My Series and Sequences Questions (1 Viewer)

[velox]

Wrong answers for most of these...

8. Find the 7th term of the arithmetic progression whose 5th term is m and whose 11th term is n.

16. Find the sum of all integers between 20 and 50 that are divisible by 3.

17. A body, falling freely from a height, travels 4.9 metres in the first second, 14.7m in the second second and 24.5m in the third. How far has it fallen after 6s?
I tried using the 'sum to n terms' formula but that yielded the incorrect answer of 205.8m unless i haven't realised my mistake. Then i tried using the term formula (tn = a + (n - 1)d) for the 6 individual terms, and then added them up, and that worked. Any reason why the sum didnt work properly?

20. Show that the sum of the first n odd natural numbers is a perfect square.

22. The third term equals -2. The 9th term equals 28. How many terms of this arithmetic sequence, beginning with the first term are required to give this a sum of 1092?

Thanks guys

 

[Xayma]

T₅=a+4d=m
T₁₁=a+10d=n

T₁₁-T₅=6d=n-m

T₇=a+6d
∴T₇=a+n-m

16. It is an arithmetic progression, first time 21, last term 48, with 10 terms.

S₁₀=5(21+48)
=345

 

[Slidey]

20. Show that the sum of the first n odd natural numbers is a perfect square.

t_1=1
t_2=3
t_3=5
a=1
d=2
S_n=(n/2)(2+(n-1)2)
S_n=(n/2)(2(1+n-1))=n*n=n^2

 

[Slidey]

22. The third term equals -2. The 9th term equals 28. How many terms of this arithmetic sequence, beginning with the first term are required to give this a sum of 1092?

t_3=-2
t_9=28
S_n=1092

-2=a+2d
28=a+8d

30=6d
d=5
a=-12
S_n=1092
S_n=(n/2)(2a+(n-1)d)
1092*2=n(-24+(n-1)5)=-24n+5n(n-1)=5n^2-29n
5n^2-29n-2184=0
n=24 or -91/5
n=24

 

[Slidey]

17. A body, falling freely from a height, travels 4.9 metres in the first second, 14.7m in the second second and 24.5m in the third. How far has it fallen after 6s?
I tried using the 'sum to n terms' formula but that yielded the incorrect answer of 205.8m unless i haven't realised my mistake. Then i tried using the term formula (tn = a + (n - 1)d) for the 6 individual terms, and then added them up, and that worked. Any reason why the sum didnt work properly?

t_1=4.9
t_2=14.7
t_3=24.5
a=4.9
d=9.8
S_6=3*(9.8+5*9.8)=3*6*9.8=176.4

 

[JamiL]

8)A7=(2n-m)/3
16)Sum=345
17)im guessing u did sumthin rong... lol, i got 176.4 meters
20)a=1
d=2
therefore Sn=n/2*(2a+(n-1)d)
=[n(2+2n-2)]/2
=2n^2/2
=n^2
therefore its a perfect square.
22)24 terms
they are the answers, i did em on paper compare with ur results cos the book might be rong.

 

[JamiL]

T₅=a+4d=m
T₁₁=a+10d=n

T₁₁-T₅=6d=n-m

T₇=a+6d
∴T₇=a+n-m

you have 2 be more specific with ur answer... wot is a???? taht prob be 1 or 2 marks at best ull get half marks

 

[withoutaface]

I think what 'a' is is implied by the change between T₅ and T₁₀, as well as it being the standard pronumeral used to mean the initial term.

 

[velox]

that answer isn't right anyway. Jamil's is right

 

[velox]

Slide, how did you get 30 = 6d for question 22? I did it by trial and error and got the same....Just wondering how you did it. Damn i just realised i was right, i got to the stage 5n^2 - 29n - 2184 = 0. But i put it into quadratic formula wrong...I still dont get question 8??

 

[Xayma]

that answer isn't right anyway. Jamil's is right

True, you solve for a above and get the same answer, but Im sure you could have done that yourself .

 

[Slidey]

8. Find the 7th term of the arithmetic progression whose 5th term is m and whose 11th term is n

t_5=a+4d=m
t+11=a+10d=n
6d=n-m
5a+20d=5m
2a+20d=2n
a=(5m-2n)/3
t_7=(5m-2n)/3 + 6d
t_7=(5m-2n)/3 + n-m
t_7=(5m-2n + 3n - 3m)/3
t_7=(2m+n)/3

So you're OK for Q22?

 

[Slidey]

And here's 16 just for completeness:

16. Find the sum of all integers between 20 and 50 that are divisible by 3.

sum of integers divisible by 3 from 21 to 48 inclusive
3,6,9
a=3
d=3
So we want S_16-S_6
S_16=8(2*3+15*3)=8*17*3=24*17=(20+4)(20-3)=400+20-12=408
S_6=3(2*3+5*3)=3*3*7=63
So, S_16 - S_6 = 408 - 63 =348-3=345

I hate calculators.

 

[JamiL]

And here's 16 just for completeness:

16. Find the sum of all integers between 20 and 50 that are divisible by 3.

sum of integers divisible by 3 from 21 to 48 inclusive
3,6,9
a=3
d=3
So we want S_16-S_6
S_16=8(2*3+15*3)=8*17*3=24*17=(20+4)(20-3)=400+20-12=408
S_6=3(2*3+5*3)=3*3*7=63
So, S_16 - S_6 = 408 - 63 =348-3=345

I hate calculators.

i think u mean a=21 as a is the 1st number???? but u got it right so im confused.... lol
nah now i see wot u did u did the sum off all of em 2 48 n then minused the sum off all of em to 18... that the hard way jus do it n=10 (21, 24, 27..., 48) and start with a=21 its so much easyer

 

[Slidey]

It only took me as long as the others, so I don't mind.

But yeah, I'll keep that in mind. THAT was the first method I tried but I got the wrong answer because I used S_10 but with a being 3.