CAT 2017 Slot 2 — QA Question 27
Arithmetic ProgressionEasy
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
Answer & solution
Answer: 51
Solution
Since 2a3 is the highest number in the new sequence, the sequence of 5 even numbers starting with the lowest, when expressed (in terms of a3) is
(2a3 - 8), (2a3 - 6), (2a3 - 4), (2a3 - 2), 2a3
Sum of these 5 numbers = 450
⇒ 2a3 - 8 + 2a3 - 6 + 2a3 - 4 + 2a3 - 2 + 2a3 = 450
⇒ 10a3 – 20 = 450
⇒ 10a3 = 470
⇒ a3 = 47
Now in the original sequence of 5 odd numbers, a3 = 47
⇒ a5 = a3 + 4
⇒ a5 = 47 + 4 = 51
Hence, 51.