Question

плиз 67 и 68__//////////

Answer 1

$67)1) a_{1}+a_{3}+...+a_{19}=\frac{a_{1} +a_{19} }{2}* 10=\frac{a_{1}+a_{1}+18d}{2} *10=\frac{2a_{1} +18d}{2}*10=(a_{1}+9d)*10\\\\2)a_{2} +a_{4}+...+a_{20}=\frac{a_{20}+a_{20}}{2}*10=\frac{a_{1} +d+a_{1}+19d }{2}*10=\frac{2a_{1}+20d }{2}*10=(a_{1}+10d)*10\\\\3)(a_{1}+9d)*10=(a_{1}+10d)* 10+10\\\\10a_{1}+90d-10a_{1}-100d=10\\\\-10d=10\\\\d=-1$

$68)\frac{a+2a+3a+...+na}{n^{2}-2n-3 }-\frac{3a}{2(n-3)}=\frac{\frac{(a+na)*n}{2} }{(n-3)(n+1)}-\frac{3a}{2(n-3)} =\frac{an(1+n)}{2(n-3)(n+1)}-\frac{3a}{2(n-3)}=\frac{an}{2(n-3)}-\frac{3a}{2(n-3)} =\frac{an-3a}{2(n-3)}=\frac{a(n-3)}{2(n-3)}=\frac{a}{2}$