Arithmetic Progression and Term Difference Summation

To find which of the given options are true, let us first find the general term $T_n$ and the sum $\sum_{k=1}^n T_k$.

Step 1: Determine the general term $a_n$

The sequence $a_1, a_2, a_3, \ldots$ is an Arithmetic Progression with first term $a_1 = 7$ and common difference $d = 8$. Therefore, the $n$-th term is:
$a_n = a_1 + (n - 1)d = 7 + (n - 1)8 = 8n - 1$

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Step 2: Determine the general term $T_n$

We are given that $T_1 = 3$ and $T_{n+1} - T_n = a_n$ for $n \ge 1$.

By telescoping the series for $n \ge 2$:
$T_n = T_1 + \sum_{k=1}^{n-1} a_k$

Substituting $a_k = 8k - 1$:

$$\begin{aligned} T_n &= 3 + \sum_{k=1}^{n-1} (8k - 1) \\ &= 3 + 8 \cdot \frac{(n-1)n}{2} - (n-1) \\ &= 3 + 4n(n-1) - (n-1) \\ &= 3 + 4n^2 - 5n + 1 \\ &= 4n^2 - 5n + 4 \end{aligned}$$

This formula holds for all $n \ge 1$ (for $n=1$, $T_1 = 4(1)^2 - 5(1) + 4 = 3$).

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Step 3: Evaluate the options

Check Option (A):

$T_{20} = 4(20)^2 - 5(20) + 4 = 4(400) - 100 + 4 = 1600 - 100 + 4 = 1504 \neq 1604$
Thus, Option (A) is FALSE.

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Check Option (B):

The sum of the first $n$ terms of $T_k$ is given by:
$\sum_{k=1}^n T_k = \sum_{k=1}^n (4k^2 - 5k + 4) = 4\sum_{k=1}^n k^2 - 5\sum_{k=1}^n k + 4\sum_{k=1}^n 1$

For $n = 20$:
$\sum_{k=1}^{20} k^2 = \frac{20 \cdot 21 \cdot 41}{6} = 2870$
$\sum_{k=1}^{20} k = \frac{20 \cdot 21}{2} = 210$
$\sum_{k=1}^{20} 1 = 20$

Therefore:
$\sum_{k=1}^{20} T_k = 4(2870) - 5(210) + 4(20) = 11480 - 1050 + 80 = 10510$
Thus, Option (B) is TRUE.

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Check Option (C):

$T_{30} = 4(30)^2 - 5(30) + 4 = 4(900) - 150 + 4 = 3600 - 150 + 4 = 3454$
Thus, Option (C) is TRUE.

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Check Option (D):

For $n = 30$:
$\sum_{k=1}^{30} k^2 = \frac{30 \cdot 31 \cdot 61}{6} = 9455$
$\sum_{k=1}^{30} k = \frac{30 \cdot 31}{2} = 465$
$\sum_{k=1}^{30} 1 = 30$

Therefore:
$\sum_{k=1}^{30} T_k = 4(9455) - 5(465) + 4(30) = 37820 - 2325 + 120 = 35615 \neq 35610$
Thus, Option (D) is FALSE.

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Conclusion:

The correct options are B and C.