For a positive integer $n$,let $a(n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2^n - 1}$. Then:

The sum of three consecutive terms in a geometric progression is $14$. If $1$ is added to the first and the second terms and $1$ is subtracted from the third,the resulting new terms are in arithmetic progression. Then the lowest of the original terms is

Five numbers are in an $AP$ with a common difference $d \neq 0$. If the $1^{st}$, $3^{rd}$, and $4^{th}$ terms are in a $GP$, then:

Let $a, b, c > 1$. If $a^3, b^3, c^3$ are in $A.P.$ and $\log_a b, \log_c a, \log_b c$ are in $G.P.$,and the sum of the first $20$ terms of an $A.P.$ with first term $\frac{a+4b+c}{3}$ and common difference $\frac{a-8b+c}{10}$ is $-444$,then $abc$ is equal to:

If $a$ is the arithmetic mean of $b$ and $c$ and $G_1, G_2$ are the two geometric means between them,then $G_1^3 + G_2^3 = $

If $t_n$ is the $n^{th}$ term of an arithmetic progression and $t_7 = 9$,what is the value of the common difference $d$ that minimizes the product $t_1 t_2 t_7$?