In the sequence $a_1, a_2, ..., a_n$ each term after the first term is equal to the preceding term plus a constant $c$ . $a_1 + a_{11} + a_{21} = 99$ . What is the value of $a_3 + a_{19}$ ?

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Comments

PatrickJul 20

Wait, is the sequence an arithmetic progression? The phrasing "each term after the first is equal to the preceding term plus a constant c" sounds like AP, right? Just want to make sure I'm not misreading.

DanielJul 22

Yes, that's exactly an arithmetic sequence with common difference c.

AhmedAug 2

Yep, AP. Common difference c.

AlbertoMay 22

I got stuck at first because I tried to find a1 and c separately, but then I realized a1+a11+a21=3a1+30c=99, so a1+10c=33. Then a3+a19 = 2a1+20c = 2(a1+10c) = 66. Nice little trick, but I definitely didn't see it immediately.

GregoryMay 29

Oh that's clean! I was about to set up two equations, way more work.

JahnviMay 28

Same, took me a minute to notice the substitution. Good catch.

JerryAug 13

Is this 555-605? It felt pretty straightforward once you notice the terms can be expressed in terms of a1 and c.

ChandrashekharSep 4

For anyone new to sequences: try writing everything in terms of a1 and c, or use the middle term idea. Here a11 is the average of a1 and a21, which makes it even faster.

TanyaSep 18

The average trick is what clicked for me. a11 = 33, then a3 and a19 are symmetric around a11, so sum is just 2a11.

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.

In the sequence a1,a2,...,ana_1, a_2, ..., a_na1​,a2​,...,an​ each term after the first term is equal to the preceding term plus a constant ccc. a1+a11+a21=99a_1 + a_{11} + a_{21} = 99a1​+a11​+a21​=99. What is the value of a3+a19a_3 + a_{19}a3​+a19​?