Insert 6 arithmetic means between 20 and 76. | vedclass

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(C) Let the six arithmetic means be $A_1, A_2, A_3, A_4, A_5, A_6$ between $20$ and $76$.Then,the sequence $20, A_1, A_2, A_3, A_4, A_5, A_6, 76$ form

Vedclass · Accepted Answer

$28, 36, 44, 52, 60, 68$

Vedclass · Answer

(C) Let the six arithmetic means be $A_1, A_2, A_3, A_4, A_5, A_6$ between $20$ and $76$.Then,the sequence $20, A_1, A_2, A_3, A_4, A_5, A_6, 76$ forms an arithmetic progression $(AP)$ with $n = 8$ terms.The first term $a = 20$ and the last term $a_8 = 76$.The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.Substituting the values,$76 = 20 + (8 - 1)d$.$76 - 20 = 7d \Rightarrow 56 = 7d \Rightarrow d = 8$.The arithmetic means are:$A_1 = 20 + 8 = 28$$A_2 = 28 + 8 = 36$$A_3 = 36 + 8 = 44$$A_4 = 44 + 8 = 52$$A_5 = 52 + 8 = 60$$A_6 = 60 + 8 = 68$Thus,the arithmetic means are $28, 36, 44, 52, 60, 68$.

Insert $6$ arithmetic means between $20$ and $76$.

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