need help-series/sequence ! (1 Viewer)

[dawma88]

a question from maths in focus:

Q: the 20th term of an A.P is 131 & the sum of the 6th to 10th terms inclusive is 235. find the sum of the first 20 terms

Answer: 1290

Q: find the sum of all integers between 1 & 100 that are not mutiples of 6

Answer: 4234

thanx guys

 

[Sober]

We have two simultaneous equations:

(1) T20 = a+19d = 131
a = 131 - 19d

(2) S10 - S5 = 5(2a + 9d) - 2.5(2a+4d) = 235

Substitute (1) into (2):

5(262-19d) - 2.5(262 - 34d) = 235

Solving for d = 7, then a = -2 (from (1))

S20 = 10(2a + 19d) = 1290

-------------------------------

There are floor(100/6) = 16 integers divisible by 6 less than or equal to 100, to find their sum we add the numbers 1 to 16 then multiply the sum by 6:

(1) 6 x [ (16/2) (2+15) ] = 816

Then we find the sum of 1 to 100:

(2) (100/2) * (2 + 99) = 5050

Subtract (1) from (2) = 4234

 

[dawma88]

hey thanx 4 ur help buddy !

 

[Sober]

No problem.

 

[DraconisV]

Isnt series and sequences a 2 unit topic??? why is it posted here guys

 

[Riviet]

Isnt series and sequences a 2 unit topic??? why is it posted here guys

It is a 2 unit topic indeed, not sure why it was posted here though. It doesn't really matter anyway.

 

[Lexie1001]

my 3u textbook smooshes all the 2u and 3u stuff together so you cant tell which is which unless u look at the syllabus or something...maybe thats why.

 

[dawma88]

Isnt series and sequences a 2 unit topic??? why is it posted here guys

farr out man dont u understand ?

i posted in this forum cos of the smart ppl..the 2u ppl cant do it ...thats y u leave it to the 3 unit/4unit ppl

makes sense ??

 

[Riviet]

farr out man dont u understand ?

i posted in this forum cos of the smart ppl..the 2u ppl cant do it ...thats y u leave it to the 3 unit/4unit ppl

makes sense ??

Whoah chill out man, we 3/4 uniters also visit the 2 unit forums, so it doesn't really matter which one you post in.