Question:
If $(2 n+1)+(2 n+3)+(2 n+5)+\ldots+(2 n+47)=5280,$ then what is the value of $1+2+3+\ldots+n$ ?
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
$\Rightarrow \left( 2n+1 \right)+(m-1)(2)=2n+47$
$\Rightarrow m=24$
Also,
$(2 n+1)+(2 n+3)+(2 n+5)+\ldots+(2 n+47)=5280,$
$=\frac{24}{2}\left[ 2\left( 2n+1 \right)+\left( 24-1 \right)\times 2 \right]$
$=24\left( 2n+1+23 \right)=48\left( n+12 \right)$
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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