Find how many two-digit numbers are divisible by 7?

OR

If the sum of first n terms of an AP is n², then find its 10^th term.

The two-digit numbers that are divisible by 7 are 14, 21,….. ,98

Here, First term = a = 14

Common difference d = 21 – 14 = 7,

Last term = a_n = 98

Since, n^th term is given by:

a_n = a + (n - 1) d

⇒ 98 = 14 + (n – 1)7

⇒ 98 = 14 + 7n – 7

⇒ 98 = 7 + 7n

⇒ 91 = 7n

⇒ n = 13

There are 13 two-digit numbers divisible by 7.

OR

We know sum of n terms of an AP is:

S_n = n²

For n = 1,

S₁ = 1²

= 1

For n = 2,

S₂ = 2²

= 4

For n = 3,

S₃ = 3²

= 9

Now S₁ = a₁

a₁ = 1

S₂ – S₁ = a₂

4 – 1 = a₂

3 = a₂

Now, d = a₂ – a₁
= 3 -1

= 2

We know, a_n = a + (n – 1)d

For n = 10,

a₁₀ = 1 + (10– 1)2

= 1 + 9(2)

= 1 + 18

= 19

So, 10^th term of the AP is 19.