Slot -2 – Quantitative Aptitude – Modern Maths – Progressions – Let a(base1), a(base2), …a(base52)
Let a(base1), a(base2), … , a(base52) be positive integers such that a(base1) < a(base2) < … < a(base52). Suppose, their arithmetic mean is one less than the arithmetic mean of a(base2), a(base3), …, a(base52). If a(base52) = 100, then the largest possible value of a(base1) is?
Answer: d) 23
We want to maximize the value of a1, subject to the condition that
(a2+ a3 +⋯….+ a52 )/51 -(a1+ a2 +⋯….+ a52 )/52 = 1
Since a(base52) = 100 and all the numbers are positive integers, maximizing a1 entails maximizing a2, a3 ….a51. The only way to do this is to assume that a2, a3…. A52 are in an AP with a common difference of 1.
Let the average of a2, a3, …, a52 i.e.
(a2+ a3 +⋯….+ a52 )/51= a27 =A ( using the average of an odd number of terms in an Arithmetic Progression is equal to the value of the middle-most term)
So a52 = a27 + 25*1 = a27 + 25 and
given a52 = 100
=> a27 = A = 100 – 25 = 75
a2 + a3 + … + a52 = 75×51 = 3825
Given a1 + a2 +… + a52 = 52(A – 1) = 3848
Hence a1 = 3848 – 3825 = 23
Ans : 23
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