The function $f$ is defined as $f(x+1)=f(x)+5$ . If $f(0)=37$ , what is the sum of $f(3)+f(4)+\cdots+f(10)$ ?

Answer Choices

Correct answer marked below

A. 476

B. 516

C. 556

Correct

D. 596

E. 636

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Comments

MohammadApr 18

Thanks! I got it by realizing each step adds 5, so f(1)=42, f(2)=47, etc. Then sum from f(3) to f(10). A bit tedious but straightforward.

LavanyaApr 30

Same here. Did you list them all or use a formula?

AyushApr 22

I listed them to be safe, but there's probably a faster way.

Malothu5d ago

Wait, why is f(3) not 52? If f(0)=37, then f(1)=42, f(2)=47, f(3)=52. So sum from 52 to 87 step 5. That's 8 terms, average (52+87)/2=69.5, times 8 = 556. Is that right?

Kshitiz1d ago

Yes, that's what I got too.

AvisekJun 18

This was medium difficulty. The function notation threw me off at first, but once I saw it's just an arithmetic sequence, it was fine. I used the sum formula for arithmetic series.

AshleyJun 26

What formula did you use?

RockstarJun 21

n/2 * (first + last) where n is number of terms.

What this question tests

What it tests

Your ability to evaluate and compose functions, including unusual defined operations.

Common trap

Applying function operations in the wrong order or ignoring domain restrictions.

The function fff is defined as f(x+1)=f(x)+5f(x+1)=f(x)+5f(x+1)=f(x)+5. If f(0)=37f(0)=37f(0)=37, what is the sum of f(3)+f(4)+⋯+f(10)f(3)+f(4)+\cdots+f(10)f(3)+f(4)+⋯+f(10)?