Let a₁, a₂, a₃,… be an arithmetic progression with a₁ = 7 and common difference 8. Let T₁, T₂, T₃,… be such that T₁ = 3 and T_n+1 – T_n = a_n for n ≥ 1. Then, which of the following is/are TRUE ?

The correct options are:
$\sum _{k=1}^{20}{T}_k=10510$ and T₃₀ = 3454

Step 1: Understand the given sequences and recurrence relationship.
We are given an arithmetic progression (A.P.) defined by:
First term, a₁ = 7
Common difference, d = 8
The n-th term of this arithmetic progression, a_n, is given by the formula:

$a_n=a_1+(n-1)d=7+(n-1)\times 8=8n-1$

We are also given a sequence T_n with T₁ = 3 and the relation:

$T_{n+1}-T_n=a_n=8n-1$

for n ≥ 1.

Step 2: Find the general term T_n of the sequence.
We can express T_n as a telescoping sum:

$T_n=T_1+\sum _{k=1}^{n-1}(T_{k+1}-T_k)$

Substituting T₁ = 3 and T_k+1 - T_k = a_k = 8k - 1:

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

Splitting the summation:

$T_n=3+8\sum _{k=1}^{n-1}k-\sum _{k=1}^{n-1}1$

Using standard summation formulas:

$T_n=3+8\times \frac{(n-1)n}{2}-(n-1)$

Simplifying the terms:

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

Thus, the general term is T_n = 4n² - 5n + 4.

Step 3: Evaluate T₂₀ and T₃₀.
For n = 20:

$T_{20}=4(20{)}^2-5(20)+4=4(400)-100+4=1600-100+4=1504$

(So, T₂₀ = 1604 is FALSE).

For n = 30:

$T_{30}=4(30{)}^2-5(30)+4=4(900)-150+4=3600-150+4=3454$

(Thus, T₃₀ = 3454 is TRUE).

Step 4: Find the sum of the first N terms, S_N = $\sum _{k=1}^N{T}_k$.
Using the formula for T_k:

$S_N=\sum _{k=1}^N(4k^2-5k+4)=4\sum _{k=1}^N{k}^2$

Substituting standard summation formulas:

$S_N=4\times \frac{N(N+1)(2N+1)}{6}-5\times \frac{N(N+1)}{2}+4N$

Now, calculate S₂₀ for N = 20:

$S_{20}=4\times \frac{20\times 21\times 41}{6}-5\times \frac{20\times 21}{2}+4\left(20\right)$

Simplifying each term:
Term 1: $4\times 2870=11480$
Term 2: $5\times 210=1050$
Term 3: $4\times 20=80$
Combining the values:

$S_{20}=11480-1050+80=10510$

(Thus, $\sum _{k=1}^{20}{T}_k=10510$ is TRUE).

Conclusion:
The correct statements are:
1. $\sum _{k=1}^{20}{T}_k=10510$
2. T₃₀ = 3454