Series question

[cosmic]

Hi guys, I have the following question:

A bank account which pays no interest is set up with an initial deposit of $10000. After a year has passed, $54 is removed from the account. After each subsequent year an amount of money is removed from the account that is $3 more than the amount removed the previous year. Work out how much money is left after 66 year.

The first term is clearly 10000 the second term is 10000-54=9946 the third terms is 9946-57=9889 the fourth term is 9889-60=9829 and so on. I just can't link them together to find the 66th term of the series. Any help here would be hugely appreciated.

 

[cosmic]

Deducted will be the sum of arithmetic series:
1st term = 54
66th term = ?
difference = 3

Thanks Denis.

So am I correct in assuming that the 66th term is 249? And to get the final answer I just need to find the sum of the sequence?

 

[Ishuda]

Thanks Denis.

So am I correct in assuming that the 66th term is 249? And to get the final answer I just need to find the sum of the sequence?

Not Denis but seems reasonable to me. Of course you could just get the the sum of the sequence up to the 65^th term, subtract that to get the ending balance in year 65, then subtract the 249

 

[cosmic]

Not Denis but seems reasonable to me. Of course you could just get the the sum of the sequence up to the 65^th term, subtract that to get the ending balance in year 65, then subtract the 249

Thanks.

I managed to work out the 66th term to be 9999 so I guess there would only be a $1 after 66 years.

 

[cosmic]

66th term is 249; sum of series is 9999; typo?
$1 remaining is CORRECT; good work.

Last term = n
a + d(n - 1) : TATTOO that on your wrist
54 + 3(66 - 1) = 249

54 + 57 + 60 + .... + 246 + 249 = 9999 ; ok?

Yeah, sorry it was a typo. The sum of the series is 9999 which leaves $1 remaining.

Thanks Denis, much appreciated.