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?

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 Tₙ₊₁ - Tₙ = aₙ for n ≥ 1. Then, which of the following is/are TRUE?

$a_n = 7 + (n-1)\cdot 8 = 8n - 1$

$T_{n+1} - T_n = a_n = 8n - 1, \qquad T_1 = 3$

$\text{Check: } T_{20},\ \sum_{k=1}^{20} T_k,\ T_{30},\ \sum_{k=1}^{30} T_k$

$T_n = T_1 + \sum_{i=1}^{n-1} a_i = 3 + \sum_{i=1}^{n-1} (8i - 1)$

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

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

$\Rightarrow (1)\ \text{is FALSE}$

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

$= \dfrac{2}{3}n(n+1)(2n+1) - \dfrac{5}{2}n(n+1) + 4n$

$\sum_{k=1}^{20} T_k = \dfrac{2}{3}(20)(21)(41) - \dfrac{5}{2}(20)(21) + 80$

$= 11480 - 1050 + 80 = 10510 \Rightarrow (2)\ \text{is TRUE}$

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

$\Rightarrow (3)\ \text{is TRUE}$

$\sum_{k=1}^{30} T_k = \dfrac{2}{3}(30)(31)(61) - \dfrac{5}{2}(30)(31) + 120$

$= 37820 - 2325 + 120 = 35615 \Rightarrow (4)\ \text{is FALSE}$

$(2)\ \text{and}\ (3)\ \text{are TRUE}$

$(1)\ \text{and}\ (4)\ \text{are FALSE}$