请求推导等比数列题：已知前5项和与第6项首项差求首项及公比

Let's break this down step by step to help you spot any calculation errors you might have made. First, let's define our variables and recall the core formulas for a G.P.:

• Let the first term be (a) (this is what we need to find)
• Let the common ratio be (r) (the other value we need to find)

Key Formulas #

1. Sum of the first (n) terms of a G.P. (when (r \neq 1)):
   [
   S_n = a \cdot \frac{r^n - 1}{r - 1}
   ]
   (Note: If (r = 1), the sum would be (S_n = n \cdot a). We can quickly rule out (r=1) here: if (r=1), the first 5 terms sum to (5a=80), so (a=16), but then the 6th term minus the first term would be (16-16=0 \neq 5). So (r \neq 1).)

2. (n)-th term of a G.P.:
   [
   T_n = a \cdot r^{n-1}
   ]
   So the 6th term (T_6 = a \cdot r^5).

Set Up the Equations #

From the problem statement:

1. Sum of the first 5 terms is 80:
   [
   a \cdot \frac{r^5 - 1}{r - 1} = 80 \tag{1}
   ]
2. The difference between the 6th term and the first term is 5:
   [
   a r^5 - a = 5 \implies a(r^5 - 1) = 5 \tag{2}
   ]

Solve the System of Equations #

Here's the clever substitution that simplifies things: notice that equation (1) has (a(r^5 - 1)) in the numerator, which is exactly the left-hand side of equation (2). Let's substitute (a(r^5 - 1) = 5) into equation (1):

[
\frac{5}{r - 1} = 80
]

Now solve for (r):
[
r - 1 = \frac{5}{80} = \frac{1}{16} \implies r = 1 + \frac{1}{16} = \frac{17}{16}
]

Now plug (r = \frac{17}{16}) back into equation (2) to find (a):
[
a\left( \left(\frac{17}{16}\right)^5 - 1 \right) = 5
]

First calculate (\left(\frac{17}{16}\right)^5):
[
\left(\frac{17}{16}\right)^5 = \frac{17^5}{165} = \frac{1419857}{1048576}
]

Subtract 1:
[
\left(\frac{17}{16}\right)^5 - 1 = \frac{1419857 - 1048576}{1048576} = \frac{371281}{1048576}
]

Now solve for (a):
[
a = 5 \div \frac{371281}{1048576} = 5 \times \frac{1048576}{371281} = \frac{5242880}{371281} \approx 14.12
]

Checking for Alternative Interpretations #

If the problem meant the difference between the first term and the 6th term is 5 (i.e., (a - a r^5 =5)), we'd follow similar steps:

• Equation (2) becomes (a(1 - r^5) =5)
• Substitute into equation (1): (\frac{-5}{r -1} =80 \implies r -1 = -\frac{1}{16} \implies r=\frac{15}{16})
• Then (a = \frac{5}{1 - \left(\frac{15}{16}\right)^5} = \frac{5242880}{289201} \approx 18.13)

But based on the problem's wording ("the difference between the 6th term and the first term is 5"), the first solution ((r=\frac{17}{16}), (a=\frac{5242880}{371281})) is the intended one.

Common Mistakes to Check #

• Mixing up the order of subtraction (6th term minus first vs first minus 6th)
• Errors in simplifying the substitution step (forgetting the denominator in the sum formula)
• Miscalculating (r^5) (it's easy to slip up with exponents of fractions)

内容的提问来源于stack exchange，提问作者Blessing Benjamin