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You save $20 the first month, $24 the second month, $28 the third month and so on. How long will it take you to save a total of $800?
How would you solve this using the Sn equation?
I know that Sn=n/2(t1+tn)
so 800=n/2(20+tn)
What would tn be?
Thanks!
Njuice8
be more careful when you interptert formulas in math...The problem you mentioned above is a typical problem of summing an arithmetic sequence with first term a=20 and ratio t=4 the formula that gives the sum of n terms is (n/2)[2a+(n-1)t] and not the one you wrote ..equate this to 800 and solve a quadratic for n to find n =16 .
Good Luck
MINOAS
I would suggest using common difference for common ratio. In AP we use common difference and for GP we use common ratio.
As Minoan Man said, that formula requires knowing tn, the last deposit, which requires knowing n, whicy is what you want to find!
In general, for an "arithmetic series", with first value a and "common difference" (not "common ratio") we have a+ (a+ d)+ (a+ 2d)+ (a+ 3d)+... + (a+ nd) That is, we are adding a n+1 times which sums to (n+ 1)a. And we can write d+ 2d+ 3d+ ...+ nd as d(1+ 2+ 3+ ...+ n).
There is a formula for 1+ 2+ 3+... + n, but a cute trick gives it very easily:
1+ 2+ 3+ ...+ n
n+ n-1+(n-2)+...+1
where I have reversed the sum in the second line. Now add "vertically": 1+ n= n+1. 2+ (n-1)= n+1, 3+ (n-2)= n+1, ..., n+1. That is, every such sum is n+ 1 and there are n of them so the total of those two lines is n(n+1). But because we have summed twice, we really have that 1+ 2+ 3+ ...+ n= n(n+1)/2.
So d+ 2d+ 3d+ ...+ nd= nd(n+1)/2.
Then we have a+ (a+ d)+ (a+ 2d)+ ...+ (a+ nd)= (n+1)a+ nd(n+1)/2.
Now, in your problem, you started by depositing $20 so a= 20. And each month after that is $4 more than the previous month so d= 4. You want that to sum to $800 so you want to solve (n+1)(20)+ 4n(n+1)/2= 800. Solve that for n. I don't think that has integer solutions so the first integer larger than the positive root will give slightly more than $800.