Answer :

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} = n^{2}

For n = 1,

S_{1} = 1^{2}

= 1

For n = 2,

S_{2} = 2^{2}

= 4

For n = 3,

S_{3} = 3^{2}

= 9

Now S_{1} = a_{1}

a_{1} = 1

S_{2} – S_{1} = a_{2}

4 – 1 = a_{2}

3 = a_{2}

Now, d = a_{2} – a_{1}

= 3 -1

= 2

We know, a_{n} = a + (n – 1)d

For n = 10,

a_{10} = 1 + (10– 1)2

= 1 + 9(2)

= 1 + 18

= 19

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

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