Partial Sums of an Ap Series Help! Harder questions.. (1 Viewer)

[sloveni]

Hi people, hope you are able to help me out. It will be much appriciated thank you. .
Hope its practice
13. The sum of 12 terms of an
arithmetic series is 186 and the
20th term is 83. Find the sum
of 40 terms.

14. How many terms of the series
2 0 + 18 + 16 +… give a sum of
104?
15. The sum of the fi rst 4 terms of
an arithmetic series is 42 and
the sum of the 3rd and 7th term
is 46. Find the sum of the fi rst
20 terms.

16. (a) Show that
( x + 1) + (2x + 4) + (3x + 7) +… is
an arithmetic series.
(b) Find the sum of the fi rst 50
terms of the series.

17. The 20th term of an arithmetic
series is 131 and the sum of the
6th to 10th terms inclusive is 235.
Find the sum of the fi rst 20 terms.

18. The sum of 50 terms of an
arithmetic series is 249 and the
sum of 49 terms of the series is
233. Find the 50th term of the
series.
20. Find the sum of all integers
between 1 and 100 that are not
multiples of 6.

Answers
13. 3420
14. 8 and 13 terms 15. 1010
16. (a) =x +3
(b) 25(51x + 149)

17. 1290 18. 16

20. 4234

 

[Timothy.Siu]

I don't think theres much of a point of me doing all of these questions, since then you won't be learning much.

I think the thing to have in mind, is that for an arithmetic series (or progression), each term differs to the previous (or next) one by a fixed constant. Also you will need to know the formula for the sum of an arithmetic series to do these questions. So if the first term is "a" the next term will be "a+d" where d is the common difference, and so, we can write for example, the 13th term as "a+12d" which may help for some of these questions. Personally, for sums i use the formula S=n(Ti+Tf)/2 which is the initial term plus the final term divided by two (so, the average) multiplied by the number of terms.

13. 12(T1+T12)/2=186
well, if we let Tn=a+(n-1)d where n is the nth term (note: you can really set it as anything u want, e.g. Tn=a+nd)
Then, 12(a+a+11d)/2=186 (note: there are a few different formulas u can use for arith. sum, any if fine)
so, 6(2a+11d)=186
2a+11d=31
also, T20=83=a+19d
So from this, we can solve simultaenously for a and d, giving, d=5 and a=-12
we want to find S40=40(T1+T40)/2
S40=20(a+a+39d)=20(2a+39d)=20(-24+195)=3420

again, there is more than one formula for arith. sum, and setting what Tn is equal to, but all ways will get the same answer. This is just one way.

I won't do any more of these, unless you really need me to, after trying.
hints: always write out the information given to you in some sort of equation, and you will usually need both the arithmetic sum equation and the general term equation.
16. if u can't see how to sum it up arithmetically, try splitting the term into two
20. There are two ways to do this one, either sum up all the integers from 1 to 100, and subtract from this, the sum of the multiple of 6 integers, or the other (slower but more obvious) way would be to sum up 5 arith. series with common different 6, starting at 1,2,3,4,5.

Hope this helps. Any questions, feel free to ask.

 

[sloveni]

Thanks a lot mate

 

[random-1005]

they are way to easy, all it reduces too is solving simulataneous eqns

 

[bouncing]

14) a = 20, d = 18-20=-2
the formula for a sum of an arithmetic series is:
Sn=n/2[2a+(n-1)d]
in this case here Sn =104
104=n/2[40+(n-1)(-2)]
=n/2[40-2n+2]
208=n[42-2n]
208=42n-2n^2
rearranging this to form a quadratic equation we get:
2n^2-42n+208=0
n^2-21n+104=0
(n-8)(n-13)=0
.'. n=8 and n=13
.'. 8 and 13 terms

15) T3+T7=46
[a+2d]+[a+6d]=46
2a+8d=46
a+4d=23
.'.a=23-4d ---- [1]

S4=42
42=4/2[2a+(3)d]
substituting [1] into this equation we get:
42=2 [2(23-4d)+3d]
42=2[46-8d+3d]
21=46-5d
5d=25
.'. d=5

substitute d=5 into [1] to find a:
a=23-4(5)
=23-20
.'. a=3

now to find the sum of the first 20 terms we use the same formula from part 14)
Sn=n/2[2a+(n-1)d]
S20=20/2[6+19(5)]
=10[6+95]
=10(101)
=1010

16)a) For a series to be arithmetic T3-T2=T2-T1
knowing this you can easily use the given terms to formulate an equation and then solving for x

T1=(x+1)
T2= (2x+4)
T2=(3x+7)
and just sub into the above

b) since you'll have x ---> just sub the values and then you'll be able to find your a and d