Let A_1, A_2, dots, A_{39} be 39 arithmetic m | vedclass

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(D) Let the arithmetic means be $A_1, A_2, \dots, A_{39}$ between $a = 59$ and $b = 159$.The common difference $d$ is given by $d = \frac{b - a}{n + 1

Vedclass · Accepted Answer

$134$

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(D) Let the arithmetic means be $A_1, A_2, \dots, A_{39}$ between $a = 59$ and $b = 159$.The common difference $d$ is given by $d = \frac{b - a}{n + 1} = \frac{159 - 59}{39 + 1} = \frac{100}{40} = 2.5$.The $k$-th arithmetic mean is $A_k = a + k \cdot d = 59 + 2.5k$.We need to find the mean of $A_{25}, A_{28}, A_{31}, A_{36}$:$	ext{Mean} = \frac{A_{25} + A_{28} + A_{31} + A_{36}}{4} = \frac{(59 + 25d) + (59 + 28d) + (59 + 31d) + (59 + 36d)}{4}$.$	ext{Mean} = \frac{4 \cdot 59 + (25 + 28 + 31 + 36)d}{4} = 59 + \frac{120d}{4} = 59 + 30d$.Substituting $d = 2.5$,we get $	ext{Mean} = 59 + 30(2.5) = 59 + 75 = 134$.

Let $A_1, A_2, \dots, A_{39}$ be $39$ arithmetic means between the numbers $59$ and $159$. Then the mean of $A_{25}, A_{28}, A_{31}$ and $A_{36}$ is equal to:

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