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# Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.

Let a be the first term and d be the common difference.

Given: a4 = 9

a6 + a13 = 40

Now, Consider a4 = 9

a + (4 - 1)d = 9

a + 3d = 9 ……….(1)

Consider a6 + a13 = 40

a + (6 - 1)d + a + (13 - 1)d = 40

2a + 17d = 40 ………..(2)

Subtracting twice of equation (1) from equation (2), we get,

11d = 22

d = 2

Common difference = d = 2

Now from equation (1),we get

a = 9 - 3d

= 9 - 6

= 3

AP is a, a + d, a + 2d, a + 3d, …

i.e. AP is 3, 5, 7,9, 11…..

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