Question 91
The 29th term of the series in AP is given as 609, and it is also provided that the first term is the same as the common difference in that series. What is the sum of 18 terms of the series?
Solution
Let the first term of the series be a, and the common difference of the series be d. It is given that a = d and the 29th term of the series is given as 609.
$$a_{29}\ =\ a\ +\ \left(29\ -1\right)d\ =\ a\ +\ 28a\ =\ 29a\ =\ 609$$
$$a\ =\ 21\ =\ d$$
The sum of 18 terms can be calculated as,
$$S_{18}\ =\ \dfrac{n}{2}\left(2a\ +\ \left(n\ -\ 1\right)d\right)\ =\ \dfrac{18}{2}\left(2\times\ 21\ +\ \left(18\ -\ 1\right)21\right)\ =\ 9\left(19\ \times\ 21\right)\ =\ 3591\ $$
The correct answer is option C.
