The sum of the first five terms of the series $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots$ will be

The difference between the fourth term and the first term of a Geometric Progression is $52$. If the sum of its first three terms is $26$,then the sum of the first six terms of the progression is

If $a, b, c$ and $d$ are in $G.P.$,show that:
$(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2})=(ab+bc+cd)^{2}$

$\frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^2}{3^{23}} + \ldots + \frac{10 \cdot 2^{24}}{3}$ is equal to

In a $GP$ series consisting of positive terms,each term is equal to the sum of the next two terms. Then,the common ratio of this $GP$ series is

Let $x_1, x_2, \ldots, x_6$ be the roots of the polynomial equation $x^6+2 x^5+4 x^4+8 x^3+16 x^2+32 x+64=0$. Then,