MGMAT EIV - Chapter 4 Sequence Problems Method

[Rosy]

Hi,

On page 60 in the Equations, Inequalities, and VIC's book there is an alternative method explained to solve sequence problems.

Here is the problem:
If each number in a sequence is three more than the previous number, and the sixth number is 32, what is the 100th number?

Answer: From the sixth to the one hundredth there are 94 jumps of 3.
94 x3 = 282.
32 + 282 = 314.

To get the 94 jumps I'm guessing you subtract 100-6 = 94.
I thought we were supposed to add 1 to make it inclusive of the 100th term. Can you explain why we don't add 1 in this case?

Thanks!

Rosy

[esledge]

Great question!

Answer: From the sixth to the one hundredth there are 94 jumps of 3.
94 x3 = 282.
32 + 282 = 314.

To get the 94 jumps I'm guessing you subtract 100-6 = 94.

This method counts the jump between terms, so you can just subtract the starting term from the ending term.

The add-one-before-you're-done rule applies when you are counting the terms themselves. If you subtract the starting term from the ending term, you would fail to count that starting term, so you add 1 to account for it.

A real life example: Imagine you are climbing stairs, and each step is numbered with a consecutive integer. You stand on step 1 to begin, and will climb to step 10.

How many steps up will you take? (i.e. how many "jumps" between terms?) 10-1 = 9.

How many steps did you touch? (i.e. how many terms in the sequence?) 10-1+1 = 10. This makes sense, as you touched steps 1 and 10 (and all in between), so 10 steps total.

[Rosy]

Thank you Emily! It makes sense now. Great example with the steps. =)

[JonathanSchneider]

: )