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
\[\begin{array}{l}
\dfrac{n}{2}\left\{ {2a + (n - 1)d} \right\} = 63...........................(i)\\
a + (n - 1)d = 18.....................................(ii)
\end{array}\]
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.