The arithmetic progression (an) is set by a formula an=10-4n a) make a formula for calculation of the sum of the first n of members of this progression b) using this formula find the sum of the first thirty members of this progression c) how many members of this progression starting with the first put if in the sum-120 turned out?

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an arithmetic progression (an) is set by a formula an=10-4n a) make a formula for calculation of the sum of the first n of members of this progression of Sn=(a1+an) · n/2, a1=10-4=6 Sn=(6+10-4n) · n/2= (8-2) n b) using this formula find the sum of the first thirty members of this progression S13 == (how many members of this progression starting with the first put 8-2·13)13=-18·13=-234 c) if in the sum-120 ngt turned out; 0 Sn= (8-2) · n=-120 ⇔ 2n²-8n-120=0 ⇔ n²-4n-60=0 ⇔ n1=10 n2=-6 n1=10
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Answer add