# Arithmetic Progression

## NCERT Exercise 5.2

### Part 4

Question: 8 – An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution: Given, a3 = 12 and a50 = 106

a_3 = a + 2d = 12

a_(50) = a + 49d = 106

Subtracting 3rd term from 50th term, we get;

a + 49d – a – 2d = 106 – 12

Or, 47d = 94

Or, d = 2

Substituting the value of d in 12th term, we get;

a + 2 xx 2 = 12

Or, a + 4 = 12

Or, a = 8

Now, 29th term can be calculated as follows:

a_(29) = a + 28d

= 8 + 28 xx 2

= 8 + 56 = 64

Question: 9 – If the 3rd and the 9th term of an AP are 4 and – 8 respectively. Which term of this AP is zero?

Solution: Given, a_3 = 4 and a_9 = - 8

a_3 = a + 2d = 4

a_9 = a + 8d = - 8

Subtracting 3rd term from 9th term, we get;

a + 8d – a – 2d = - 8 – 4 = - 12

Or, 6d = - 12

Or, d = - 2

Substituting the value of d in 3rd term, we get;

a + 2(-2) = 4

Or, a – 4 = 4

Or, a = 8

Now; 0 = a + (n – 1)d

Or, 0 = 8 + (n – 1)(- 2)

Or, (n – 1)(- 2) = - 8

Or, n – 1 = 4

Or, n = 5

Thus, 5th term of this AP is zero.

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

Solution: Tenth and seventeenth terms of this AP can be given as follows:

a_(10) = a + 9d

a_(17) = a + 16d

Subtracting 10th term from 17th term, we get;

a + 16d – a – 9d = 7

Or, 7d = 7

Or, d = 1

Question: 11 – Which term of the AP: 3, 15, 27, 39, …. will be 132 more than its 54th term.

Solution: Here, a = 3, d = 15 – 3 = 12

54th term can be given as follows:

a_(54) = a + 53d

= 3 + 53 xx 12

= 3 + 636 = 639

So, the required term = 639 + 132 = 771

Or, 771 = a + (n – 1)d

Or, 771 = 3 + (n -1)12

Or, (n – 1)12 = 771 – 3 = 768

Or, n – 1 = 64

Or, n = 65

Thus, the required term is 65th term

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