Given a sequence of $4$ numbers,the first three of which are in $G.P.$ and the last three are in $A.P.$ with a common difference of $6$. If the first and last terms in this sequence are equal,then the last term is:

Given an $A.P.$ and a $G.P.$ with positive terms, where the first and second terms of both progressions are equal. If $a_n$ and $b_n$ are the $n^{\text{th}}$ terms of the $A.P.$ and $G.P.$ respectively, then:

Let $b_i > 1$ for $i = 1, 2, \ldots, 101$. Suppose $\log _e b_1, \log _e b_2, \ldots, \log _e b_{101}$ are in Arithmetic Progression $(A.P.)$ with the common difference $\log _e 2$. Suppose $a_1, a_2, \ldots, a_{101}$ are in $A.P.$ such that $a_1 = b_1$ and $a_{51} = b_{51}$. If $t = b_1 + b_2 + \cdots + b_{51}$ and $s = a_1 + a_2 + \cdots + a_{51}$,then:

The sum of two numbers is $6$ times their geometric mean. Show that the numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$.

If $a, b, c$ are in $A.P.$ and $a, b, d$ are in $G.P.$,then $a, a - b, d - c$ will be in

If $a, b, c$ are in $A.P.;$ $b, c, d$ are in $G.P.$ and $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in $A.P.,$ prove that $a, c, e$ are in $G.P.$