S15=20; S20=15; S35-?
2a1+14d
S15 = ----------- • 15 = (a1+7d)•15
2
2a1+19d
S20 = ------------ • 20 = (a1+9,5d)•20
2
{(a1+7d)•15=20
{(a1+9,5d)•20=15
{15a1+105d=20 |:5
{20a1+190d=15 |:5
{3a1+21d=4 |•4
{4a1+38d=3 |•(-3)
{12a1+84d=16
{–12a1–114d=–9
–30d=5
d=–1/6
3a1+21•(-1/6)=4
3a1=7,5
a1=2,5
2a1+34d
S35 = ------------ • 35 = (a1+17d)•35 =
2
= (2,5+17•(-1/6))•35 = (5/2–17/6)•35 =
= (-1•35)/3 = –35/3 = –11 2/3