Определите число членов и сумму четырех последних членов арифм. прогрессии [latex]a_{n}[/latex], в которой [latex]a_{1}[/latex]=10, d=2.5, [latex]a_{n}[/latex]=27,5Сумма члена [latex] a_{3} [/latex] и [latex] a_{9} [/latex]в ар...

a1=10  d=2,5  an=27,5 10+2,5(n-1)=27,5 2,5(n-1)=27,5-10=17,5 n-1=17,5:2,5=7 n=7+1=8 S=S(8)-S(4)=150-55=95 S(8)=(2*10+2,5*7)*8/2=(20+17,5)*4=37,5*4=150 S(4)=(2*10+2,5*3)*4/2=(20+7,5)*2=27,5*2=55 a3+a9=8 a1+2d+a1+8d=2a1+10d=8 S(11)=(2a1+8d)*11/2=8*11/2=44

[latex]1) \ a_1 = 10, \ d = 2.5, \ a_n = a_1 + (n-1)d\\\\ 27.5 = 10 + (n - 1)2.5\\\\ 17.5 = (n - 1)2.5 \ | \ : \ 2.5\\\\ 7 = n - 1\\\\ \boxed{n = 8}[/latex] [latex] a_5+a_6+a_7+a_8 = 4a_1 + (4+5+6+7)d = 4a_1 + 22d =\\\\ = 40 + 55 = \boxed{95}[/latex] [latex]2) \ a_3 + a_9 = 8\\\\ a_3 = a_1 + 2d, \ a_9 = a_1 + 8d\\\\ 2a_1 + 10d = 8\\\\ S_{11} = a_1 + a_2 + a_3 + ... + a_{11} = a_1 + a_1 + d + ... + a_1 + 10d =\\\\= (a_1 + a_1 + 10d) + ... + (a_1 + 4d + a_1 + 6d) + (a_1 + 5d) =\\\\= 5*(2a_1 + 10d) + \frac{1}{2}(2a_1 + 10d)= 40 + 4 = \boxed{44}[/latex]