The sequence $a_1, a_2, a_3, ..., a_n$ is defined such that $a_n=a_{n-1}+9+n$ for all $n>1$ . If $a_1=10$ , what is the value of $a_{11}$ ?

Answer Choices

Correct answer marked below

150

155

160

165

Correct

170

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Comments

YariandSep 7

Is there a faster way to do this? I just listed out a2 through a11, but that took a while.

YatharthSep 20

You can group the +9 and the +n separately. The 9 adds up quickly across 10 steps, and the n part is 2+3+...+11.

AdityaSep 19

Yeah, telescoping is the move here. Rewrite a11 as a1 plus the sum of all the increments.

JohnAug 27

Wait, does the +n part start at n=2? I keep second-guessing whether the first increment uses 2 or 1.

JacobSep 3

Yes, n>1, so the first step from a1 to a2 uses n=2.

VageeshaSep 10

Got 165 after summing the increments, pretty straightforward once you set it up right.

RaymondSep 12

Same here. The 9s alone give 90, then the triangular part is the rest.

AnjelinaSep 23

Nice, that matches what I got by plugging into the formula.

ChiragSep 24

This one felt medium. The tricky part is not losing track of which n corresponds to which step.

What this question tests

What it tests

Your ability to spot and extend arithmetic and geometric sequences and work with recursive definitions.

Common trap

Off-by-one errors in indexing (a₀ vs a₁) or confusing an arithmetic difference with a geometric ratio.

The sequence a1,a2,a3,...,ana_1, a_2, a_3, ..., a_na1​,a2​,a3​,...,an​ is defined such that an=an−1+9+na_n=a_{n-1}+9+nan​=an−1​+9+n for all n>1n>1n>1. If a1=10a_1=10a1​=10, what is the value of a11a_{11}a11​?