NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 7

Question:

Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Given:

11^th term (a₁₁) = 38
16^th term (a₁₆) = 73

To Find:

31^st term (a₃₁)

Formula:

a_n = a + (n-1)d where 

a is the first term and 

d is the common difference.

Solution:

a₁₁ = a + 10d = 38 ......(1)

a₁₆ = a + 15d = 73 ......(2)

Subtracting (1) from (2):
(a + 15d) - (a + 10d) = 73 - 38 

⇒ a + 15d - a - 10d = 73 - 38 

⇒ 5d = 35 

⇒ d = 7

Substituting d = 7 in (1):
a + 10(7) = 38 

⇒ a + 70 = 38 

⇒ a = 38 - 70 

⇒ a = -32

Now, 

a₃₁ = a + 30d 

⇒ a₃₁ = -32 + 30(7) 

⇒ a₃₁= -32 + 210 

⇒ a₃₁ = 178

Result:

The 31^st term of the AP is 178.

Next question solution:

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 8

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