+470 Let $a_1,$ $a_2,$ $a_3,$ $\dots,$ $a_{10},$ $a_{11},$ $a_{12}$ be an arithmetic sequence. If $a_1 + a_3 = -2$ and $a_2 + a_4 = 5$, then find $a_1$.
+177 We are given that a1+a3=−2 and a2+a4=5. In any arithmetic sequence, the common difference between terms is constant. Let d be the common difference.
We can rewrite the first equation as a1+(a1+d)=−2 which combines the first and third terms. Simplifying the left side gives 2a1+d=−2. Similarly, we can rewrite the second equation as a2+(a2+d)=5 , which combines the second and fourth terms.
Simplifying the left side gives 2a2+d=5.
Since each equation represents the sum of consecutive terms in the arithmetic sequence, both equations must have the same common difference, d.
Subtracting the first equation from the second equation gives us (2a2+d)−(2a1+d)=5−(−2), which combines the information about d from both equations.
Simplifying both sides gives 2a2−2a1=7.
We are solving for a1, so we want to manipulate this equation to get a1 by itself. Noticing that both terms on the left-hand side of the equation have a coefficient of 2, we can rewrite it as 2(a2−a1)=7. Dividing both sides by 2 gives us a2−a1=27.
Now we can use the fact that the arithmetic sequence has a common difference of d. The difference between a2 and a1 is the same as the common difference d. Thus, d=27.
We can now substitute this value for d in our first equation, 2a1+d=−2. Substituting 27 for d gives us 2a1+27=−2. Subtracting 27 from both sides gives us 2a1=−211. Dividing both sides by 2 gives us a1=−411.
