NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question2(ii)

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

Question:

Find the sums given below : 34 + 32 + 30 + . . . + 10

Given:

An arithmetic progression: 34, 32, 30, ..., 10

To Find:

The sum of the given 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:

Here, a = 34, 

d = 32 - 34 = -2, and 

the last term l = 10.

First, let's find the number of terms (n): l = a + (n-1)d

⇒ 10 = 34 + (n - 1)(-2) 

⇒ 10 = 34 -2n + 2 

⇒ 2n = 26 

⇒ n = 13

Now, we can find the sum using the formula: Sn = n 2  (2a + (n - 1)d)

⇒ Sn = 13 2  (2(34) + (13 - 1)(-2)) 

⇒ Sn = 13 2  (68 - 24) 

⇒ Sn = 13 2  (44) 

⇒ Sn = 13 × 22 

⇒ Sn = 286

Result:

The sum of the arithmetic progression is 286.

© Kaliyuga Ekalavya. All rights reserved.

Comments

Post a Comment