a: \(S_n=260\)
=>\(n\cdot\dfrac{\left[2u_1+\left(n-1\right)d\right]}{2}=260\)
=>\(n\cdot\dfrac{2\cdot2+\left(n-1\right)\cdot3}{2}=260\)
=>\(n\left(4+3n-3\right)=520\)
=>\(3n^2+n-520=0\)
=>\(3n^2-39n+40n-520=0\)
=>(n-13)(3n+40)=0
=>\(\left[{}\begin{matrix}n=13\left(nhận\right)\\n=-\dfrac{40}{3}\left(loại\right)\end{matrix}\right.\)
b:
Số số hạng trong dãy là 80-60+1=20+1=21(số)
\(S=u_{60}+u_{61}+...+u_{80}\)
\(=u_1+59d+u_1+60d+...+u_1+79d\)
\(=21u_1+d\left(59+60+...+79\right)\)
\(=21\cdot2+3\cdot\dfrac{\left(79+59\right)\cdot21}{2}\)
=42+4347
=4389