Insert $6$ numbers between $3$ and $24$ such that the resulting sequence is an $A.P.$

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Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ and $A_{6}$ be six numbers between $3$ and $24$ such that $3, A _{1}, A _{2}, A _{3}, A _{4}, A _{5}, A _{6}, 24$ are in $A.P.$ Here, $a=3, b=24, n=8$

Therefore, $24=3+(8-1) d,$ so that $d=3$

Thus ${A_1} = a + d = 3 + 3 = 6;\quad $

${A_2} = a + 2d = 3 + 2 \times 3 = 9$

${A_3} = a + 3d = 3 + 3 \times 3 = 12;\quad $

${A_4} = a + 4d = 3 + 4 \times 3 = 15$

${A_5} = a + 5d = 3 + 5 \times 3 = 18;\quad $

${A_6} = a + 6d = 3 + 6 \times 3 = 21$

Hence, six numbers between $3$ and $24$ are $6,9,12,15,18$ and $21$

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