Question

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

• A. 4851
• B. 1458
• C. 4718
• D. 4378

Answer 1

Answer: (A)

4851

Answer 2

The sequence $(2n+1)+(2n+3)+(2n+5)+…+(2n+47)=5280$ is an arithmetic progression with the first term $(a)$ as $2n+1,$ the common difference $(d)$ as 2, and the last term $(t_n​)$ as $2n+47.$

Let 'm' be the number of terms in this sequence.
The last term of the arithmetic progression is given by $a+(n−1)d:$

$(2n+1)+(m−1)(2)=2n+47$
$⇒m=24$

Also,
$(2n+1)+(2n+3)+(2n+5)+…+(2n+47)=5280$
$=24×2[2(2n+1)+(24−1)×2]$
$=24(2n+1+23)=48(n+12)$

Therefore,
$48(n+12)=5280$
$⇒n=98$

Hence,
$1+2+3+…+n=2n(n+1)​=298×99​=4851$