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

Question:

In an AP: a = 5, d = 3, a_n = 50, find n and S_n

Given:

a = 5, d = 3, a_n = 50

To Find:

n and S_n

Formula:

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

Sum of an arithmetic progression = S_n = n 2 (2a + (n - 1)d)

Solution:

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

⇒ 50 = 5 + (n - 1)3

50 = 5 + 3n - 3 

⇒ 50 = 2 + 3n 

⇒ 48 = 3n 

⇒ n = 48 3  

⇒ n = 16

Therefore, n = 16

Now, using the formula S_n = n2(2a + (n - 1)d), we have:

⇒ S₁₆ = 16 2 (2(5) + (16 - 1)3)

S₁₆ = 8(10 + 45) = 8(55) = 440

Result:

n = 16, S₁₆ = 440

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NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question3(ii)

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