Sequence Within A Sequence 2

We have $\begin{aligned} T_1&=T_1 \\ T_2&=T_1+d_1 \\ T_3&=T_1+d_1+d_2 \\ &\vdots \\ T_n&=T_1+d_1+d_2+\ldots+d_{n-1} \end{aligned}$ Add all these equations together, we have $S_n=nT_1+\displaystyle \sum_{i=1}^{n-1}{\sum_{j=1}^i {d_j}}$ From the question we know that $d_1, d_2, \ldots, d_{n-1}$ is an A.P. (Arithmetic Progression), suppose $d_k-d_{k-1}=D, ~~k=2,3,\ldots,n-1$ , applying the sum of A.P. formula, we have $\sum_{j=1}^i {d_j}=\frac {i}{2}[2d_1+(i-1)D]$ Hence, $\begin{aligned} \sum_{i=1}^{n-1}{\sum_{j=1}^i {d_j}}&=\sum_{i=1}^{n-1}{\frac {i[2d_1+(i-1)D]}{2}} \\ &=\frac {1}{2}\sum_{i=1}^{n-1}{[(2d_1-D)i+Di^2]} \\ &=\frac{1}{2}\left [\frac {(2d_1-D)n (n-1)}{2}+\frac {Dn (n-1)(2n-1)}{6}\right] \\ &=\frac {n (n-1)}{12}[6d_1-3D+2nD-D] \\ &=\frac {n (n-1)[3d_1+(n-2)D]}{6} \end{aligned}$ $\therefore S_n=nT_1+\frac {n (n-1)[3d_1+(n-2)D]}{6}$ Now, substitute $n=153$ , $T_1=5$ , $d_1=3$ , $D=1$ , we could get $S_{153}=620925$ .