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CAT 2019 Quant Question with Solution 64

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

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Correct Answer: 4851

The sequence $(2 n+1)+(2 n+3)+(2 n+5)+\ldots+(2 n+47)=5280,$ is in arithmetic progression with first term (a) = 2n+1, common difference (d) = 2 and last term ($t_{n}$)=2n+47.

Let ‘m’ be the number of terms in this sequence.

The last term of A.P. is given by a+(n-1)d

$\Rightarrow \left( 2n+1 \right)+(m-1)(2)=2n+47$

$\Rightarrow m=24$


$(2 n+1)+(2 n+3)+(2 n+5)+\ldots+(2 n+47)=5280,$

$=\frac{24}{2}\left[ 2\left( 2n+1 \right)+\left( 24-1 \right)\times 2 \right]$

$=24\left( 2n+1+23 \right)=48\left( n+12 \right)$

Therefore, $48\left( n+12 \right)=5280\Rightarrow n=98$

Hence, $1+2+3+\ldots +n=\frac{n(n+1)}{2}=\frac{98\times 99}{2}=4851$

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