In a sequence $$\{a_n\}$$, if $$a_1=-60$$, $$a_{n+1}=a_n+3$$, then $$|a_1|+|a_2|+\dots+|a_{10}|=$$

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Uniteasy Admin · Accepted Answer

$$\because  a_{n+1}-a_n=3$$
$$	herefore   \{a_n\}$$ is an arithmetic sequence with common difference 
$$d = 3$$
$$|a_1|+|a_2|+\dots+|a_{10}|$$
$$= |a_1| +|a_1+d| + |a_1+9d|$$
$$=60+57+54+51+...+33$$
$$|a_n|$$ is an arithmetic sequence with common difference -3 and initial term 60.
$$|a_1|+|a_2|+\dots+|a_{10}|$$
$$=\dfrac{60+33}{2}	imes 10   $$
$$=93	imes 5  $$
$$=465$$

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