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

Question

In an AP: given a₃ = 15, S₁₀ = 125, find d and a₁₀.

Given

a₃ = 15, S₁₀ = 125

To Find

d and a₁₀

Formula

The nth term of arithmetic progression is given by a_n = a₁ + (n-1)d
Sum of an arithmetic progression = S_n = n 2 (2a₁ + (n-1)d)

Solution

a₃ = a₁ + 2d = 15 

⇒ a₁ = 15 - 2d

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

S₁₀ = 10 2 (2a₁ + 9d) = 125

⇒ 5(2a₁ + 9d) = 125

⇒ 2a₁ + 9d = 25

Substituting a₁ = 15 - 2d:

2(15 - 2d) + 9d = 25

⇒ 30 - 4d + 9d = 25

⇒ 5d = -5

⇒ d = -1

a₁ = 15 - 2(-1) = 17

a₁₀ = a₁ + 9d = 17 + 9(-1) = 8

Result

d = -1, a₁₀ = 8

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

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