NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 4

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 4

Question:

How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Given:

Arithmetic Progression (AP): 9, 17, 25, ...
Sum of terms (Sn) = 636

To Find:

Number of terms (n) to be taken to get a sum of 636.

Formula:

The sum of n terms of an AP is given by: Sn = n 2 [2a + (n - 1)d] where 

a is the first term, 

d is the common difference, and 

n is the number of terms.

Solution:

Here, 

a = 9, 

d = 17 - 9 = 8, and 

Sn = 636.

Substituting the values in the formula, Sn = n 2 [2a + (n - 1)d] we get:
⇒ 636 = n 2 [2(9) + (n - 1)8]

⇒ 1272 = n[18 + 8n - 8]
⇒ 1272 = n(8n + 10)
⇒ 1272 = 8n2 + 10n
⇒ 8n2 + 10n - 1272 = 0
⇒ 4n2 + 5n - 636 = 0

For a quadratic equation ax2 + bx + c = 0, the roots are given by the quadratic formula:

                                                    x = -b ± √(Δ) 2a

                                                    Î” = b2 - 4ac

Solving the quadratic equation 4n2 + 5n - 636 = 0 using the quadratic formula:

⇒ n = -5 ± √(52 - 4(4)(-636)) 2(4)
⇒ n = -5 ± √(25 + 10176) 8  

⇒ n = -5 ± √10201 8  

⇒ n = -5 ± 101 8

Therefore, 

n = -5 + 101 8  or  -5 - 101 8

⇒ n = -96 8   or  -106 8

⇒ n = 12 or  -13.25

Since the number of terms cannot be negative, n = 12.

Result:

12 terms of the AP must be taken to give a sum of 636.

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