Step 1: Find the general term of the arithmetic progression $a_n$.

Given an arithmetic progression with the first term $a_1 = 7$ and common difference $d = 8$. The general term $a_n$ is given by the formula $a_n = a_1 + (n-1)d$. Substituting the given values: $a_n = 7 + (n-1)8 = 7 + 8n - 8 = 8n - 1$.

Step 2: Find the general term of the sequence $T_n$.

We are given $T_1 = 3$ and the recurrence relation $T_{n+1} - T_n = a_n$ for $n \geq 1$. We can express $T_n$ in terms of $T_1$ and the sum of terms of $a_k$ by writing out the recurrence for consecutive values: $T_2 - T_1 = a_1$ $T_3 - T_2 = a_2$ $T_4 - T_3 = a_3$ $....$ $T_n - T_{n-1} = a_{n-1}$

Summing these equations, we get a telescoping series on the left side: $(T_n - T_{n-1}) + ... + (T_3 - T_2) + (T_2 - T_1) = a_{n-1} + ... + a_2 + a_1$ $T_n - T_1 = \sum_{k=1}^{n-1} a_k$ $T_n = T_1 + \sum_{k=1}^{n-1} a_k$

Now, we calculate the sum $\sum_{k=1}^{n-1} a_k = \sum_{k=1}^{n-1} (8k-1)$. $\sum_{k=1}^{n-1} (8k-1) = 8 \sum_{k=1}^{n-1} k - \sum_{k=1}^{n-1} 1$ Using the formula for the sum of the first $m$ integers, $\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$, with $m = n-1$: $\sum_{k=1}^{n-1} (8k-1) = 8 \frac{(n-1)(n-1+1)}{2} - (n-1)$ $= 4(n-1)n - (n-1)$ $= (n-1)(4n-1) = 4n^2 - 4n - n + 1 = 4n^2 - 5n + 1$.

Now substitute this sum back into the expression for $T_n$: $T_n = T_1 + (4n^2 - 5n + 1)$ Given $T_1 = 3$, we have: $T_n = 3 + 4n^2 - 5n + 1$ $T_n = 4n^2 - 5n + 4$. This formula is valid for $n \geq 2$. For $n=1$, $T_1 = 4(1)^2 - 5(1) + 4 = 3$, which matches the given condition.

Step 3: Evaluate each option.

Option A: $T_{20}=1604$ Using the formula for $T_n$ with $n=20$: $T_{20} = 4(20)^2 - 5(20) + 4 = 4(400) - 100 + 4 = 1600 - 100 + 4 = 1504$. So, $T_{20} = 1504$, which is not equal to 1604. Thus, option A is FALSE.

Option C: $T_{30}=3454$ Using the formula for $T_n$ with $n=30$: $T_{30} = 4(30)^2 - 5(30) + 4 = 4(900) - 150 + 4 = 3600 - 150 + 4 = 3454$. So, $T_{30} = 3454$. Thus, option C is TRUE.

Option B: $\sum_{k=1}^{20} T_{k}=10510$ We need to compute the sum $\sum_{k=1}^{20} T_k = \sum_{k=1}^{20} (4k^2 - 5k + 4)$. $\sum_{k=1}^{N} T_k = 4 \sum_{k=1}^{N} k^2 - 5 \sum_{k=1}^{N} k + \sum_{k=1}^{N} 4$. We use the standard summation formulas: $\sum_{k=1}^{N} k = \frac{N(N+1)}{2}$ and $\sum_{k=1}^{N} k^2 = \frac{N(N+1)(2N+1)}{6}$. For $N=20$: $\sum_{k=1}^{20} k = \frac{20(21)}{2} = 210$. $\sum_{k=1}^{20} k^2 = \frac{20(21)(41)}{6} = 10 \cdot 7 \cdot 41 = 2870$. $\sum_{k=1}^{20} 4 = 4 \cdot 20 = 80$. Substituting these values: $\sum_{k=1}^{20} T_k = 4(2870) - 5(210) + 80 = 11480 - 1050 + 80 = 10430 + 80 = 10510$. So, $\sum_{k=1}^{20} T_{k}=10510$. Thus, option B is TRUE.

Option D: $\sum_{k=1}^{30} T_{k}=35610$ For $N=30$: $\sum_{k=1}^{30} k = \frac{30(31)}{2} = 15 \cdot 31 = 465$. $\sum_{k=1}^{30} k^2 = \frac{30(31)(61)}{6} = 5 \cdot 31 \cdot 61 = 9455$. $\sum_{k=1}^{30} 4 = 4 \cdot 30 = 120$. Substituting these values: $\sum_{k=1}^{30} T_k = 4(9455) - 5(465) + 120 = 37820 - 2325 + 120 = 35495 + 120 = 35615$. So, $\sum_{k=1}^{30} T_{k}=35615$, which is not equal to 35610. Thus, option D is FALSE.

Conclusion

The correct statements are B and C.