Cho cấp số cộng $(u_n)$ thỏa mãn $\{\begin{array}{l}{u}_2+u_5=35\\ u_3+u_{10}=53\end{array}$.

a) Sai. $\{\begin{array}{l}{u}_2+u_5=35\\ u_3+u_{10}=53\end{array}\iff \{\begin{array}{l}2u_1+5d=35\\ 2u_1+11d=53\end{array}\iff \{\begin{array}{l}{u}_1=10\\ d=3\end{array}$.
b) Sai. Ta có: $u_{100}=10+99.3=307$.
c) Đúng. $\frac{3}{u_1{u}_2}+\frac{3}{u_2{u}_3}+\frac{3}{u_3{u}_4}+....+\frac{3}{u_{24}.u_{25}}\Rightarrow \frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{79}-\frac{1}{82}=\frac{1}{10}-\frac{1}{82}=\frac{18}{205}$.
d) Đúng. $\{\begin{array}{l}{u}_2+u_5=35\\ u_3+u_{10}=53\end{array}\iff \{\begin{array}{l}2u_1+5d=35\\ 2u_1+11d=53\end{array}\iff \{\begin{array}{l}{u}_1=10\\ d=3\end{array}$.
a) Sai. $\{\begin{array}{l}{u}_2+u_5=35\\ u_3+u_{10}=53\end{array}\iff \{\begin{array}{l}2u_1+5d=35\\ 2u_1+11d=53\end{array}\iff \{\begin{array}{l}{u}_1=10\\ d=3\end{array}$.
b) Sai. Ta có: $u_{100}=10+99.3=307$.
c) Đúng. $\frac{3}{u_1{u}_2}+\frac{3}{u_2{u}_3}+\frac{3}{u_3{u}_4}+....+\frac{3}{u_{24}.u_{25}}\Rightarrow \frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{79}-\frac{1}{82}=\frac{1}{10}-\frac{1}{82}=\frac{18}{205}$.
d) Đúng. $\{\begin{array}{l}{u}_2+u_5=35\\ u_3+u_{10}=53\end{array}\iff \{\begin{array}{l}2u_1+5d=35\\ 2u_1+11d=53\end{array}\iff \{\begin{array}{l}{u}_1=10\\ d=3\end{array}$.