SOLUTION: Each week for 52 weeks, an amount is deposited into an account ($1 first week, $2 second week, $3 third week, etc.) Determine balance at end of the 52 weeks and interest (not in

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Question 1205476: Each week for 52 weeks, an amount is deposited into an account ($1 first week, $2 second week, $3 third week, etc.)
Determine balance at end of the 52 weeks and interest (not indicated if interest is calculated daily, weekly, monthly, etc.)
Not sure how to solve.
Called The "Money Challenge".

Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13216) About Me  (Show Source):
You can put this solution on YOUR website!


The statement of the problem gives no information about interest rates, so we can only determine the total amount of the deposits.

The amounts form an arithmetic progression: 1, 2, 3, 4, ..., 52.

The sum of the terms of an arithmetic progression is

(number of terms) times (average of terms)

In an arithmetic progression, the average of all the terms is just the average of the first and last terms, so...

number of terms: 52
average of terms (1+52)/2

The total dollar amount deposited is

52%28%281%2B52%29%2F2%29=26%2A53=1378

ANSWERS:
deposited: $1378
interest: ??

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Formula for a spreadsheet for the sum of the integers from 1 to N:

N%28N%2B1%29%2F2

If you are using excel, note that it doesn't recognize "N(N+1)/2" -- you need to enter the formula as "N*(N+1)/2".


Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

The interest cannot be calculated since the interest rate isn't stated.

But we can determine the sum 1+2+3+...+50+51+52.

There are n = 52 values being added up.
Because n is even, we can form n/2 = 52/2 = 26 pairs.

1st pairs with last
2nd pairs with 2nd to last
3rd pairs with 3rd to last
and so on

Add up each pair.
1+52 = 53
2+51 = 53
3+50 = 53
...
25+28 = 53
26+27 = 53
Each pair sums to 53.

We have 26 copies of 53 added together, so the grand total sum is 26*53 = 1378

Therefore,
1+2+3+...+50+51+52 = 1378

An alternative is to use the formula that tutor greenestamps mentioned.