Determine the sum value of the first terms of that sequence.

Question

Consider a geometric progression in which the common ratio is a non-zero natural number, knowing that the logarithm of the nth term at the base equal to the common ratio of the sequence is equal to 6, that the logarithm of the product of the first n terms at the base equal to that of the sequence is equal to 20 and that the product between the first and the nth term of the sequence is equal to 243, determine the sum value of the first terms of that sequence.

I imagine I have to use the logarithm formulas but I can't equalize them

$ S_n = a_1 \cdot \frac{q^{n}-1}{q-1}\\ P_n = a_1^n . q^{\frac{n(n-1)}{2}}\\ $

Answer 1

Constraints:

• Assume that $a = a_1$, which is the first term in the sequence.

• The sequence is geometric, with common ratio $q \in \Bbb{Z^+}.$

• $\log_q (a_n) = 6.$

• $\log_q (a_1 \times a_2 \times \cdots \times a_n) = 20.$

• $a_1 \times a_n = 243.$

Goal: Compute $(a_1 + a_2 + \cdots + a_n)$.

---

$\displaystyle q^6 = a_n = aq^{(n-1)} \implies a = q^{7-n}.$

$\displaystyle \sum_{k=0}^{n-1} (k) = \frac{(n-1)n}{2}.$

$\displaystyle \prod_{k=1}^n a_k = (a^n) \times q^{[0 + 1 + \cdots + (n-1)]} ~=~ a^n \times q^{\frac{(n-1)n}{2}}.$

Therefore,
$\displaystyle q^{20} = a^n \times q^{\frac{(n-1)n}{2}} ~=~ q^{[n(7-n)]} \times q^{\frac{(n-1)n}{2}} ~=~ q^{[n(7-n)] + \frac{(n-1)n}{2}}.$

Therefore,
$\displaystyle 20 = [n(7-n)] + \frac{(n-1)n}{2} = (n) \left[(7-n) + \frac{(n-1)}{2}\right] \implies $

$\displaystyle 40 = (n) [(14 - 2n) + (n-1)] = (n) (13 - n)] \implies $

$\displaystyle n^2 - 13n + 40 = 0 \implies n = \frac{1}{2}\left[13 \pm\sqrt{9}\right] \implies n \in \{5,8\}.$

Therefore, there are two candidate values for $(n)$. It remains to explore each candidate value, to see which one (if any) generates an integer value for $(q)$. Then, if/when $(n,q,a)$ are successfully determined, the problem's goal can be completed.

$\underline{\text{Case 1:} ~n = 5}$
Since $\displaystyle q^6 = a_n = aq^{(n-1)} = aq^4,~~ a = q^2.$

Therefore
$\displaystyle 3^5 = 243 = (a_1)(a_n) = (q^2)[(q^2)(q^4)] = q^8.$

Since $\displaystyle \sqrt[8]{243}$ is not an integer, and $q$ is required to be an integer, the case of $(n=5)$ must be rejected.

$\underline{\text{Case 2:} ~n = 8}$
Since $\displaystyle q^6 = a_n = aq^{(n-1)} = aq^7,~~ a = q^{(-1)}.$

Therefore
$\displaystyle 3^5 = 243 = (a_1)(a_n) = (q^{(-1)})[q^{(-1)}(q^7)] = q^5.$

Therefore, $\displaystyle 3^5 = q^5 \implies q = 3 \in \Bbb{Z^+}.$

Verifying that all the constraints are satisfied:

• $\displaystyle (q,a,n) = \left(3, 3^{(-1)}, 8\right).$

• $\displaystyle a_n = aq^{(n-1)} = 3^{(-1)} 3^7 = 3^6 \implies a_n = q^6.$

• $\displaystyle (a_1 \times \cdots \times a_n) = 3^{(-8)} 3^{(28)} = 3^{(20)} = q^{(20)} \implies \log_q \{a_1 \times \cdots \times a_n\} = 20.$

• $\displaystyle 3^5 = 243 = \left(3^{(-1)}\right) \times \left[\left(3^{(-1)}\right) 3^7\right] = \left(q^{(-1)}\right) \times \left[\left(q^{(-1)}\right) q^7\right] = a_1 \times a_n$.

Therefore, $\displaystyle (q,a,n) = \left(3, 3^{(-1)}, 8\right)$ is verified.

$\displaystyle \sum_{k=0}^{n-1} q^k = \frac{q^n - 1}{q - 1}.$

Therefore, $\displaystyle \sum_{k=0}^{n-1}a_k = \sum_{k=0}^{7}aq^k = a \times \sum_{k=0}^{7}q^k = 3^{(-1)} \times \sum_{k=0}^{7} 3^k.$

This equals $\displaystyle ~~3^{(-1)}~~\frac{3^8 - 1}{3 - 1} = \frac{1}{3} \times \frac{6560}{2} = \frac{3280}{3}. $