Question:
Let $A_1, A_2, A_3, \dots, A_{39}$ 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 :
Let $A_1, A_2, A_3, \dots, A_{39}$ 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 :
Updated On: Apr 12, 2026
- 129
- 136
- 131.50
- 134
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
When $n$ arithmetic means are inserted between $a$ and $b$, they form an A.P. with $n+2$ terms. The common difference is $d = \frac{b-a}{n+1}$.
: Key Formula or Approach:
$a = 59, b = 159, n = 39$.
$d = \frac{159 - 59}{39 + 1} = \frac{100}{40} = 2.5$.
The $k$-th mean is $A_k = a + kd$.
Step 2: Detailed Explanation:
The required mean is:
Mean $= \frac{A_{25} + A_{28} + A_{31} + A_{36}}{4}$
Mean $= \frac{(59 + 25d) + (59 + 28d) + (59 + 31d) + (59 + 36d)}{4}$
Mean $= \frac{4 \times 59 + (25 + 28 + 31 + 36)d}{4} = 59 + \frac{120}{4}d = 59 + 30d$.
Mean $= 59 + 30 \times 2.5 = 59 + 75 = 134$.
Step 3: Final Answer:
The mean of the specified arithmetic means is 134.
When $n$ arithmetic means are inserted between $a$ and $b$, they form an A.P. with $n+2$ terms. The common difference is $d = \frac{b-a}{n+1}$.
: Key Formula or Approach:
$a = 59, b = 159, n = 39$.
$d = \frac{159 - 59}{39 + 1} = \frac{100}{40} = 2.5$.
The $k$-th mean is $A_k = a + kd$.
Step 2: Detailed Explanation:
The required mean is:
Mean $= \frac{A_{25} + A_{28} + A_{31} + A_{36}}{4}$
Mean $= \frac{(59 + 25d) + (59 + 28d) + (59 + 31d) + (59 + 36d)}{4}$
Mean $= \frac{4 \times 59 + (25 + 28 + 31 + 36)d}{4} = 59 + \frac{120}{4}d = 59 + 30d$.
Mean $= 59 + 30 \times 2.5 = 59 + 75 = 134$.
Step 3: Final Answer:
The mean of the specified arithmetic means is 134.
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