# Arithmetic Progression

## Exercise 5.2 Part 3

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

Solution: Given, a = 3, d = a2 – a1 = 8 – 3 = 5, an = 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

Solution: Here, a = 7, d = 6, an = 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

Solution: 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, ……………

Solution: Here, a = 11, d = 8 – 11 = - 3, an = - 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

Solution: Given, a11 = 38 and a16 = 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