right) = \left( {13 - 8} \right) = 5$

Let us suppose number of terms in AP is n

Then, According to AP property

${a_n} = {a_1} + \left( {n - 1} \right)d$

$ \Rightarrow 248 = 3 + \left( {n - 1} \right)5$

$ \Rightarrow 248 - 3 = \left( {n - 1} \right)5$

$ \Rightarrow \left( {n - 1} \right) = \dfrac{{245}}{5} = 49$

$ \Rightarrow n = 49 + 1 = 50$

So 248 is 50th term of an AP

NOTE: Whenever we face such a problem the key concept is that we have to remember all the formulas of an AP, it will give you a lot of help in finding your desired answer.

“Complete step-by-step answer:”

Given AP series is 3, 8, 13……

Now you have to find out which term is 248

So, let us suppose the last term of an AP, ${a_n} = 248$

First term of an AP, ${a_1} = 3$

Common difference of an AP, $d = \left( {8 - 3} \right) = \left( {13 - 8} \right) = 5$

Let us suppose number of terms in AP is n

Then, According to AP property

${a_n} = {a_1} + \left( {n - 1} \right)d$

$ \Rightarrow 248 = 3 + \left( {n - 1} \right)5$

$ \Rightarrow 248 - 3 = \left( {n - 1} \right)5$

$ \Rightarrow \left( {n - 1} \right) = \dfrac{{245}}{5} = 49$

$ \Rightarrow n = 49 + 1 = 50$

So 248 is 50th term of an AP

NOTE: Whenever we face such a problem the key concept is that we have to remember all the formulas of an AP, it will give you a lot of help in finding your desired answer.

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