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

Question

In an AP: given a = 2, d = 8, S_n = 90, find n and a_n.

Given

a = 2, d = 8, S_n = 90

To Find

n and a_n

Formula

Sum of an arithmetic progression = S_n = n 2 (2a + (n - 1)d)
The nth term of arithmetic progression is given by a_n = a + (n - 1)d

Solution

Using the formula for the sum of an AP:

90 = n 2 (2(2) + (n - 1)8) 

⇒ 180 = n(4 + 8n - 8) 

⇒ 180 = n(8n - 4) 

⇒ 180 = 8n² - 4n 

⇒ 8n² - 4n - 180 = 0 

⇒ 2n² - n - 45 = 0

Solving the quadratic equation 2n² - n - 45 = 0:

(2n + 9)(n - 5) = 0

⇒ n = 5 or n = - 9 2

Since n cannot be negative, n = 5

Now, finding a_n (a₅):

a₅ = a + (5 - 1)d 

= 2 + 4(8) 

= 2 + 32 

= 34

Result

n = 5, a₅ = 34

Next question solution

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

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