Arithmetic Progression Sum

Let first term be $a$ and common difference be $d$. $T_4 + T_7 + T_{10} = (a+3d) + (a+6d) + (a+9d) = 3a + 18d = 99 \implies a + 6d = 33$. $S_{14} = \frac{14}{2}(2a + 13d) = 7(2a + 13d) = 497 \implies 2a + 13d = 71$. Solving these equations: From $a = 33 - 6d$, substitute into second: $2(33 - 6d) + 13d = 71 \implies 66 + d = 71 \implies d = 5$. Then $a = 33 - 30 = 3$. Sum of first 5 terms $S_5 = 3 + 8 + 13 + 18 + 23 = 65$.