Let a 1 , a 2 , a 3 , … be an arithmetic progression with a 1 = 7 and common difference 8 . Let T 1 ,…

Given,

$a_n=7+\left(n-1\right)8$ and $T_1=3$

Also $T_{n+1}=T_n+a_n$

Now $T_n=T_{n-1}+a_{n-1}$

Putting $n=1$ we get, 

$T_2=T_1+a_1$

Now putting $n=2$ we get,

$T_3=T_2+a_2\Rightarrow T_3=T_1+a_1+a_2$

And so on we get,

$\Rightarrow T_{n+1}=T_1+a_1+a_2+\dots +a_n$

$\Rightarrow T_{n+1}=T_1+\frac{n}{2}\left[2\left(7\right)+\left(n-1\right)8\right]$

$\Rightarrow T_{n+1}=T_1+n\left(4n+3\right) \cdots \left(1\right)$

Now for $n=19$ we get,

$T_{20}=3+\left(19\right)\left(79\right)=1504$

And for $n=29$ we get,

$T_{30}=3+\left(29\right)\left(119\right)=3454$ {option C is correct}

Now finding sum  we get,

$\sum _{k=1}^{20}{T}_k=3+\sum _{k=2}^{20}{T}_k=3+\sum _{k=1}^{19}\left(3+4n^2+3n\right)$

$=3+3\left(19\right)+\frac{3\left(19\right)\left(20\right)}{2}+\frac{4\left(19\right)\left(20\right)\left(39\right)}{6}$

$=3+10507=10510$ {option B is correct}

And Similarly $\sum _{k=1}^{30}{T}_k=3+\sum _{k=1}^{29}\left(4n^2+3n+3\right)=35615$