Q. 285.0( 2 Votes )

Insert f

Answer :


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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