Arithmetic series/progression problems

Did new topic in class today which i found so so hard. I have been given some questions to do but no idea how to do it.


Can anybody please help me?

No. 1:


"A bank account is opened at the beginning of the year 2013 by making a deposit of £1000. Further deposits of £500 are made at the beginning of the subsequent year. The rate of interest is 2.5% and the interest is paid at the end of each year. Find the total amount gathered in the account at the end of year 2013, just after the interest is paid"

How to solve this? it says it is 15 marks so there must be lots of calculations. i dont know how to do it. anyone can please help?

No. 2:

Find the minimum no. of terms of the geometric sequence 1.5, 2.75, 3.375, ... which will make the terms more than 1000? (6)

Any idea of how to do this aswell? I am so stuck!

Are you sure you have copied the question correctly. If you have then the adding of £500 at th beginning of each year does not make sense because there are no further years.If the latter date should have been later than 2013 t5rhe proceed as follows

I would start by listing it year by year
i.e.
Amount at end of year 1==?
Interest added=?
Amount at start of year 2=?
Amount at end of year 2=?
Interest added=?
Amount at start of year 3=? etc and see if you can see any patterns.

Please can someone explain to me how to do this? 15 marks is alot and i dont know what answers and working out is neeeded

Beg of 2013:1000
End of 2013: 1000*(1-0.025)=1000*0.975 (1st year end)
Beg of 2014: 1000*0.975+500
End of 2014: (1000*0.975+500)*0.975=1000*0.975^2 + 500*0.975 (2nd year end)
beg of 2015: 1000*0.975^2 + 500*0.975+500
End of 2014: (1000*0.975^2 + 500*0.975+500)*0.975=1000*0.975^3+500(0.975+0975^2)
........
......
End of the n th year :
$\displaystyle 1000 \cdot 0.975^n +500\cdot (0.975 +0.975^2 + ... + 0.975^{n-1})=$
$\displaystyle =1000 \cdot 0.975^n +500\cdot 0.975 \cdot (1 +0.975 + ... + 0.975^{n-2})$

Use the formula for sum of first (n-1) terms of geometric sequence. When q is the quotient
and a_1 is the first term, then
$\displaystyle s_{n-1}=a_1\cdot \frac{q^{n-1}-1}{q-1}$

So what are the figures for Sn, Q and n. I know a is 1000. Still confused about how to work out total amount gathered

Can somebody please explain where i am going wrong with question 2?
What do i do from this point?




1st term is 1.5 and common ratio is also 1.5 i think. What do i do next? Did i make mistake?

I saw this example on google to find out where i am going wrong




I don't know how he got from the 1st to the 2nd line, or from 2nd to 3rd.

I saw this example on google to find out where i am going wrong




I don't know how he got from the 1st to the 2nd line, or from 2nd to 3rd.

Can somebody please explain where i am going wrong with question 2?
What do i do from this point?




1st term is 1.5 and common ratio is also 1.5 i think. What do i do next? Did i make mistake?

Is that 1.5 x 1.5 the next line and if so how does that come about? Do i do 2.25n divided by 1.5n too?

What do i do with the 1000 on the RHS? Cheers mate

Sorry i should have made myself clearer.

You wrote that $1.5^n\times1.5=2.25^n$. This is a mistake.

$3^2\times3\not=9^2[br]3^2\times3=3^3$

See where you've gone wrong with that bit?

Yes i think so mate, but not sure what the answer should be if it's not -2.25. I am struggling to see it, and what to do next after with the 1000 on the other side

$3^2\times3=3^{2+1}$

Any clearer on where we should be?

Need to get it in a form we can proceed with before you even need to worry about the 1000.

So long as what we're doing to the LHS doesn't actually change it the 1000 stays as it is.

Would it be$1.5^{n+1}$

Yep. $1.5^{n+1}$.

So now we have $\dfrac{1.5-1.5^{n+1}}{-0.5}>1000$

We can now get rid of the fraction.
For the sake of keeping the inequality simple we can multiply the LHS by $\dfrac{-1}{-1}$ to bring us to a point where we can easily move forward.

Any idea where to go next?