The sum and the number of terms in an arithmetic sequence between two numbers

Question

I’m troubled with this question—Given the arithmetic sequence 110,116,122,128, ..., obtain the number of terms between 400 and 600, and their sum.

I’ve been given some instruction, but I can’t wrap my head around it. I can’t figure out how to find the terms which match the domain of where the sequence should lie. I’ve tried converting the sequence into a general term and using systems of equations from there but I can’t necessarily get anything out of it without more values. I’m practically stuck.

All help is very much appreciated—I just don’t have the patience to wrap my head around it lmao 😂. well, thanks for your time—appreciate it!

Answer 1

110, 116, 122, 128,...

128+6n>400

6n>272

n>45.3

n=46

46*6=276

128+276=404

404 is the first number between 400 and 600

404+6n<600

6n<196

n<32.6

n=32

32*6=192

404+192=596

596 is the last number between 400 and 600

there are 33 numbers in the sequence between 400 and 600 (32+1=33; 404 was the first !!)

or 600-400=200, 200/6=33.3, 33 numbers in the sequence between 400 and 600

check:

404,410,416,422,428,434,440,446,452,458,464,470,476,482,488,494,500,506,512,518,524,530,536,542, 548,554,560,566,572,578,584,590,596

now find the sum...

S=k(a₁+a_k)/2

S=33(404+596)/2

S=33(1000)/2

S=33,000/2

S=16,500