NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question1(iii)

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 1(iii)

Question:

Find the sum of the following APs: 0.6, 1.7, 2.8, . . ., to 100 terms.

Given:

Arithmetic progression: 0.6, 1.7, 2.8, ...

Number of terms (n) = 100

To Find:

The 100th term (a100) of the arithmetic progression.

Formula:

The sum of an arithmetic progression is given by: Sn = 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:

First term (a) = 0.6

Common difference (d) = 1.7 - 0.6 = 1.1

n = 100

Sum to 100 terms can be found by substituting a, d and n in the formula Sn = n 2 [2a + (n – 1)d]
S100 = 100 2 [2(0.6) + (100 – 1)1.1]
S100 = 50 x 
[1.2 + 99 x 1.1]
S100 = 50 x (1.2 + 108.9 ) = 5505

Result:

The sum to 100th term of the given arithmetic progression is 5505

© Kaliyuga Ekalavya. All rights reserved.

Comments

Post a Comment