Sum of n Terms of an Arithmetic progression (A.P)

To find the sum of natural numbers, we need to know the formula to find it.

The sum of “n” terms of an Arithmetic progression (A.P) can be easily found out using a simple formula given as;

S_n = \( \frac{n}{2}  \scriptsize (a + L) \)

where L = last term

OR

S_n = \( \frac{n}{2} \left [ \scriptsize 2a + \left (n-1 \right)d \right]\)

Example 2.1.1:

The first and last terms of an A.P are 21 and -47 respectively. If the sum of the series is given as -234. Calculate:

(i) the number of terms in the A.P
(ii) the common difference
(iii) the sum of the first 18 terms

Solution:

(i) recall S_n = \( \frac{n}{2}  \scriptsize (a + L) \)

where L = last term = – 47

a = 1st term = 21

and S_n = -234

Substitute the values of a, l, S_n into the equation

-234 = \( \frac{n}{2} \left [ \scriptsize 21 + \left (-47 \right) \right]\)

multiply both sides by 2

2 × (-234) = n(21 – 47)

-468 = -26n

divide both sides by -26

⇒ \( \frac{-468}{-26}= \frac{-26n}{-26} \)

n = \( \frac{-468}{-26} \)

n = 18

(ii) T_n = a + (n – 1)d