# Insert five number between 8 and 26 such that the resulting sequence is an A.P.

Let the five A.M’s be A1, A2, A3, A4, A5

Therefore 8, A1, A2, A3, A4, A5, 26 are in A.P. Let d be the common difference

Here T1 = 8

T7 = 26

⇒a + ( 7 – 1)d = 26

or a + 6d = 26

or 8 + 6d = 26

or 6d = 18

or d = 3

Therefore A1 = 8 + 3 = 11
A2 = 11 + 3 = 14
A3 = 14 + 3 = 17
A4 = 17 + 3 = 20
A5 = 20 + 3 = 23

Thus five required numbers are

11, 14, 17, 20 and 23

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