How would I find the sum of the first 25 terms for the sequence an=3n+13
\[(3(1)+13)+(3(2)+13)+(3(3)+13)+\cdots+(3(25)+13)\\~\\=[3(1)+3(2)+3(3)+\cdots3(25)] +[13\cdot25]\\~\\=3(1+2+3+4+\cdots+25)+13\cdot25\]
That would find my 25th term, I have my first 5 terms from the 1st part of the question, and I have the two formulas I need to figure this out... sn=n/2[2a1+(n-1)d] and an-a1/n1=D I just don't know what to plug in where...
I would calculate the first term, assuming n=1, a1=3(3)+13, term 2 I calculate as a2=3(2)+13, a3 = 3(3)+13,..... Next, I would check if there exist a common difference between the terms or whether a common ratio? If the former, then I say its an arithmetic progression (AP) else a geometric progression (GP). Then I use either the AP formula for first 25 terms as S25 = n/2(2*a1 + (n-1)*d), where d = common difference for an AP. U can do the exhaustive summation by calculating a1, a2, a3, ....a25 which is time consuming!
@StudyGurl14 ^ @geerky42 I have to show my work, can yhu please tell me the formula yhu used?
I didn't use formula. I just applied commutative property of addition ( 1+2 = 2+1 ) and distributive property ( 2(a+b)=2a+2b ).
why would you apply commutative and distributive property? U need to find the sum of n=25 terms quickly using a formula or exhaustive means. I wonder why u need commutativity or distributivity?
Once you find the sum of first 25 term of \(a_n\), you will end up adding 13 25 times, hence \(13\cdot25\), and I factored out 3, from \(3(1)+3(2)+3(3)+\cdots3(25) = 3(1+2+3+\cdots+25)\)
The first 5 terms are 16, 19, 22, 25, and 28. They all go up by 3, so would 3 be D?
@da_ScienceMan Problem didn't say we should "quickly using a formula or exhaustive means"
Good @Ria23 it means d common difference is d = 3. a = 16 and the S25 = 25/2(2*16+(25-1)*3) and check if its not same with summing individual terms?
@geerky42 I see what yhu did. c: I understand it. My teacher wants the formulas she taught tho. :c I understand how yhu got that tho. c: And thank yhu~ @da_ScienceMan so, I would complete that formula and that would give me the sum of the first 25 terms?
@Ria23 yes hopefully. It will be correct if you guys have done stuffs related to arithmetic progression else I would be blabbing! Please check and understand it too.
I got S25= 1300 I'm not really sure how I would check that...
@da_ScienceMan
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