Question

6) Find the sum of the positive terms of the arithmetic sequence 85, 78, 71, ... 1

Answer 1

Answer:a=first term

d=common difference

here,a=85,d=-7

tn=a+(n-1)d

tn=85+(n-1)-7

tn=85+(7n+7)

92-7n=0

7n=92

n=92/7

n=13.14

n=13

Sn=(n/2)(2a+(n-1)d)

sn=13/2(2*85+(13-1)(-7)

sn=(170+(12*-7)

sn=170-84

sn=13/2*86

sn=559

Step-by-step explanation:looks like it is 7 minus previous number. So, unfortunately 7 does not divide into 85 nicely. 7*12 = 84. So, how every many numbers there are — does it stop at positive numbers only? If not, then it is negative infinity… .

85+78+71+64+57+50+43+36+29+22+15+8+1 = 559

p.s. I tried to be clever with summations (there are 13 numbers) and ran into trouble. Probably the ‘1’ at the end or the 85 at the beginning ... … 7*(summation of 1 to 13) plus an offset to account for the not divisible by 12 exactly. Didn’t work, and it was easier to just do the subtractions and add them together.