# Arithmetic Progression

## NCERT Exercise 5.2

### Part 3

Question: 4 – Which term of the AP: 3, 8, 13, 18, ……………… is 78?

**Answer:** Given, a = 3, d = a_{2} – a_{1} = 8 – 3 = 5, a_{n} = 78, n = ?

We know that `a_n = a + (n – 1)d`

Or, `78 = 3 + (n – 1)5`

Or, `(n – 1)5 = 78 – 3 = 75`

Or, `n – 1 = 15`

Or, `n = 15 + 1 = 16`

Thus, 78 is the 16th term of given AP.

Question: 5 – Find the number of terms in each of the following APs

(i) 7, 13, 18, ……, 205

**Answer:** Here, a = 7, d = 6, a_{n} = 205, n = ?

We know that `a_n = a + (n – 1)d`

Or, `205 = 7 + (n – 1)6`

Or, `(n – 1)6 = 205 – 7 = 198`

Or, `n – 1 = 33`

Or, `n = 34`

Thus, 205 is the 34th of term of this AP.

(ii) 18, 15.5, 13, …………, - 47

**Answer:** Here, a = 18, d = 15.5 – 18 = - 2.5

We know that `a_n = a + (n – 1)d`

Or, `- 47 = 18 + (n – 1)(- 2.5)`

Or, `(n – 1)(- 2.5) = - 47 – 18 = - 65`

Or, `n – 1 = 26`

Or, `n = 27`

Thus, - 47 is the 27th term of this AP.

Question: 6 – Check whether – 150 is a term of the AP; 11, 8, 5, 2, ……………

**Answer:** Here, a = 11, d = 8 – 11 = - 3, a_{n} = - 150, n = ?

We know that `a_n = a + (n – 1)d`

Or, `- 150 = 11 + (n – 1)(- 3)`

Or, `(n – 1)(-3) = - 150 – 11 = - 161`

Or, `n – 1 = (161)/(3)`

It is clear that 161 is not divisible by three and we shall get a fraction as a result. But number of term cannot be a fraction.

Hence, - 150 is not a term of the given AP.

Quesiton: 7 – Find the 31st term of an AP whose 11th term is 38 and 16th term is 73

**Answer:** Given, a_{11} = 38 and a_{16} = 73

We know that `a_n = a + (n – 1)d`

Hence, `a_(11) = a + 10d = 38`

And, `a_(16) = a + 15d = 73`

Subtracting 11th term from 16th term, we get following:

`a + 15d – a – 10d = 73 – 38`

Or, `5d = 35`

Or, `d = 7`

Substituting the value of d in 11th term we get;

`a + 10 xx 7 = 38`

Or, `a + 70 = 38`

Or, `a = 38 – 70 = - 32`

Now 31st term can be calculated as follows:

`a_(31) = a + 30d`

`= - 32 + 30 xx 7`

`= - 32 + 210 = 178`