NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 2(iii)

Question:

Find the sums given below : –5 + (–8) + (–11) + . . . + (–230)

Given:

Arithmetic progression: –5, –8, –11, ..., –230

To Find

The sum of the given arithmetic progression.

Formula:

Sum of an arithmetic progression = n 2 [2a + (n – 1)d] where
'n' is the number of terms,
'a' is the first term, and
'd' is the common difference.

Solution:

Here, a = –5,
d = –8 – (–5) = –3, and
last term l = –230.

First, find the number of terms (n): l = a + (n – 1)d
⇒ –230 = –5 + (n – 1)(–3)
⇒ –225 = (n – 1)(–3)
⇒ n – 1 = 75
⇒ n = 76

Now, find the sum: S_n = n 2 [2a + (n – 1)d]
⇒ S_n = 76 2 [2(–5) + (76 – 1)(–3)]
⇒ S_n = 38[–10 – 225]
⇒ S_n = 38(–235)
⇒ S_n = –8930

Alternatively, using the formula S_n = n/2 (a+l):
⇒ S_n =  76 2 (-5 + -230)
⇒ S_n = 38(-235)
⇒ S_n = -8930

Result:

The sum of the arithmetic progression is –8930.

Next question solution:

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 3(i)

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