In the arithmetic sequence $\{a_n\}$, it is known that $a_1 + a_2 + a_3 = 6$

Key takeaway: Arithmetic sequences become easy once you write each term in terms of the first term and common difference.

Let the first term be $a_1=a$ and the common difference be $d$.

Then

$a_2=a+d,\quad a_3=a+2d,\quad a_5=a+4d.$

From $a_1+a_2+a_3=6$:

$a+(a+d)+(a+2d)=6$ $3a+3d=6$ $a+d=2.$

Also, $a_5=7$ gives

$a+4d=7.$

Subtract the two equations:

$(a+4d)-(a+d)=7-2$ $3d=5$ $d=\frac53.$

Then

$a=2-d=2-\frac53=\frac13.$

Now find $a_8$:

$a_8=a+7d=\frac13+7\cdot\frac53=\frac13+\frac{35}{3}=12.$

So

$\boxed{12}$

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Pitfalls the pros know 👇 Do not treat $a_1+a_2+a_3=6$ as if it directly gives one term. In an arithmetic sequence, you should first express each term using $a$ and $d$. Also avoid mixing up $a_5$ with $a_8$ when shifting by common differences.

What if the problem changes? If instead you were asked for $a_n$ in general, you could use $a_n=a+(n-1)d=\frac13+\frac{5}{3}(n-1)$. That formula can produce any term once $a$ and $d$ are known.

Tags: arithmetic sequence, common difference, nth term