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