CAT 2019 Set-2 | Question: 96

Question

If $(2n+1)+(2n+3)+(2n+5)+\dots+(2n+47)=5280,$ then what is the value of $1+2+3+\dots+n$ _______

Answer 1

Given that, $ (2n+1) + (2n+3) + (2n+5)+ \dots + (2n+47) = 5280$

Here, the first term $a = 2n+1$

And, common difference $d = (2n+3) – (2n+1)$

$\Rightarrow d =  2n+3-2n-1 = 2 $

Let $’x’$ be the number of terms.

The last term $l=a+(x-1)d$

$ \Rightarrow 2n+47 =2n+1+(x-1)2$

$ \Rightarrow 47=1+2x-2$

$ \Rightarrow 2x =48$

$ \Rightarrow \boxed {x=24}$

Now, the sum of A.P. $= \frac {x}{2} \left [ {a+l} \right]$

$ \Rightarrow 5280 = \frac {24}{2} \left [{2n+1+2n+47 } \right] $

$ \Rightarrow 12(4n +48) = 5280 $

$ \Rightarrow 12 \times 4 (n+12) =5280$

$ \Rightarrow 48 (n+12) =5280 $

$ \Rightarrow n+12 = \frac {5280}{48} $

$ \Rightarrow n+12=110$

$ \Rightarrow \boxed {n=98}$

Now, the value of $1+2+3+ \dots +98$

$ \quad = \frac {98}{2} \left [{1+98} \right]$

$ \quad = 49 \times 99 = 4851.$

Correct Answer $:4851$