SOLVING WORD PROBLEMS INVOLVING ARITHMETIC SEQUENCE

Problem 1 :

The 17^th term of an AP exceeds its 10^th term by 7. Find the common difference.

Solution :

a₁₇  =   a₁₀ + 7

a + 16d  =  a + 9d  + 7

a - a + 16d - 9d  =  7

7d  =  7

d  =  7/7

d  =  1

So, the common difference is 1.

Problem 2 :

Which term of the AP :

3, 15, 27, 39,.......... will be 132 more than its 54th term?

Solution :

a_n = 132 + a₅₄

a + (n - 1) d = 132 + a + 53d

 a = 3    d = 15 - 3  =  12

3 + (n - 1) 12  =  132 + 3 + 53 (12)

3 + 12 n - 12  =  135 + 636

-9 + 12 n  =  771

12 n  =  771 + 9

12 n  =  780

n  =  780/12

n  =  65

Problem 3 :

Two APs have the same common difference. The difference between their 100th term is 100, What is the difference between their 1000th terms?

Solution :

Let the first two terms of two APs a₁ and a₂ respectively and the common difference of these two A.Ps be d

for first A.P

a₁₀₀ = a₁ + (100 - 1) d

  =  a₁ + 99 d

a₁₀₀₀ = a₁ + (1000 - 1) d

  =  a₁ + 999 d

For second AP

a₁₀₀  =  a₂ + (100 - 1) d

  =  a₂ + 99 d

a₁₀₀₀  =  a₂ + (1000 - 1) d

=  a₂ + 999 d

given that,

difference between two  100^th term of two APs = 100

(a₁ + 99 d) - (a₂ + 99 d) = 100

a₁ + 99 d - a₂ - 99 d = 100

a₁ - a₂ = 100

Difference between 1000th term of two APs

   = (a₁ + 999 d) - (a₂ + 999 d)

   = a₁ + 999 d - a₂ - 999 d

   = a₁ - a₂

   =  100

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