Quantitative Aptitude – Modern Maths – Progressions – Let a(base1), a(base2)

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?

a) 45
b) 20
c) 48
d) 23

Answer: d) 23

Solution:
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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