Question

The sum of $10$ terms of the series $2\cdot 5+5\cdot 8+8\cdot 11+\dots$ is

• A. $3610$
• B. $3630$
• C. $3620$
• D. $3600$

Answer 1

The correct option is A $3610$

Let
$S=2\cdot 5+5\cdot 8+8\cdot 11+\dots \dots$
Now,
$2,5,8,\dots$ are in A.P.
$T_n=2+(n-1)\times 3=3n-1$
$5,8,11,\dots$ are in A.P.
$T_n=5+(n-1)\times 3=3n+2$

So, the general term of the series is
$T_r=(3r-1)(3r+2)\Rightarrow T_r=9r^2+3r-2$

Now,
$S_n=\sum _{r=1}^n{T}_r=9\sum _{r=1}^n{r}^2+3\sum _{r=1}^{n}r-\sum _{r=1}^{n}2=9\left(\frac{n(n+1)(2n+1)}{6}\right)+3\left(\frac{n(n+1)}{2}\right)-2n=\frac{3n(n+1)}{2}[2n+1+1]-2n=3n(n+1{)}^2-2n\therefore S_{10}=3610$