A lvl H2 Maths: APGP

AJC Prelims 2000

The n^th term of an infinite geometric series is equal to one fourth of the sum of all the terms after ( but not including ) the n^th term. Show that the sum to infinity of the geometric series is five times its first term.

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Answer:

T_n = 1/4 (S_∞ - S_n)




4 - 4r = r
5r = 4
r = 4/5

Hence, S_∞ = a / (1 - r) = 5a (shown)

A lvl H2 Maths: APGP

AJC Prelims 1999 

An arithmetic sequence has first term a and common difference d. It is given that the sum of the first four terms is less than the sum of the next four terms by 8. Also, the first, third and sixth term of the sequence are three consecutive terms of a geometric progression. Find the exact values of a and d.

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Answer:
Given first term a and common difference d

S₄ < S₈ - S₄
(4/2) (2a + 3d) < (8/2) (2a + 7d) - (4/2) (2a + 3d)
4 (2a + 3d) < 4 (2a + 7d)
4d > 0
d > 0 ------- (1)

T₁, T₃, T₆ are in G.P.
(T₃)² = T₁ * T₆
(a + 2d)² = (a)(a + 5d)
4d² - ad = 0
d (4d - a) = 0
d = 0 (reject since d > 0) or d = a/4

Choose a = 4 and d = 1

A lvl H2 Maths: APGP

Question from http://www.sgforums.com/forums/2297/topics/348959

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 a³ - a² - 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.

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Answer:

Let first term of AP = a
Thus Common ratio of GP = a
Let first term of GP = b

First 3 terms of GP = b, ba, ba²
First 3 terms of AP = a, a+d, a+2d

Hence,
b + a = -1 ---(1)
ba + a + d = -2 ---(2)
ba² + a + 2d = 6 --- (3)

From (1), b = -1 - a ---(4)
From (2), a + d = -2 - ba
2a + 2d = -4 - 2ba ---(5)

Sub (5) into (3)
ba² - a + (-4 - 2ba) = 6
ba² - a - 2ba = 10 ---(6)

Sub (4) into (6)
(-1 - a)a² - a - 2(-1 - a)a = 10
-a² - a³ - a + 2a + 2a² - 10 = 0
-a³ + a² + a - 10 = 0
a³ - a² - a + 10 = 0

Thus, the 1st term of the AP is a root of the equation a³ - a² - a +10 = 0 (shown)

When a = -2,
(-2)³ - (-2)² - (-2) +10 = -8 - 4 + 2 + 10 = 0
Thus, a = -2 is a root of this equation (verified)

Sub a = -2 into (1)
b = -1 -a = 1
Sub b = 1, a = -2 into (2)
d = -2 - a - ba = -2 + 2 + 2 = 2

So,
nth term of S = nth term of AP + nth term of GP
= (-2) + (n - 1)(2) + (1)(-2)^(n-1)
= 2n - 4 + (-2)^(n-1)

A lvl H2 Maths: APGP

From http://www.sgforums.com/forums/2297/topics/348847

1) In an AP , 8th term is twice the 4th term and the 20th term is 40. Find the common difference and the sum of the terms from the 8th to the 20th term.

2) The r^th term of an AP is (1+4r). Find in terms of n, the sum of the first n terms.

3) An arithmetic series has first term 1000 and common difference -1.4. Calculate the 1st negative term and the sum of all the positive terms.

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Answer:

1) Given: U₈ = 2U₄, U₂₀ = 40
Apply nth term formula:
a + 7d = 2(a +3d)
a - d = 0 ---(1)
Also, a + 19d = 40 ---(2)
solving simultaneously, d = 2
a = 2

Sum of terms = S₂₀-S₇
= (20/2)(2(2) + (19)(2)) - (7/2)(2(2) + (6)(2))
= (10)(42) - (7/2)(16)
= 364


2) U_r= 1 + 4_r
When r = 1, U₁ = 5
When r = 2, U₂ = 9
When r = 3, U₃ = 13
Thus, common difference = 4

Thus, S_n = (n/2)(2(5) + (n-1)(4))
S_n = (n)(2n + 3)


3) T_n = a + (n-1)d
a = 1000, d = -1.4
Thus, when T_n <> 1001.4
n > 715.286

Hence, the first negative n term is when n = 716

First negative n term = 1000 + (716 - 1)(-1.4) = -1

First negative term is when n = 716
so the last positive term is when n = 715
So, sum of all positive numbers
= (715/2) (2000 + 714 * -1.4)
= 357643

A lvl H2 Maths: APGP

2002 A lvl Maths P1 Q10
Question from http://www.sgforums.com/forums/2297/topics/348847

An AP in which the common difference is twice the 1st term is called a special AP. Show that the sum of the first n terms of a special AP with 1st term a is n²a.

The (M-2)th term of a special AP is 663 and the Mth term is 731. Find M and find also the sum of the first M terms.

The sum of the first N terms of another special AP is S. Find in the terms of S, the sum of the next N terms.

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Answer:

Let first term = a, common difference = 2a

S_n = (n/2)*(2a + (n-1)(2a))
= (n/2)*(2a + 2an - 2a)
= (n/2)*(2an)
= n²a (shown)


Mth term - (M-2)th term = 2 * common difference
731 - 663 = 2 * 2a
a = 17, d = 34

731 = 17 + (M-1)(34)
M = 22

Sum of first M terms =22² * 17 = 8228


S_n = n²a
S_2n = 4n²a

Difference of next N terms = S_2n - S_n
= 3n²a
= 3S

A lvl H2 Maths: APGP

H2 Maths 2008 P1 Q10

(i) A student saves $10 on 1 January 2009. On the first day of each subsequent month she saves $3 more than in the previous month, so that she saves $13 on 1 February 2009, $16 on 1 March 2009, and so on. On what date will she first have saved over $2000 in total? [5]

(ii) A second student puts $10 on 1 January 2009 into a bank account which pays compound interest at a rate of 2% per month on the last day of each month. She puts a further $10 into the account on the first day of each subsequent month.

(a) How much compound interest has her original $10 earned at the end of 2 years? [2]
(b) How much in total is in the account at the end of 2 years? [3]
(c) After how many complete months will the total in the account first exceed $2000? [4]

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Answer:

(i)
first term = 10
2nd term = 13
3rd term = 16
so this is an AP, first term 10, common difference 3

When sum of AP ≥ 2000
n/2 (2*10 + (n-1)*3) = 2000
n (20 + 3n - 3) = 4000
3n² + 17n - 4000 = 0
Use your graphic calculator: n=33.79 or -39.46 (reject)
So on the 34th month, she will save over $2000
Date is 1st Oct 2011


(ii)
(a) end of 2 years = 24 months
Amount of compound interest = 10(1.02)24 - 10 = $6.08

(b) end of 2nd month = 10(1.02)2 + 10(1.02)
end of 3rd month = 10(1.02)3 + 10(1.02)2 + 10(1.02)
end of 4th month = 10(1.02)4 + 10(1.02)3 + 10(1.02)2 + 10(1.02)

So end of 24th month = 10(1.02)24 + 10(1.02)23 + ... + 10(1.02)2 + 10(1.02)
= sum of GP with first term 10(1.02) and common ratio 1.02 for 24 terms
= $10(1.02) (1.0224 - 1) / (1.02 - 1)
= $310.30

(c) Supposed at the end of n months, the account will first exceed $2000

sum of GP with first term 10(1.02) and common ratio 1.02 for n terms ≥ 2000
10(1.02) (1.02n - 1) / (1.02 - 1) ≥ 2000
(1.02n - 1) ≥ 3.92157
1.02n ≥ 4.92157
n ≥ lg (4.92157) / lg (1.02)
n ≥ 80.475

End of 81st month, amount = 10(1.02) (1.0281 - 1) / (1.02 - 1) = $2026.20
Start of 81st month = $2026.2 / 1.02 = $1986.47 (not needed, show for fun only)

The account first exceed $2000 after 81 complete months.

A lvl H2 Maths: APGP

The radii of n circular discs, labelled 1,2,...,n, from smallest to largest, follow an arithmetic progression.
The radius of disc 4 is 11 units. The sum of the radii of the first 4 discs is 26 units.
Disc 2 to disc n are coloured with (n-1) different colours such that disc k is coloured with colour k, for k = 2,3,...,n. The n discs are then stacked in decreasing order of radii with disc n at the bottom and disc 1 on top such that a vertical line passes through the centres of all the n discs. The figure below shows a stack with 4 discs when viewed from the top.

It is given that Uk, k = 2,...,n, represents the area of region with colour k in the stack with n discs when viewed from the top.

(i) Show that
(a) the radius of the disc k is (3k-1) units,
(b) Uk = 3π (6k -5 )

(ii) Show that the sequence U2, U3, U4, ..., Un is an arithmetic progression.

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Answer:


(i)
(a)
Disc 4 = a + 3d = 11 units -------- (1)
Sum of radii of first 4 discs = 2 (2a + 3d)
= 4a + 6d = 26 units --------- (2)

(1) × 2: 2a + 6d = 22 ------ (3)
(2) - (3): 2a = 4
a = 2

thus, 2 + 3d = 11
3d = 9
d = 3

Thus, radius of disc k = 2 + (k-1) × 3
= 2 + 3k - 3
= (3k - 1) units (shown)


(b)
Hence, radius of disc k-1 = (3(k-1) - 1)
= 3k - 4

Uk = π(3k - 1)² - π(3k - 4)²
= π (9k² - 6k + 1) - π (9k² - 24k + 16)
= π (18k -15)
= 3π (6k - 5) (shown)



(ii)
Uk = 18π k - 15π
Uk-1 = 18π (k-1) - 15π = 18π k - 33π
Uk - Uk-1 = 18π = constant
Hence, the sequence U2, U3, U4, ..., Un is an arithmetic progression (shown)

A lvl H2 Maths: APGP

A geometric series has first term a and common ratio r, where |r| < 1. The sum to infinity of the series is S. The sum to infinity of the series obtained by adding all the odd-number terms (i.e. first term + third term + fifth term + ...) is 4S/3. Find the value of r.

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Answer:

Sum to infinity = a / (1-r)

Sum to infinity of odd-number terms = a / (1-r²) => odd numbered terms are a, ar², ar⁴, ...

Sum to infinity / Sum to infinity of odd-number terms
= {a / (1-r)} / {a / (1-r²)}
= (1-r²) / (1-r)
= (1+r)(1-r) / (1-r)
= 1+r

But Sum to infinity / Sum to infinity of odd-number terms
= S / {4S/3}
= 3/4

Hence, 1+r = 3/4
r = -0.25