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Question 27 : Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3. If the sum of the numbers in the new sequence is 450, then a5 is [TITA]


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Explanatory Answer

Method of solving this CAT Question from Progressions

Let a1, a2, a3, a4, a5 be n - 4 , n - 2 , n , n + 2 ,n + 4 respectively which is the sequence of consecutive odd numbers, where n = a3.
Let the new sequence of five consecutive even numbers be m - 4, m - 2, m, m + 2, m + 4.
Where the last term m + 4 = 2a3 = 450
Average of any number of terms in AP is the middle term
5 terms in AP ⟹ 5 × (middle term) = 450
⟹ Middle term (m) = 90
⟹ m - 4, m - 2, m, m + 2, m + 4 is equal to 86, 88, 90, 92, and 94 respectively.
⟹ 2a3 = 94
⟹ a3 = \\frac{94}{2}) = 47
Hence a1, a2, a3, a4, a5 is equal to 43, 45, 47, 49, and 51 respectively.
Therefore a5 = 51

The question is "Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3. If the sum of the numbers in the new sequence is 450, then a5 is [TITA]"

Hence, the answer is 51

 

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