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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

Correct answer: 51

Answer: 51

Solution

Easy

The new even sequence ends at 2a32a_3, so its five terms are 2a3−8,…,2a32a_3-8,\dots,2a_3. Sum them to find a3a_3, then use the constant gap of consecutive odd numbers to get a5a_5.

1

Write the even sequence in terms of a3a_3. Five consecutive even numbers ending at 2a32a_3:

2a3−8,  2a3−6,  2a3−4,  2a3−2,  2a3\begin{aligned} &2a_3-8,\ \ 2a_3-6,\ \ 2a_3-4,\ \ 2a_3-2,\ \ 2a_3 \end{aligned}
2

Set the sum equal to 450450. Add the five terms.

(2a3−8)+(2a3−6)+(2a3−4)+(2a3−2)+2a3=450⇒ 10a3−20=450⇒ 10a3=470⇒ a3=47\begin{aligned} &(2a_3-8)+(2a_3-6)+(2a_3-4)+(2a_3-2)+2a_3=450\\ &\Rightarrow\ 10a_3-20=450\\ &\Rightarrow\ 10a_3=470\\ &\Rightarrow\ a_3=47 \end{aligned}
3

Find a5a_5. In five consecutive odd numbers, consecutive terms differ by 22, so a5=a3+4a_5=a_3+4.

a5=a3+4=47+4⇒ a5=51\begin{aligned} &a_5=a_3+4=47+4\\ &\Rightarrow\ a_5=51 \end{aligned}
a5=51a_5=51

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CAT 2017 Slot 2 QA Q27: Let a 1 , a 2 , a 3 , a 4 , a 5 be a sequence of five consecutive odd numbers. Consider a new sequence of five — Solution | TheCATExam