Let $$a_1, a_2, a_3, \ldots$$ be an arithmetic progression with $$a_1 = 7$$ and common difference 8. Let $$T_1, T_2, T_3, \ldots$$ be such that $$T_1 = 3$$ and $$T_{n+1} - T_n = a_n$$ for $$n \geq 1$$. Then, which of the following is/are TRUE?

The arithmetic progression $$a_1,a_2,a_3,\ldots$$ has first term $$a_1 = 7$$ and common difference $$8$$, therefore

$$a_n \;=\; 7 + 8(n-1) \;=\; 8n - 1\;,\; n \ge 1$$

The sequence $$T_1,T_2,T_3,\ldots$$ satisfies $$T_1 = 3$$ and $$T_{n+1}-T_n = a_n$$. Add the first $$n-1$$ relations of this form:

$$T_n - T_1 = \sum_{k=1}^{\,n-1} a_k$$

Hence

$$T_n = 3 + \sum_{k=1}^{\,n-1} (8k-1)$$

First find the partial sum of $$a_k$$. For any positive integer $$m$$,

$$\sum_{k=1}^{\,m} (8k-1) = 8\sum_{k=1}^{\,m} k - \sum_{k=1}^{\,m} 1 = 8\cdot\frac{m(m+1)}{2} - m = 4m(m+1) - m = 4m^{2}+3m$$

Put $$m = n-1$$:

$$\sum_{k=1}^{\,n-1} a_k = 4(n-1)^2 + 3(n-1) = 4n^{2}-8n+4 + 3n-3 = 4n^{2} - 5n + 1$$

Therefore

$$T_n = 3 + (4n^{2} - 5n + 1) = 4n^{2} - 5n + 4$$

Check each option
Option A: $$T_{20} = 4(20)^2 - 5(20) + 4 = 1600 - 100 + 4 = 1504 \neq 1604$$ (False).
Option C: $$T_{30} = 4(30)^2 - 5(30) + 4 = 3600 - 150 + 4 = 3454$$ (True).

To evaluate the required sums, note that

$$T_k = 4k^{2} - 5k + 4$$

so

$$\sum_{k=1}^{\,n} T_k = 4\sum_{k=1}^{\,n} k^{2} - 5\sum_{k=1}^{\,n} k + 4\sum_{k=1}^{\,n} 1$$

Use the standard formulas $$\sum k = \frac{n(n+1)}{2}$$ and $$\sum k^{2} = \frac{n(n+1)(2n+1)}{6}$$:

$$\sum_{k=1}^{\,n} T_k = 4\cdot\frac{n(n+1)(2n+1)}{6} - 5\cdot\frac{n(n+1)}{2} + 4n$$

Simplify:

$$\sum_{k=1}^{\,n} T_k = \frac{2n(n+1)(2n+1)}{3} - \frac{5n(n+1)}{2} + 4n$$

Sum up to 20 terms
For $$n = 20$$:

First term: $$\frac{2\cdot20\cdot21\cdot41}{3} = 11480$$
Second term: $$\frac{5\cdot20\cdot21}{2} = 1050$$
Third term: $$4\cdot20 = 80$$

$$\sum_{k=1}^{20} T_k = 11480 - 1050 + 80 = 10510$$

Option B is therefore True.

Sum up to 30 terms
For $$n = 30$$:

First term: $$\frac{2\cdot30\cdot31\cdot61}{3} = 37820$$
Second term: $$\frac{5\cdot30\cdot31}{2} = 2325$$
Third term: $$4\cdot30 = 120$$

$$\sum_{k=1}^{30} T_k = 37820 - 2325 + 120 = 35615$$

Thus Option D (which states 35610) is False.

Hence the correct statements are:
Option B $$\left(\sum_{k=1}^{20} T_k = 10510\right)$$ and
Option C $$\left(T_{30} = 3454\right)$$.