Maths Question:
$\begin{align} & \text{Find the sum of the first }2n\text{ terms of the series } \\ & a+3b+2a+6b+3a+9b+\cdot \cdot \cdot \\\end{align}$
Maths Solution:
$\begin{align} & \text{Sum of }{{S}_{n}}=(a+3b)+(2a+6b)+(3a+9b)+--- \\ & a=(a+3b),\text{ }d=(2a+6b)-(a+3b)=a+3b \\ & {{S}_{n}}=\frac{n}{2}\left[ 2a+(n-1)d \right] \\ & {{S}_{2n}}=\frac{2n}{2}\left[ 2(a+3d)+(2n-1)(a+3d) \right] \\ & {{S}_{2n}}=n\left[ (a+3d)(2+2n-1) \right] \\ & {{S}_{2n}}=n(a+3d)(2n-1) \\\end{align}$
University mathstopic: