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Let $a_1,$ $a_2,$ $a_3,$ $\dots,$ $a_{10},$ $a_{11},$ $a_{12}$ be an arithmetic sequence. If $a_1 + a_3 + a_5 = 22$ and $a_2 + a_4 = 17$, then find $a_1$.

 Jun 16, 2024
 #1
avatar+129733 
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a1 + a3 + a5 =  22

a1 + (a1 + d) + (a1 + 2d)   =22

3a1 + 3d =  22   

 

a2 + a4 =  17

(a1 + d) + (a1 + 3d)  = 17

2a1 + 4d = 17

 

3a1 + 3d  = 22   mult through by 4  →   12a1 + 12d  =  88

2a1 + 4d =  17   mult through by -3 →     -6a1 - 12d  = -51       add these

 

6a1  =  37

 

a1  = 37 / 6

 

cool cool cool

 Jun 16, 2024

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