Recover terms from partial sums and compare a geometric tail

$u_n=S_n-S_{n-1}=n^2+6n-[(n-1)^2+6(n-1)]=2n+5$, with $S_0=0$ also giving $u_1=7$. For $n\ge2$, $u_n-u_{n-1}=2$, independent of n, so the terms form an AP.

The odd-index subsequence has $25$ terms, first term $7$ and common difference $4$. Its sum is $25[2(7)+24(4)]/2=1375$.

$u_1=7$, $u_{49}=2(49)+5=103$. Hence the sum is $25(7+103)/2=1375$.

The tail starts at $T_{n+1}=ar^n$, not $T_n$. Therefore $ar^{n-2}=ar^n/(1-r)$.

$T_{n+1}+T_{n+2}+\cdots=S_\infty-S_n=a/(1-r)-a(1-r^n)/(1-r)=ar^n/(1-r)$. Equating this with $ar^{n-2}$ and dividing by the nonzero factor gives $1-r=r^2$, hence $r^2+r-1=0$.

The roots are $(-1\pm\sqrt5)/2$. The negative root is below $-1$, so it violates $|r|<1$. Thus $r=(\sqrt5-1)/2\approx0.618$ (3 s.f.), matching the source. The rendered source does include the modulus bars in its rejection condition.