Question

В прогрессии S3= 3, S5= - 5. Найдите сумму первых семи членов прогрессии.

Answer 1

$S_{3}=3\\\\ a_{1}+ a_{2}+ a_{3}=3\\\\ a_{1}+ a_{1}+d+ a_{1}+2d=3\\\\3 a_{1}+3d=3\\\\ a_{1} +d=1\\\\ a_{1} =1-d\\\\\\ S_{5}= \frac{2 a_{1} +4d}{2} *5\\\\ S_{5} =( a_{1}+2d)*5\\\\(1-d+2d)*5=-5\\\\1+d=-1\\\\d=-2\\\\ a_{1}=1-(-2)=3\\\\ S_{7}= \frac{2 a_{1}+6d }{2} *7=( a_{1}+3d)*7=(3+3*(-2))*7=(3-6)*7=$$=-3*7=-21$