The sequence $a_1, a_2, a_3, ...$ is such that $a_n = 2a_{n-1} - 40$ for all $n > 1$ . If $a_6 = 8$ , what is the value of $a_2$ ?

Correct answer marked below

Correct

You can see the correct answer above. Sign in for free to unlock the complete worked solution.

Get detailed analytics, unlock full explanations, and move up difficulty tiers as you practice.

Explanation

Create a free account to reveal the correct answer, see the step-by-step explanation, and start tracking your GMAT progress.

Comments

AnusreeJul 21

Thanks, this one clicked pretty fast once I wrote out the recurrence backwards from a6.

VaivaJul 29

Same here, working backward made the numbers way less scary.

VishalAug 4

Yeah, forward substitution also works but takes longer.

RadhikaMay 3

Quick question: when I go from a6 to a5, do I solve 8 = 2a5 - 40? I got a5 = 24, but then I second-guessed myself.

LangMay 13

Yes, that setup is right. Then keep going the same way.

Amisha6d ago

The -40 threw me off at first — kept wanting to add 40 before dividing. Once I fixed the order it was fine.

Jwarkjust now

Classic trap. Isolate the 2a term first, then divide.

TamJun 15

Medium-ish for me. The arithmetic isn't hard, but it's easy to make a silly slip when you chain four steps. Might just plug choices next time.

PrateekJun 23

Plugging choices works here, but backward substitution is cleaner IMO.

What this question tests

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

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

The sequence a1,a2,a3,...a_1, a_2, a_3, ...a1​,a2​,a3​,... is such that an=2an−1−40a_n = 2a_{n-1} - 40an​=2an−1​−40 for all n>1n > 1n>1. If a6=8a_6 = 8a6​=8, what is the value of a2a_2a2​?