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

The Correct Option is A

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,
\(1+2+3+…+n=2n(n+1)​=298×99​=4851\)