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Question

(A) Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ be six numbers between $3$ and $24$ such that $3, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, 24$ are

Vedclass · Accepted Answer

$6, 9, 12, 15, 18, 21$

Vedclass · Answer

(A) Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ 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,the first term $a = 3$,the last term $l = 24$,and the total number of terms $n = 6 + 2 = 8$.The formula for the $n^{th}$ term is $a_{n} = a + (n - 1)d$.Substituting the values: $24 = 3 + (8 - 1)d$.$24 - 3 = 7d \implies 21 = 7d \implies d = 3$.The numbers are:$A_{1} = a + d = 3 + 3 = 6$$A_{2} = a + 2d = 3 + 6 = 9$$A_{3} = a + 3d = 3 + 9 = 12$$A_{4} = a + 4d = 3 + 12 = 15$$A_{5} = a + 5d = 3 + 15 = 18$$A_{6} = a + 6d = 3 + 18 = 21$Thus,the six numbers are $6, 9, 12, 15, 18, 21$.

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

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