Question

For an A.P. the sum of its terms is 63, common difference is 2 and last term is 18. Find the number of terms in an A.P.

Answer 1

- Hint:The sum of all terms in an A.P is {S_n} = \dfrac{n}{2}\left\\{ {2a + (n - 1)d} \right\\} and the general term is given by ${a_n} = a + (n - 1)d$ , where a, n, and d are the first term, total number of terms and the common difference between two terms respectively.

Complete step-by-step solution -
First of all let us see what we are given here
So we are given the value of {S_n} = \dfrac{n}{2}\left\\{ {2a + (n - 1)d} \right\\} = 63 the value of $d = 2$ and the last term, that will be ${a_n}$ in this case it is 18.
Now let us try to form 2 equations using this value of ${a_n}\& {S_n}$ and that will be

\dfrac{n}{2}\left\\{ {2a + (n - 1)d} \right\\} = 63...........................(i)\\\ a + (n - 1)d = 18.....................................(ii) \end{array}$$ Now let us simply the equations (i) and (ii) by putting the value of d $$\begin{array}{l} \therefore \dfrac{n}{2}\left\\{ {2a + (n - 1)d} \right\\} = 63\\\ \Rightarrow \dfrac{n}{2}\left\\{ {2a + (n - 1)2} \right\\} = 63\\\ \Rightarrow \dfrac{n}{2} \times 2\\{ a + (n - 1)\\} = 63\\\ \Rightarrow n\\{ a + (n - 1)\\} = 63 \end{array}$$ Now the simplified equation of (i) looks like $$n\\{ a + (n - 1)\\} = 63$$ Let us see the equation (ii) now, it is $$\begin{array}{l} \therefore a + (n - 1)d = 18\\\ \Rightarrow a + (n - 1) \times 2 = 18\\\ \Rightarrow a + (n - 1) = 9 \end{array}$$ Clearly we can put the value of equation (ii) in equation (i), So we will get it as $$\begin{array}{l} \therefore n\\{ a + (n - 1)\\} = 63\\\ \Rightarrow n \times 9 = 63\\\ \Rightarrow n = \dfrac{{63}}{9}\\\ \Rightarrow n = 7 \end{array}$$ _So the value of n is 7. Which we were told to find._ Note: A lot of students make mistake while putting the value, a lot of them usually, just took the value of n in terms of a and then put it in the previous equation to solve the problem, but that will be a lengthier method, there’s a lot of chances of making silly mistake but in the method discussed in the solution, it becomes a one-liner problem just by putting the correct value.