Insert six A.M.s between 15 and −13.

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Solution

Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ be the 6 A.M.s between 15 and $-$ 13.
Then, 15, $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ and $-$ 13 are in A.P. whose common difference is as follows:

d = $\frac{- 13 - 15}{6 + 1}$
= $- 4$

$A_{1} = 15 + d = 15 + (- 4) = 11 A_{2} = 15 + 2 d = 15 + (- 8) = 7 A_{3} = 15 + 3 d = 15 + (- 12) = 3 A_{4} = 15 + 4 d = 15 + (- 16) = - 1 A_{5} = 15 + 5 d = 15 + (- 20) = - 5 A_{6} = 15 + 6 d = 15 + (- 24) = - 9$
Hence, the required A.M.s are $11, 7, 3, - 1, - 5, - 9$ .

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