Multiple Choice Question (MCQ)

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

  1. ×

    -495

  2. ×

    765

  3. 465

  4. ×

    3150

Collected in the board: Arithmetic sequence

Steven Zheng posted 1 month ago


Answer

  1. \because a_{n+1}-a_n=3

    \therefore \{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}\times 10

    =93\times 5

    =465

Steven Zheng posted 1 month ago

Scroll to Top