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UGCNET-June-2019-I-22

For all integers $y>1, \: \langle y \rangle = 2y+(2y-1)+(2y-2) + \dots +1$.

What is the value of $\langle 3 \rangle \times \langle 2 \rangle $? Where $\times$ is a multiplication operator?

  1. $116$
  2. $210$
  3. $263$
  4. $478$
in Numerical Ability by Veteran (416k points)
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2 Answers

+2 votes
according to the above function for <2> = 4+3+2+1= 10

& for <3> = 6+5+4+3+2+1=21

so the final value is 21*10 = 210

please verify it.
by Boss (13.1k points)
+2 votes
$ \langle y \rangle = 2y+(2y-1)+(2y-2) + \dots +1$

we can write like this

 $ \langle y \rangle = 1+2+3 +\dots +(2y-2) +(2y-1) +2y $

It is arithmetic progression whose first term is $a = 1$ and difference $d =1$ and last terms $l = 2y$

Sum of the AP series $S = \dfrac{n}{2}\left(a+ l \right)$

 Now, find the sum  $ \langle y \rangle = \dfrac{2y}{2}(1+2y)$

$ \implies \langle y \rangle = y(1+2y) $

$ \implies \langle y \rangle  =  y + 2y^{2}\rightarrow(1)$

Now find the value of  $ \langle 3 \rangle = 3 + 18 = 21$ and $ \langle 2 \rangle = 2 + 8 = 10$

The value of $ \langle 3 \rangle \times  \langle  2 \rangle  = 21 \times 10 = 210$

Option $(B)$ is the correct answer.
by Boss (45.8k points)

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