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Help again sorry!! C1 maths

help?
(edited 9 years ago)
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Reply 1
Avatar for TeeEm
TeeEm
19
Original post by kandykissesxox
help?


I cannot see anything tried for this question and maybe that is why people are not responding.

look at a similar example in the link

http://madasmaths.com/archive_maths_booklets_basic_topics_various.html


FILE NAME arithmetic_series_exam_questions

similar questions 49 and 56
Reply 2
Avatar for Zacken
Zacken
22
Original post by kandykissesxox
help?


It is basically saying that the largest term u16=6u_{16} = 6 and that the sum of all the terms is 72: S16=72S_16 = 72

So set up your equation for
Unparseable latex formula:

u_{16} = u_1 + (16-1)d \Righarrow 6 = u_1 + 15d

and the sum

162[2u1+(16−1)d]=72\frac{16}{2}[2u_1 + (16-1)d] = 72 Put u_1 in terms of d and substitute into your other equation, post your working! :smile:
Original post by Zacken
It is basically saying that the largest term u16=6u_{16} = 6 and that the sum of all the terms is 72: S16=72S_16 = 72

So set up your equation for
Unparseable latex formula:

u_{16} = u_1 + (16-1)d \Righarrow 6 = u_1 + 15d

and the sum

162[2u1+(16−1)d]=72\frac{16}{2}[2u_1 + (16-1)d] = 72 Put u_1 in terms of d and substitute into your other equation, post your working! :smile:


this is stressing me out!! i can't do the simultaneous equation:frown:
Reply 4
Avatar for Zacken
Zacken
22
Original post by kandykissesxox
this is stressing me out!! i can't do the simultaneous equation:frown:


Calm down, not to worry. It's simple! :smile:

You have u16=u1+(16−1)du_{16} = u_1 + (16-1)d

u16=6u_{16} = 6 from the question, so:

6=u1+15d6 = u_1 + 15d

6−15d=u16 - 15d = u_1 (subtract 15d from both sides)

You also have: 162[2u1+(16−1)d]=72\displaystyle \frac{16}{2}[2u_1 + (16-1)d] = 72

You know that you can write u1u_1 as 6−15d6 -15d so you can substitute that in:

162[2(6−15d)+15d]=72\displaystyle \frac{16}{2}[2(6-15d) + 15d] = 72

8(12−30d+15d)=72⇒12−15d=9⇒d=0.28(12-30d + 15d) = 72 \Rightarrow 12 - 15d = 9 \Rightarrow d = 0.2

Now you know the value of d, and you have an equation involving d and u_1, substitute d into that and find u_1. :smile:
Original post by Zacken
Calm down, not to worry. It's simple! :smile:

You have u16=u1+(16−1)du_{16} = u_1 + (16-1)d

u16=6u_{16} = 6 from the question, so:

6=u1+15d6 = u_1 + 15d

6−15d=u16 - 15d = u_1 (subtract 15d from both sides)

You also have: 162[2u1+(16−1)d]=72\displaystyle \frac{16}{2}[2u_1 + (16-1)d] = 72

You know that you can write u1u_1 as 6−15d6 -15d so you can substitute that in:

162[2(6−15d)+15d]=72\displaystyle \frac{16}{2}[2(6-15d) + 15d] = 72

8(12−30d+15d)=72⇒12−15d=9⇒d=0.28(12-30d + 15d) = 72 \Rightarrow 12 - 15d = 9 \Rightarrow d = 0.2

Now you know the value of d, and you have an equation involving d and u_1, substitute d into that and find u_1. :smile:


why did i get d as 5
Reply 6
Avatar for Zacken
Zacken
22
Original post by kandykissesxox
why did i get d as 5


Post a picture of your working and let us have a look. :smile:
Original post by Zacken
Calm down, not to worry. It's simple! :smile:

You have u16=u1+(16−1)du_{16} = u_1 + (16-1)d

u16=6u_{16} = 6 from the question, so:

6=u1+15d6 = u_1 + 15d

6−15d=u16 - 15d = u_1 (subtract 15d from both sides)

You also have: 162[2u1+(16−1)d]=72\displaystyle \frac{16}{2}[2u_1 + (16-1)d] = 72

You know that you can write u1u_1 as 6−15d6 -15d so you can substitute that in:



162[2(6−15d)+15d]=72\displaystyle \frac{16}{2}[2(6-15d) + 15d] = 72

8(12−30d+15d)=72⇒12−15d=9⇒d=0.28(12-30d + 15d) = 72 \Rightarrow 12 - 15d = 9 \Rightarrow d = 0.2

Now you know the value of d, and you have an equation involving d and u_1, substitute d into that and find u_1. :smile:

never mind, i divided it wrong! haha xx
Original post by kandykissesxox
never mind, i divided it wrong! haha xx


so now do i have to find the shortest length?
Reply 9
Avatar for Zacken
Zacken
22
Original post by kandykissesxox
never mind, i divided it wrong! haha xx


Yay! Finding your own mistakes is a critical part of Maths, that's great! :biggrin:

What did you get for your value of u1u_1? :smile:
Reply 10
Avatar for Zacken
Zacken
22
Original post by kandykissesxox
so now do i have to find the shortest length?


Yes, your shortest length is the first term in the arithmetic series, as we labelled it: u1u_1 :smile:
Original post by Zacken
Yes, your shortest length is the first term in the arithmetic series, as we labelled it: u1u_1 :smile:


3?
Reply 12
Avatar for Zacken
Zacken
22
Original post by kandykissesxox
3?


Yay! :biggrin:

Correctemundo. :smile:
will you please help me on one last thing?
FullSizeRender-7.jpg532.1KB
Reply 14
Avatar for Zacken
Zacken
22
Original post by kandykissesxox
will you please help me on one last thing?


First, I'd like you to write down what your two equations would be and post them here. Then I'll guide you from there. :smile:
Original post by Zacken
First, I'd like you to write down what your two equations would be and post them here. Then I'll guide you from there. :smile:

Screen Shot 2014-10-23 at 23.19.46.png316.8KB
Reply 16
Avatar for Zacken
Zacken
22
Original post by kandykissesxox
x


You have your 10th term given by: u10=a+9du_{10} = a + 9d

You know that the 10th term however is 110, so you get: 110=a+9d⇒a=110−9d110 = a + 9d \Rightarrow a = 110 - 9d

The sum of the first 3(THREE) is 66, so

32[2a+2d]=66\displaystyle \frac{3}{2}[2a + 2d] = 66

Can you substitute and solve from here? :smile:
(edited 9 years ago)
Original post by Zacken
You have your 10th term given by: u10=a+9du_{10} = a + 9d

You know that the 10th term however is 110, so you get: 110=a+9d⇒a=110−9d110 = a + 9d \Rightarrow a = 110 - 9d

The sum of the first 3(THREE) is 66, so

32[2a+2d]=66\displaystyle \frac{3}{2}[2a + 2d] = 66

Can you substitute and solve from here? :smile:


do i put 110-9d in place of 2d?
Reply 18
Avatar for Zacken
Zacken
22
Original post by kandykissesxox
do i put 110-9d in place of 2d?


You have a=110−9da = 110 - 9d and

32[2a+2d]=66\displaystyle \frac{3}{2}[2a + 2d] = 66

You can replace the a in the above equation with a=110−9da = 110 - 9d.
So it becomes:

32[2(110−9d)+2d]=66\displaystyle \frac{3}{2}[2(110-9d) + 2d] = 66

Basically, if you have a = something, you can rewrite the a in another equation to be the something.

In this case a = 110-9d, so you can replace the a in the other equation with 110-9d, but only the a, don't forget to keep the 2 as well, you're multiply a by 2 in the second equation. :smile:
Reply 19
Avatar for Zen-Ali
Zen-Ali
8
Original post by kandykissesxox
...


The terms in an arithmetic progression are given by

a,a+d,a+2d,⋯ ,a+(n−1)da, a + d, a + 2d, \cdots, a + (n - 1)d

If the sum of the first three terms is 66, you can say what?

Hint



The nthn^{\text{th}} term of an AP is given by:

a+(n−1)da + (n - 1)d

So, what can you deduce about n=10n = 10 in relation to the nth term?

Hint



Then, solve the equations you've derived simultaneously.
(edited 9 years ago)

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