# arithmetic sequence

• May 20th 2008, 04:34 AM
sux@math
arithmetic sequence
Meredith starts a saving fund for her son. She deposits $10 on his first birthday,$20 on his second, $30 on his third and so on....for 40 years (wow, is he still going to be living with her, too? Lol *ahem*....). Use $t_{n}=a+(n+1)d$ to determine how much she will have deposited in total (ignore interest at this point). I don't understand how to plug this into the formula, example: $t_{3}=10+(30+1)(10)$, which equals$320, but logic tells me that $10 +$20 +30, only equals $60. What am I missing? • May 20th 2008, 05:11 AM Soroban Hello, sux@math! Quote: Meredith starts a saving fund for her son. She deposits$10 on his 1st birthday,
$20 on his 2nd,$30 on his 3rd, and so on for 40 years.

Use $t_{n}=a+(n+1)d$ . . . . No!

That's not the formula . . . In fact, it's not any formula!

The sum of the first n terms of an arithmetic series is:
. . $S_n \;=\;\frac{n}{2}\bigg[2a + (n-1)d\bigg]$

Now try it with: . $n = 40,\;a = 10,\;d = 10$

• May 20th 2008, 05:20 AM
sux@math
Hi Soroban!

That first formula, could it be a typo and it should read n-1? That's the closest thing I can find in my book and it is the one on my pre-test sheet that they have given me....

BUT the formula you gave is RIGHT! It works on every term I have tried! THANK YOU!
• May 20th 2008, 05:38 AM
sux@math
And Soroban, can you help with part 2?

If Meredith's first deposit of \$10 earns 5% interest, compounded monthly, what will the final deposit be?

Will it work if I tack on the end of that formula you gave me:

$+0.05^{n}$, where n=40?
• May 20th 2008, 05:41 AM
sux@math
Actually, n=480, correct? Because it will be 12 x 40?