Find the Sum of a Sequence

[Moon]

The sequence $a_1, a_2,\cdots , a_{98}$ satisfies $a_{n+1} =a_n + 1$ for $n = 1, 2, \cdots , 97$ and has sum $137$.
\[a_2 + a_4 + a_6 + \cdots + a_{98}=?\]

AIME 84_2/1

[Labib]

Here's how I've worked on it!!

Soln::

If we see carefully we'll notice that $a_2+a_4+a_6+....+a_n= n\cdot a_1+\frac {n(n-1)}{2}$

so we get $98a_1=-4616=>49a_1=\frac {-4616}{2} = -2308$

again, $a_2+a_4+a_6+.....+a_2n= n\cdot a_1+ n^2$

thus the solution is=$49a_1+49^2=93$

[MathWasheefSci]

I have another technique. Given that \[a_{1}+a_{2}+a_{3}+.......+a_{98}=137
\] \[\Leftrightarrow S+a_{1}+a_{3}+a_{5}+.......+a_{97}=137\; ;[let\, \; a_{2}+a_{4}+.....+a_{98}=S]
\]\[\Leftrightarrow S+a_{1}+(a_{2}+1)+(a_{4}+1)+.......+(a_{96}+1)=137\; \: ;[\because a_{n+1}=a_{n}+1]
\]\[\Leftrightarrow 2S+a_{1}-a_{98}+48=137\; \; ;[\because \textrm{There are 48 terms}]
\]
Now \[\begin{matrix}
&a_{2} &-a_{1}=1 \\
&a_{3} &-a_{2}=1 \\
&a_{4} &-a_{3}=1 \\
&.... &.... \\
&a_{98} &-a_{97}=1
\end{matrix}\] which sum up to \[a_{98}-a_{1}=97\]
\[\therefore 2S-97+48=137\]
\[\therefore S=93\]

[Tahsin24]

the question says that the summation is 137 for n=1,2,3,...........,97. Not 98