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CAT 2023 Slot 2 — QA Question 21

Arithmetic ProgressionEasy

Let both the series a1, a2, a3, ... and b1, b2, b3, ... be in arithmetic progression such that the common differences of both the series are prime numbers. If a5 = b9, a19 = b19 and b2 = 0, then a11 equal?

Answer & solution

  • A

    86

  • B

    84

  • 79

  • D

    83

Solution

Easy

Subtract one given equation from the other so the unknown starting terms cancel, leaving a clean relation between the two common differences pp and qq. The "both prime" condition then forces unique values, and b2=0b_2=0 anchors the numbers.

Let the common difference of {an}\{a_n\} be pp and of {bn}\{b_n\} be qq, both prime. Recall am−ak=(m−k)pa_m-a_k=(m-k)p.

1

Subtract the two given equalities a19=b19a_{19}=b_{19} minus a5=b9a_5=b_9:

a19−a5=b19−b9⇒ 14p=10q⇒ pq=57\begin{aligned} &a_{19}-a_5=b_{19}-b_9\\ &\Rightarrow\ 14p=10q\\ &\Rightarrow\ \frac{p}{q}=\frac{5}{7} \end{aligned}
2

Use primality. The ratio pq=57\tfrac{p}{q}=\tfrac{5}{7} in lowest terms, and both must be prime, so:

p=5,q=7\begin{aligned} &p=5,\qquad q=7 \end{aligned}
3

Anchor with b2=0b_2=0 to find b9b_9, then a5a_5:

b9=b2+7q=0+7⋅7=49a5=b9=49(given)\begin{aligned} &b_9=b_2+7q=0+7\cdot 7=49\\ &a_5=b_9=49 \quad\text{(given)} \end{aligned}
4

Step up from a5a_5 to a11a_{11} (six steps of pp):

a11=a5+6p=49+6⋅5=79\begin{aligned} &a_{11}=a_5+6p=49+6\cdot 5=79 \end{aligned}
a11=79a_{11}=\mathbf{79}
CAT 2023 Slot 2 QA Q21: Let both the series a 1 , a 2 , a 3 , ... and b 1 , b 2 , b 3 , ... be in arithmetic progression such that the — Solution | TheCATExam