JEE Main 2026 Apr 5 Shift 2, Mathematics Q6: Arithmetic, Geometric And Harmonic Progressions

JEE Main2026Apr 5, Shift 2Mathematics

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:

A
129
B
136
C
131.50
D
134

Solution

The 41-term AP runs 59, $A_{1}, \dots, A_{39}$ , 159. Common difference $d = \frac{159 - 59}{40} = \frac{5}{2}$ .

$A_{k} = 59 + k \cdot d$ . Compute:

$A_{25} = 59 + 62.5 = 121.5$ , $A_{28} = 59 + 70 = 129$ , $A_{31} = 59 + 77.5 = 136.5$ , $A_{36} = 59 + 90 = 149$ .

Mean $= \frac{121.5 + 129 + 136.5 + 149}{4} = \frac{536}{4} = 134$ .

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JEE Main 2026 Apr 5 Shift 2, Mathematics Q6: Arithmetic, Geometric And Harmonic Progressions

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