a1+a5= 26
a2*a4=10
a1+a1+4d=26
(a1+d)(a1+3d)=10
a1+2d=13
a1= 13-2d
(13-2d+d)(13-2d+3d)=10
(13-d)(13+d)=10
169-d²=10
d²=159
поскольку прогрессия возрастающая, то d=√159
a1= 13-2d=13-2√159
S6= (2a1+5d)*6/2
S6= (2(13-2√159)+5√159)*3= (26-4√159+5√159)*3= 78+3√159
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Verified answer
a1+a5= 26
a2*a4=10
a1+a1+4d=26
(a1+d)(a1+3d)=10
a1+2d=13
a1= 13-2d
(13-2d+d)(13-2d+3d)=10
(13-d)(13+d)=10
169-d²=10
d²=159
поскольку прогрессия возрастающая, то d=√159
a1= 13-2d=13-2√159
S6= (2a1+5d)*6/2
S6= (2(13-2√159)+5√159)*3= (26-4√159+5√159)*3= 78+3√159