SOLUTION: Could you please explain the procedure in solving this problem: a1=7 an=224 Sn=441 n=?

Question 1090201: Could you please explain the procedure in solving this problem:
a1=7
an=224
Sn=441
n=?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

You haven't stated whether this is an arithmetic sequence or a geometric sequence.

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Let's assume for a moment that the sequence is arithmetic. If that's the case, then we'd use the formula
Sn = (n/2)*(a1+an)

Plug in the given info and solve for n
Sn = (n/2)*(a1+an)
441 = (n/2)*(7+224)
441 = (n/2)*231
441/231 = n/2
n/2 = 441/231
n/2 = 21/11
n = 2*(21/11)
n = 42/11
n = 3.81818181818181 (approximate)

The fact that n is NOT a whole number indicates that the sequence is NOT arithmetic.

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Let's assume that the sequence is geometric instead.

The pattern of geometric terms (a1,a2,a3,...) look like this
a1 = 7
a2 = a1*r = 7*r
a3 = a2*r = (7r)*r = 7r^2 = 7r^(3-1)
a4 = a3*r = (7r^2)*r = 7r^3 = 7r^(4-1)
a5 = a4*r = (7r^3)*r = 7r^4 = 7r^(5-1)
Each term is the result of multiplying the common ratio r by the previous term
Take note how the exponents are formed.

The pattern continues forever. The nth term is
an = a1*r^(n-1)
an = 7*r^(n-1)

Since an = 224, we can say
an = 7*r^(n-1)
224 = 7*r^(n-1)
224/7 = r^(n-1)
32 = r^(n-1)
2^5 = r^(n-1)
2^(6-1) = r^(n-1)

Matching up terms shows that
r = 2 and n = 6

So because n is a whole number, this means we do have a geometric sequence. In this case,

an = 7*r^(n-1)
an = 7*(2)^(n-1)
a1 = 7*2^(1-1) = 7
a2 = 7*2^(2-1) = 14
a3 = 7*2^(3-1) = 28
a4 = 7*2^(4-1) = 56
a5 = 7*2^(5-1) = 112
a6 = 7*2^(6-1) = 224

Adding up the six terms leads to
a1+a2+a3+a4+a5+a6 = 7+14+28+56+112+224 = 441
which confirms the answer.

Or you can use the shortcut formula
Sn = a1*(1-r^n)/(1-r)
Sn = 7*(1-2^6)/(1-2)
Sn = 7*(1-64)/(1-2)
Sn = 7*(-63)/(-1)
Sn = 7*63
Sn = 441
which is another way to confirm the answer.

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