Question

плиз 67 и 68__//////////

Answer 1

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

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