1) a1+a2+a3=21 ; d-? a1*a2*a3=231 ; a1-? 2) b5-b1=15; b1-? b1-b2=6 ; q-? ...

1)
$\left \{ {{ a_{1}+ a_{2}+ a_{3} =21} \atop a_{1}* a_{2}*a_{3} =231{}} \right. \\ \\$
{a₁+a₁+d+a₁+2d=21
{a₁(a₁+d)(a₁+2d)=231

{3a₁+3d=21
{a₁(a₁+d)(a₁+2d)=231

{a₁=7-d
{(7-d)(7-d+d)(7-d+2d)=231

{a₁=7-d
{(7-d)(7+d)=231:7

{a₁=7-d
{49-d²=33

{a₁=7-d
{d²=16

{a₁=7-4 или {a₁=7+4
{d=4 {d=-4

ответ {a₁=3 или {a₁=11
{d=4 {d=-4

2)
$\left \{ {{ b_{5} - b_{1} =15} \atop { b_{4} - b_{2} =6}} \right. \\ \\ \left \{ {{ b_{1} q^4- b_{1} =15} \atop { b_{1}q^3- b_{1} q =6}} \right. \\ \\ \left \{ {{ b_{1}(q^4-1) =15} \atop { b_{1} q(q^2-1)=6}} \right. \\ \\$

разделим

$\frac{q^2+1}{q} = \frac{15}{6} = \frac{5}{2} \\ \\ 2q^2-5q+2=0 \\ \\ D=25-16=9 \\ \\ 1)q=(5+3)/4=2; b_{1} = \frac{15}{2^4-1} =1 \\ \\ 2)q=(5-3)/4=1/2; b_{1} = \frac{15}{( \frac{1}{2} )^4-1} =-16 \\ \\$

3)
a₄+a₇+a₁₄+a₁₇=18
a₁+3d+a₁+6d+a₁13d+a₁+16d=18
4a₁+38d=18
2a₁=9-19d

19d=9-2a₁
a₂₀=a₁+19d=a₁+9-2a₁=9-a₁

S₂₀=(a₁+a₂₀)/2 *20=(a₁+9-a₁)*10=9*10=90