Question

Which term of the A.P. 3, 8, 13, … is 248?

Answer 1

Hint: Here in this question, given a sequence of arithmetic progression we have to find the term where 248 is located in the sequence. To find this by using a formula $$a_n=a+(n-1)d$$ , where $$a_n$$ is the value of nth term of A.P sequence, $$a$$ is the value of first term and $$d$$ is the common difference it can be find by $$d=a_2-a_1$$

Complete step-by-step answer:
The general arithmetic progression is of the form $$a,a+d,a+2d,...$$ where a is first term nth d is the common difference which is the same between the distance on any two numbers in sequence. The nth term of the arithmetic progression is defined as $$a_n=a+(n-1)d$$ .
Now let us consider the A.P. sequence which is given in the question 3, 8, 13, …, 248. Let us find the term i.e., $$n$$ where the last number 248 is located.
Here $$a=a_1=3$$ , $$a_2=8$$ , $$a_3=13$$ , and, $$a_n=248$$
The common difference:
 $$\Rightarrow d=a_2-a_1$$
 $$\Rightarrow d=8-3$$
 $$\Rightarrow d=5$$
Let we find the $$n$$ value by considering the formula of nth term of the arithmetic progression i.e.,
 $$\Rightarrow a_n=a+(n-1)d$$
On substituting the values, then we have
 $$\Rightarrow 248=3+(n-1)5$$
Remove a parenthesis by multiplying a 5.
 $$\Rightarrow 248=3+5n-5$$
On simplification, we have
 $$\Rightarrow 248=5n-2$$
Add 2 on both side, then
 $$\Rightarrow 248+2=5n$$
 $$\Rightarrow 250=5n$$
Divide 5 on both side, then we get
 $$\Rightarrow {\displaystyle \frac{250}{5}}=n$$
 $$\Rightarrow 50=n$$
Or
 $$\Rightarrow n=5$$
Therefore, in the A.P. 3, 8, 13, … the number 248 is a $${50}^{th}$$ term.
So, the correct answer is “ $${50}^{th}$$ term”.

Note: By considering the formula of arithmetic sequence we verify the obtained value which we obtained. We have to check the common difference for all the terms. Suppose if we check for the first two terms not for other terms then we may go wrong. So, definition of arithmetic sequence is important to solve these kinds of problems.