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Maths Question 6 – JEE-MAIN 2026

Let A1,A2,A3,...,A39 be 39 arithmetic means between the numbers 59 and 159. Then the mean of A25,A28,A31 and A36 is equal to :

The arithmetic means inserted between two numbers, along with the two numbers themselves, form a single Arithmetic Progression (AP).

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Ninja StrategySymmetry and Averaging

The mean of the given terms is the term at the average index, A30. This term lies halfway between the central term A20 (value is the average of endpoints, 109) and the conceptual term A40 (value is the endpoint, 159). Thus, the mean is (109+159)/2=134.

Step 1: Determine the Properties of the Arithmetic Progression✦ Active

The sequence 59,A1,A2,,A39,159 forms an Arithmetic Progression (AP). The first term is a=59 and the last term is b=159. Since there are 39 arithmetic means, the total number of terms in the AP is N=39+2=41.

The 41st term is T41=159. We can find the common difference d using the formula TN=a+(N1)d.

159=59+(411)d
100=40dd=10040=2.5

The formula for the k-th arithmetic mean is Ak=a+kd.

Step 2: Calculate the Mean of the Specified Terms○ Expand

We need to find the mean of A25,A28,A31, and A36.

Mean=A25+A28+A31+A364

Substitute the formula Ak=a+kd into the expression for the mean:

Mean=(a+25d)+(a+28d)+(a+31d)+(a+36d)4

Simplify the expression:

Mean=4a+(25+28+31+36)d4=a+1204d=a+30d

Now, substitute the values of a=59 and d=2.5:

Mean=59+30(2.5)=59+75=134
💡 Teacher's Secret Hint

Notice that the mean of a set of terms from an AP is equal to the term corresponding to the average of their indices. Here, the average index is (25+28+31+36)/4=30, so the mean is simply A30.

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