In an increasing geometric progression ol positive terms, the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is $49$. Then the sum of the $4^{\text {th }}, 6^{\text {th }}$ and $8^{\text {th }}$ terms is :-

If $2^{10}+2^{9} \cdot 3^{1}+28 \cdot 3^{2}+\ldots+2 \cdot 3^{9}+3^{10}=S -211$ then $S$ is equal to

The ${20^{th}}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + .......$ will be

Consider two G.Ps. $2,2^{2}, 2^{3}, \ldots$ and $4,4^{2}, 4^{3}, \ldots$ of $60$ and $n$ terms respectively. If the geometric mean of all the $60+n$ terms is $(2)^{\frac{225}{8}}$, then $\sum_{ k =1}^{ n } k (n- k )$ is equal to.

If $a, b, c$ and $d$ are in $G.P.$ show that:

$\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$

$0.5737373...... = $