APGP question

Occam's Razor SG

Each term of an AP is added to the corresponding term of a GP to form a 3rd sequence S, whose 1st 3 terms are -1, -2, 6. The common ratio of the GP is equal to the 1st term of the AP.

Prove that the 1st term of the AP is a root of the equation a^3 - a^2 - a +10 = 0. Verify that a=-2 is a root of this equation. With this value of a, find the nth term of the sequence S and obtain an expression for the sum of the 1st n terms of S.

wee_ws SG

Hi,

We start with writing the terms of AP and GP.

AP: a, a + d, a + 2d, ...

GP: b, br, br^2, ...

We are given that r = a, so GP can be re-written as

b, ba, ba^2, ...

Now, we have

a + b = 1 -- (1)

a + d + ba = -2 -- (2)

a + 2d + ba^2 = 6 -- (3)

Try to proceed from this point onwards. Thanks!

Cheers,
Wen Shih

Occam's Razor SG

Thanks, got it. =)

Occam's Razor SG

Another question: If Sn = 1-4+9-16+...+ (n)^2 (-1)^(n+1) and given that
S2r+1 - S2r-1 = 4r + 1, show that S2n+1 = (n+1)(2n+1). Thanks :)

wee_ws SG

Hi,

Given that S_(2r + 1) - S_(2r - 1) = 4r + 1, we use it repeatedly for different r values:

when r = 1, we have S_3 - S_1 = 5

when r = 2, we have S_5 - S_3 = 9

when r = n - 1, we have S_(2n - 1) - S_(2n - 3) = 4(n - 1) + 1

when r = n, we have S_(2n + 1) - S_(2n - 1) = 4n + 1

Adding all these (very useful strategy), we obtain

S_(2n + 1) - S_1 = 5 + 9 + ... + {4(n - 1) + 1} + {4n + 1}

Note that RHS is a sum of AP. Try to continue the rest on your own...

Thanks!

Cheers,
Wen Shih

Occam's Razor SG

Thanks! =D

Destined2REIGN SG

10) When a ping pong ball is dropped on a table, the time interval between and particular bounce and the following bounce is 90%. The interval between the 1st and 2nd bounce is 2 sec. Given that the nth bounce and the (n+1)th bounce is the first such interval less than 0.02 sec, find n and the total time from the first bounce to the n bounce in 4 sig fig.

ans.n=45, S=19.81

11) John loaned $5000 from a bank in jan 2009. The bank charges an interest rate of 5% of the outstanding amt at the end of Dec each year.
a) John pays back $1000 annually in Jan. Starting in 2010.
How much does he owe at the end of Dec 2010?
b)if John pays $x instead anuually from jan 2010, find x so that he can clear his debt in Jan 2018 after he makes his last payment of $x.

wee_ws SG

Hi,

For Q10, let I_i be the interval i between i-th and (i+1)-th bounce.

Now I_1 = 2

I_2 = 2(0.9)

I_3 = 2(0.9)^2

...

I_n = 2(0.9)^?

From the problem statement, we want I_n < 0.02.

Total time = I_1 + I_2 + ... + I_(n-1)

Try it, thanks!

Cheers,
Wen Shih

wee_ws SG

Hi,

For Q11, analyse the problem year by year, by considering the start and end of any year:

Jan '09: 5000

Dec '09: 5000(1.05)

Jan '10: 5000(1.05) - 1000

Dec '10: {5000(1.05) - 1000}(1.05) <-- enough for part (a)

Jan '11: {5000(1.05) - x}(1.05) - x <-- replace 1000 by x for part (b)

Dec '11: ?

Jan '18: ? - x (this amount must be zero)

Please try, thanks!

Cheers,
Wen Shih

Destined2REIGN SG

managed to complete Q4 and 5a) with your hint. But I still cant see the pattern for 5b).

BTW: is the answer for 5a) $4462.50?

wee_ws SG

Hi,

Correct for (a).

For (b), write down the expression for a few years and observe a pattern:

Jan '10: 5000(1.05) - x

Jan '11: 5000(1.05)^2 - 1.05x - x

Jan '12: 5000(1.05)^3 - (1.05^2) x - 1.05x - x

Jan '18: 5000(1.05)^? - ?

Thanks!

Cheers,
Wen Shih

Occam's Razor SG

xstryker SG

nice wall of computer language that we do not understand

Occam's Razor SG

Bleh some kind of glitch =/ try editing it but won't go away

Anyway my Q

Figure 1 shows an infinite sequence of squares where the largest (1st square) has sides of length x cm. The second square is formed by joining the mid-points of the sides of the first square and so forth. The (n + 1)th square is formed by joining the mid-points of the sides of the nth square, for all real and positive integers.

The shaded areas, as illustrated on the diagram, form an infinite spiral.
Find A_1 and state the value of A_n+1 / A_n. Hence find A_n in terms of x.

Find the least number of n such that the sum of the area of the 1st n triangles is within 0.05% of the total spiral area.

[Ans: A_1 = (x^2)/8 ; A_n+1/A_n = 1/2 ; n= 11]

Able to do A_1 but not the rest.

wee_ws SG

Hi,

To find A_2, first ask what is the length of the square. Observe that it is the hypothenuse of the triangle with area A_1.

To find A_3, ask the same question and make a similar observation.

Try to make out the pattern from the expressions of A_1, A_2, A_3. The fact that we are asked to find the ratio means that we should see a geometric progression yes?

For the last part, we want to find n such that

A_1 + A_2 + ... + A_n < 0.05% of total spiral area.

Note that the total spiral area is an infinite sum.

Try it, thanks!

Cheers,
Wen Shih

Occam's Razor SG

Ok spotted my mistake thanks again !=)

Occam's Razor SG

just to confirm the last part the Q asks 'sum of the area of the 1st n triangles is within 0.05% of the total spiral area.'

shouldn't it be Sn < 99.95% S instead of 'A_1 + A_2 + ... + A_n < 0.05% of total spiral area.'

since Sn at most is 0.05% of the total area

wee_ws SG

Hi,

Very good, you are correct! Keep it up!

Cheers,
Wen Shih