$11^2 + 12^2 + 13^2 + \dots + 20^2 = $

Let $a, b \in R$ be such that $a, a + 2b, 2a + b$ are in $A.P.$ and $(b + 1)^2, ab + 5, (a + 1)^2$ are in $G.P.$ then $(a + b)$ equals

Which term of the sequence $(-8 + 18i), (-6 + 15i), (-4 + 12i), ......$ is purely imaginary (in $^{th}$)?

The odd numbers are divided as follows:
Row $1$: $1, 3$
Row $2$: $5, 7, 9, 11$
Row $3$: $13, 15, 17, 19, 21, 23$
Then the sum of the $n^{th}$ row is:

If the $9^{th}$ term of an $A.P.$ is $99$ and $99^{th}$ term is $9,$ find the $108^{th}$ term.

If $a_1, a_2, a_3, ..., a_n$ are in $A.P.$,where $a_i > 0$ for all $i$,then the value of $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ... + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = $