There are 20 terms in an arithmetic sequence. The sum of the terms with odd numbers is equal to 220. The sum of the terms with the even numb

There are 20 terms in an arithmetic sequence. The sum of the terms with odd numbers is equal to 220. The sum of the terms with the even numbers is equal to 250. Find the two middle terms of the sequence

Mmm, the 2 middle terms are the 10th and the 11th.

I am going to split ths into 2 sequences.

odd terms

a,   a+2d,   a+4d,      ....    a+(n-1)*2d  ........  a+9*2d  

 sum=10/2(a+a+18d)=5(2a+18d)

$$\\5(2a+18d)=220\\
10(a+9d)=220\\
a+9d=22\qquad (1)\\$$

even terms

a+d,   a+3d,    a+5d,   .....    (a+d)+(n-1)*2d   ......   (a+d)+9*2d  

 sum=10/2(a+d+a+d+18d=5(2a+20d)

$$\\5(2a+20d)=250\\
10(a+10d)=250\\
a+10d=25\qquad (2)\\$$

(2) - (1)

d=3,    substitute     a= -5

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check   the fist 20 terms should add up to 470

$$\\s_{n}=(n/2)[2a+(n-1)d \\\\
s_{20}= 10[-10+19*3] = 10[-10+57]=470 \qquad correct$$

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$$\\t_{10}=-5+9*3 = -5+27 = 22\\\\
t_{11}=22+3=25$$

The 2 middle terms are 22 and 25

Nice one, Melody......!!!    [ I always like to try to figure these out  ]

Thanks Chris :)

First solution: 

arithmetic sequence:  $$\small{\text{$
-5,~-2,~1,~4,~7,~10,~\dots ~,~22,~25,~\dots ~,~46,~49,~52
$}}$$

$$\small{\text{$\mathrm{with~} d=3,\quad t_1= -5 \qquad t_{10} = 22\quad t_{11}=25$}}$$

Second solution:

reverse arithmetic sequence:  $$\small{\text{$ 52,~49,~46,~43,~40,~37,~\dots ~,~25,~22,~\dots ~,~1,~-2,~-5 $}}$$

$$\small{\text{$\mathrm{with~} d=-3,\quad t_1= 52 \qquad t_{10} = 25\quad t_{11}=22$}}$$