Algebra DPP Sequences - 1.7(pending - 34,35)

Q1. A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains 100 cans, how many rows are there?
Solution:
1 + 3 + .. (2n-1) = 100 = n/2[2.1 + (n-1).2] => 100 = n.[1 + n - 1] = n^2 => n = 10. 

Q2. The second and fourth terms of a geometric sequence are 2 and 6. Find the possible first terms.
Solution:
r^2 = 3, ar = 2 => a = -+2/sqrt(3)

Q3. Figures 0, 1, 2, and 3 consists of 1, 5, 13, and 25 nonoverlapping unit squares, respectively. If the pattern were continued, how many non-overlapping unit squares would there be in Figure 100?

Solution:
a1 = 1 + 4
a2 = 1 + 4 + 8
a3 = 1 + 4 + 8 + 12
a100 = 1 + (4 + 8 + 12 .. 100 times)
= 1 + 100/2[8 + 99.4] = 1 + 200.[2 + 99] = 1 + 101.200 = 20201

Q4. Let 1, 4, . . . and 9, 16, . . . be two arithmetic sequences. The set S is the union of the first 2004 terms of each sequence. How many distinct numbers are in S?
Solution:
Total terms = 2004 * 2 = 4008
From this we need to subtract common terms.
Let's manually check for some common terms.

First A.P.: 1,4,7,10,13,16,19,22,25,28,31,34,37.....
Second A.P.: 9,16,23,30,37....
Here we see 16 and 37 are common.

1 + (m-1).3 = 9 + (n-1).7
=> 3m - 2 = 7n + 2
=> 7n = 3m - 4
=> n = (3m - 4)/7
For n to be an integer, 3m - 4 has to be divisible by 7.
3m - 4 = 0 mod 7
=> 3m + 3 = 0 mod 7
=> 3(m+1) = 0 mod 7
=> (m+1) = 0 mod 7 since 3 isn't divisible by 7.
=> m = -1 mod 7
=> m = 6 mod 7
So all those values of m which leave remainder of 6 when divided by 7 will result in equal terms in both A.P.s.
Let's check.
1 + (6-1).3 = 16 which is present in the second A.P.
1 + (13-1).3 = 37 which is present in second A.P.
2004 divided by 7 leaves remainder of 2 and quotient = 286.
So 286 terms are common.
Answer: 4008 - 286 = 3722

Q5. Let $a_1, a_2, \ldots, a_k$ be a finite arithmetic sequence with
$a_4 + a_7 + a_{10} = 17$,
$a_4 + a_5 + a_6 + a_7 + a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13} + a_{14} = 77$,
and $a_k = 13$. What is $k$?

Solution:
a4 + a7 + a10 = 17 => a + 3d + a + 6d + a + 9d = 17
=> 3a + 18d = 17
=> 3(a + 6d) = 17_________[1]

a4 + a5 .... a13 + a14 = 77
=> (a4 + a14) + (a5 + a13) + (a6 + a12) + (a7 + a11) +  (a8 + a10) + a9 = 77
=> 5.(2a + 16d) + a + 8d = 77
=> 11a + 88d = 77
=> a + 8d = 7_______[2]

Solving 2d = 4/3 => d = 2/3 => a = 5/3
ak = 5/3 + (k-1).2/3 = 1/3(3 + 2k) = 13
=> 2k + 3 = 39 => k = 18

Q6. A sequence of three real numbers forms an arithmetic sequence whose first term is 9. If the first term is unchanged, 2 is added to the second term, and 20 is added to the third term, then the three resulting numbers form a geometric sequence. What is the smallest possible value for the third term of the geometric progression?
Solution:
9, 9+d, 9+2d
9, 11+d, 29+2d
(11+d)^2 = 9.(29 + 2d)
=> d^2 + 121 + 22d = 261 + 18d
=> d^2 + 4d - 140 = 0
=> d = -14,10
Choose d = -14 => last term of G.P. = 1

Q7. Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, . . . . For $n > 2$, the $n$th term of the sequence is the units digit of the sum of the two previous terms. Let $S_n$ denote the sum of the first $n$ terms of this sequence. What is the smallest value of $n$ for which $S_n > 10,000$?
Solution:
Let's see if the sequence repeats.
4,7,1,8,9,7,6,3,9,2,1,3,4,7,1.....
Yes it does.
After every 12 terms.
Sum of the first 12 terms = 60.
60*166 = 9960
10,000 - 9960 = 40 remaining still.
So total 166*12 = 1992 terms so far.
Now, adding first 6 terms yields 36. And adding the next one crosses 40 and hence 10,000.
So 1992 + 7 = 1999.

Q8. 

Suppose that the sequence $\{a_n\}$ is defined by
$a_1 = 2$, and $a_{n+1} = a_n + 2n$, when $n \geq 1$.
What is $a_{100}$?

Answer: 9902
Solution:
a2 = a1 + 2
a3 = a2 + 4
a4 = a3 + 6
...
a100 = a99 + 198
Rewrite as:
a2 - a1 = 2
a3 - a2 =  4
a4 - a3 = 6
...
a100 - a99 = 198
Add LHS and RHS =>
a100 - a1 = 2 + 4 + 6 ... 198 = 99/2[2 + 198] = 9900
a100 = a1 + 9900 = 9902

Q9.

The increasing sequence of positive integers $a_1, a_2, a_3, \ldots$ has the property that
$a_{n+2} = a_n + a_{n+1}$, for all $n \geq 1$. Suppose that $a_7 = 120$. What is $a_8$?

Answer: 194
Solution:
Let the first 2 terms be: a,b
a3 = a + b
a4 = a + 2b
a5 = 2a + 3b
a6 = 3a + 5b
a7 = 5a + 8b
a8 = 8a + 13b

Given a7 = 120 = 5a + 8b
=> a = (120 - 8b)/5
a is a positive integer so:
120 - 8b = 0 mod 5
-8b = 0 mod 5
2b = 0 mod 5
2 is not divisible by 5, so:
b = 0 mod 5
So b has to be divisible by 5.
From the given:
120 = 5a + 8b
b = 5 => a = 16 but a,b are in increasing order so invalid.
b = 10 => a = 8. Valid
b =15 => a = 0 but a,b are positive so invalid.
Finally,
a = 8, b = 10
a8 = 64 + 130 = 194.

Q10.  Same as question 6 but asking for largest value.
So put d = 10 and get 49 as answer.

Q11.

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In the sequence 2001, 2002, 2003, ..., each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is
$2001 + 2002 - 2003 = 2000$.

What is the 2004th term in this sequence? 
Solution:
Let's subtract 2000 from each term and just focus on the remaining.
a1 = 1, a2 = 2, a3 = 3
a4 = 1 + 2 - 3 = 0
a5 = 2 + 3 - 0 = 5
a6 = 3 + 0 - 5 = -2
a7 = 0 + 5 -(-2) = 7
a8 = 5 - 2 - 7 = -4
a9 = -2 + 7 -(-4) = 9
a10 = 7 - 4 - 9 = -6

So for the even terms the pattern is:
a4 = 0
a6 = -2
a8 = -4
a10 = -6
a_k = -2.(k/2 - 2)
a2004 = -2(1002 - 2) = -2000
And we need to add 2000 back to it.
So -2000 + 2000 = 0

Q12. Done already (Q7)
Q13.

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Let $x$ and $y$ be positive real numbers. What is the smallest possible value of

$$\frac{16}{x}+\frac{108}{y}+xy?$$

Solution:
A.M >= GM
1/3(16/x + 108/y + xy) >= {(16/x).(108/y).(xy)}^(1/3) = 12
=> 16/x + 108/y + xy >= 36
Min value  = 36 when 16/x = 108/y = xy = 12
Answer: 36

Q14.

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Find the value of the positive integer $n$ if

$$\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\cdots +\frac{1}{\sqrt{n}+\sqrt{n+1}}=5.$$Answer: 48
Solution:
Rationalize each term and then cancel to get:
sqrt(n+1) - sqrt(4) = 5
sqrt(n+1) = 7
n+1 = 49
n = 48

Q15.

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Consider a sequence of real numbers $\{a_n\}$ defined by

$$a_1 = 1 \quad \text{and} \quad a_{n+1} = \frac{a_n}{1 + n a_n} \quad \text{for } n \geq 1.$$

Find the value of

$$\frac{1}{a_{2005}}-2009000.$$
Solution:

$$a_{n+1} = \frac{a_n}{1 + n a_n}$$

Rewrite the recurrence in terms of $b_n = \frac{1}{a_n}$, yielding:

$$\frac{1}{a_{n+1}} = \frac{1 + n a_n}{a_n} = \frac{1}{a_n} + n \Rightarrow b_{n+1} = b_n + n$$

From this deduce that:

$$b_{n+1} - b_n = n$$ $$b_2 - b_1 = 1,\quad b_3 - b_2 = 2,\quad \ldots,\quad b_{n+1} - b_n = n$$

So:

$$b_{n+1} = b_1 + \sum_{k=1}^n k = 1 + \frac{n(n+1)}{2}$$

Then for $n = 2004$:

$$b_{2005} = 1 + \frac{2004 \times 2005}{2}$$

This confirms the value:

$$\frac{1}{a_{2005}} - 2009000 = \left(1 + \frac{2004 \times 2005}{2}\right) - 2009000$$

Q16.

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James calculates the sum of the first $n$ positive integers and finds that the sum is 5053. If he has counted one integer twice, which one is it?

Solution:
n * (n+1)/2 ~= 5053
n^2 + n ~= 10106
Dominant term is n^2.
n is likely to be 100.
100*101/2 = 5050 and add 3 extra.
So yeah, n = 100, and 3 was the repeated number.

Q17.

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What is the value of

$$2100-(x+1)(x+2006)[\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\cdots +\frac{1}{(x+2005)(x+2006)}]?$$
Solution:

$$\sum_{k=1}^{2005} \frac{1}{(x + k)(x + k + 1)}$$

Using partial fractions:

$$\frac{1}{(x + k)(x + k + 1)} = \frac{1}{x + k} - \frac{1}{x + k + 1}$$

This leads to a telescoping sum:

$$\left( \frac{1}{x + 1} - \frac{1}{x + 2} \right) + \left( \frac{1}{x + 2} - \frac{1}{x + 3} \right) + \cdots + \left( \frac{1}{x + 2005} - \frac{1}{x + 2006} \right)$$

Almost all terms cancel, and the result is:

$$\frac{1}{x + 1} - \frac{1}{x + 2006}$$

Now multiplying this sum by $(x + 1)(x + 2006)$, as seen in the original expression from your earlier image, gives:

$$(x + 1)(x + 2006) \left( \frac{1}{x + 1} - \frac{1}{x + 2006} \right) = (x + 2006) - (x + 1) = 2005$$

Thus, the full expression:

$$2100-(x+1)(x+2006)\left[\sum _{k=1}^{2005}\frac{1}{(x+k)(x+k+1)}\right]=2100-2005=\boxed{{\displaystyle 95}}$$
Q18.

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Suppose $a_n$ denotes the last two digits of $7^n$. For example, $a_2 = 49$, $a_3 = 43$. The value of
$a_1 + a_2 + a_3 + \cdots + a_{2007}$ is given by $x$. Find the sum of all the digits of $x$.

Solution:
a1 = 7 mod 100 = 7
a2 = 49 mod 100 = 49
a3 = 343 mod 100 = 43
a4 = 43*7 mod 100 = 301 mod 100 = 1
a5 = 1*7 mod 100 = 7

So a5 = a1 and the repetition starts.
a1 + a2 + a3 + a4 = 100
a1 + ... a2004 = 100*501 = 50100
a5 + a6 + a7 = 7 + 49 + 43 = 99
a1 + .... + a2007 = 50199
Sum of digits  = 24

Q19.

A sequence {aₙ} is defined by a₁ = 2, aₙ = (1 + aₙ₋₁) / (1 - aₙ₋₁), n ≥ 2. Find the value of (1100 + 2008 a₂₀₀₇).

Solution:
The sequence repeats every 4 terms.
a1 = 2
a2 = -3
a3 = -1/2
a4 = 1/3
a5 = 2
....
So a2007 = -1/2
Answer = 96

Q20.

If the minimum value of √(∑ₖ₌₁¹⁰⁰ |n − k|), where n ranges over all positive integers, is m, find m/50.

Solution:

First rewrite the expression without the square‐root (since √x is increasing, minimizing x and √x are equivalent):

$$S(n)=\sum_{k=1}^{100}\lvert n-k\rvert .$$

1. Where is $S(n)$ minimized?

For any finite data set, the sum of absolute deviations is minimized when $n$ is a median of the set.
The numbers $1,2,\dots,100$ have two medians: 50 and 51.
Hence the minimum of $S(n)$ occurs at $n=50$ or $n=51$.

2. Evaluate $S(50)$ (and $S(51)$)

Because the set is symmetric, the two values are equal:

$$\begin{aligned} S(50) &=\sum_{k=1}^{50}(50-k)+\sum_{k=51}^{100}(k-50)\\ &=\bigl(0+1+\dots+49\bigr)+\bigl(1+2+\dots+50\bigr)\\ &=\frac{49\cdot50}{2}+\frac{50\cdot51}{2}\\ &=1225+1275=2500. \end{aligned}$$

(The same computation with $n=51$ also gives 2500.)

3. Return to the original expression
m = sqrt(2500) = 50
So answer = 50/50 = 1

Q21.

The sum of an infinite geometric series is a positive number S, and the second term in the series is 1. What is the smallest possible value of S?
Solution:

Let's try few cases.
Since the second term is 1 and |r| < 1,
a = 2, r = 1/2 =>  S = a/1-r = 4
a = 3, r = 1/3 => S = 9/2
a = 3/2, r = 2/3 => S = 9/2
Let's try with negative values.
a = -2, r = -1/2 => S = -2/(3/2) = -4/3 But S > 0 as given hence first term and common ratio can't be < 0.
So in our limited sample a = 2 yields minimum sum. Values more or less than 2 yield bigger sums.
So S = 4 is a potential candidate.
Let's do a formal solution.
a.r =1 => a = 1/r
S = (1/r)/(1-r) = 1/r.(1-r)
It can be rewritten as:
r + 1 - r/r.(1-r) = 1/r + 1/(1-r)
By A.M. G.M. inequality:
1/2[1/r + 1/1-r] >= sqrt(1/r.(1-r)) = sqrt(1/r + 1/1-r)
=> S/2 >= sqrt(S)
=> S^2 >= 4S
=> (S-2)^2 >= 4
=> S-2 >= 2
=> S >= 4 (Answer min val of S = 4).

Q22. Let a < b < c be three integers such that a, b, c is an arithmetic progression and a, c, b is a geometric progression. What is the smallest possible value of 10c?

Solution:
b = a + d
c = a + 2d
d > 0 since a < b < c
a,b,c being integers means d is also integer.
c^2 = a.b
=> (a + 2d)^2 = a.(a + d)
=> 4d^2 + 4ad = ad
=> 4d^2 + 3ad = 0
=> d.(4d + 3a) = 0
=> d = -3a/4 since d != 0
d > 0 => a < 0
a has to be multiple of 4 so that d is also integer.
c = a + 2d = -a/2
a = -4 => c = 2 (for smallest c)
10c = 20 = answer.

Q23. The sequence S₁, S₂, S₃, ……, S₁₀ has the property that every term beginning with the third is the sum of the previous two. That is, Sₙ = Sₙ₋₂ + Sₙ₋₁ for n ≥ 3. Suppose that S₉ = 110 and S₇ = 42. What is S₄?
Solution:
One way is to set up equations in S1 and S2 and solve.
You will get S1 = 2, S2 = 4
Other way is:

S₉ = 110, S₇ = 42
S₈ = S₉ - S₇ = 110 - 42 = 68 ⇒ S₆ = S₈ - S₇ = 68 - 42 = 26 
S₅ = S₇ - S₆ = 42 - 26 = 16 ⇒ S₄ = S₆ - S₅ = 26 - 16 = 10 = Answer.

Q24. The sequence log₁₂162, log₁₂x, log₁₂y, log₁₂z, log₁₂1250 is an arithmetic progression. What is x⁄10?
Solution:
If 'd' is the common difference then log(1250/162) = 4d = 4.log(5/3) => d = log(5/3)
d = log(x/162) = log(5/3) => x = 270 => x/10 = 27 = Answer.

Q26. Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is N. The smallest possible value of N is n. What is half of n?

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Solution: Let the first two terms of the first sequence be 

$x_1$ and $x_2$ and the first two of the second sequence be $y_1$ and $y_2$.
Computing the seventh term, we see that
$5x_1 + 8x_2 = 5y_1 + 8y_2$.
Note that this means that $x_1$ and $y_1$ must have the same value modulo 8.
To minimize, let one of them be 0; WLOG assume that $x_1 = 0$.
Thus, the smallest possible value of $y_1$ is 8, and since the sequences are non-decreasing we get $y_2 \geq 8$.
To minimize, let $y_2 = 8$. Thus,

$$5y_1+8y_2=40+64=104=n\Rightarrow \frac{n}{2}=\boxed{{\displaystyle 52}}\mathrm{.}$$

Q27. The first four terms of an arithmetic sequence are $p, 9, 3p - q,$ and $3p + q$.
What is the sum of digits of the 2010th term of the sequence?

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Solution.

$3p - q$ and $3p + q$ are consecutive terms, so the common difference is

$$(3p + q) - (3p - q) = 2q.$$ $$p + 2q = 9 \Rightarrow 9 + 2q = 3p - q \Rightarrow q = 2 \Rightarrow p = 5$$

The common difference is 4. The first term is 5 and the 2010th term is

$$5 + 4(2009) = 8041.$$

Hence, sum of digits = 13.

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Q28. Let $a + ar_1 + ar_1^2 + ar_1^3 + \ldots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \ldots$ be two different infinite geometric series of positive numbers with the same first term. The sum of the first series is $r_1$, and the sum of the second series is $r_2$. What is $31(r_1 + r_2)$?

Solution. Using the formula for the sum of a geometric series we get that the sums of the given two sequences are

$$\frac{a}{1 - r_1} \quad \text{and} \quad \frac{a}{1 - r_2}$$

Hence we have

$$\frac{a}{1 - r_1} = r_1 \quad \text{and} \quad \frac{a}{1 - r_2} = r_2$$

This can be rewritten as

$$r_1(1 - r_1) = r_2(1 - r_2) = a$$

As we are given that $r_1$ and $r_2$ are distinct, these must be precisely the two roots of the equation

$$x^2 - x + a = 0$$

Using Vieta’s formulas we get that the sum of these two roots is $\boxed{1}$. So, $31(r_1 + r_2) = 31$.

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Q29. For each positive integer 

$n$, the mean of the first $n$ terms of a sequence is $n$. What will be the answer when the 2008th term of the sequence is divided by 55?

Solution: Let $S_n$ be the $n^\text{th}$ partial sum of the sequence.

$$\frac{S_n}{n} = n \quad \Rightarrow \quad S_n = n^2$$

The only possible sequence with this result is the sequence of odd integers.

$$a_n = 2n - 1 \quad \Rightarrow \quad a_{2008} = 2(2008) - 1 = 4015$$

So the answer is $\boxed{73}$.

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Q30. The geometric series $a + ar + ar^2 + \ldots$ has a sum of 7, and the terms involving odd powers of $r$ have a sum of 3. What is $10(a + r)$?

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Solution. The sum of an infinite geometric series is given by

$$\frac{a}{1 - r} \quad \text{where } a \text{ is the first term and } r \text{ is common ratio.}$$

In this series

$$\frac{a}{1 - r} = 7 \tag{1}$$

The series with odd powers of $r$ is given as

$$ar + ar^3 + ar^5 + \ldots$$

Its sum can be given by

$$\frac{ar}{1 - r^2} = 3 \tag{2}$$

Using (1) and (2):

$$ar = 3(1 - r)(1 + r)$$ $$ar = 3 \left( \frac{a}{7} \right)(1 + r) \Rightarrow \frac{7}{3}r = 1 + r \Rightarrow r = \frac{3}{4}$$ $$\Rightarrow a = 7(1 - r) = \frac{7}{4} \Rightarrow a + r = \boxed{\frac{5}{2}}$$

So, $10(a + r) = 25$.

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Q31. Find the maximum value of $\sqrt{x - 144} + \sqrt{722 - x}$.

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Solution. From AM-GM: $\sqrt{ab} \leq \frac{a + b}{2}$, we have $\sqrt{a} + \sqrt{b} \leq \sqrt{2(a + b)}$ (∵ R.M.S. ≥ A.M.).

Hence,
$\sqrt{x - 144} + \sqrt{722 - x} \leq \sqrt{2(722 - 144)} = 34$.

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Q32. If

$$S = \sum_{k=1}^{99} \frac{(-1)^{k+1}}{\sqrt{k(k+1)}(\sqrt{k+1} - \sqrt{k})},$$

find the value of $10S$.

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Solution.

$$S = \frac{\sqrt{2} + \sqrt{1}}{\sqrt{1 \times 2}} - \frac{\sqrt{3} + \sqrt{2}}{\sqrt{2 \times 3}} + \frac{\sqrt{4} + \sqrt{3}}{\sqrt{3 \times 4}} - \cdots + \frac{\sqrt{100} + \sqrt{99}}{\sqrt{99 \times 100}}$$ $$= \left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} \right) - \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} \right) + \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} \right) - \cdots + \left( \frac{1}{\sqrt{99}} + \frac{1}{\sqrt{100}} \right)$$ $$= 1 + \frac{1}{\sqrt{100}} = 1 + \frac{1}{10} = \frac{11}{10}$$

Hence, $1000S = 1100$. So, $10S = 11$.

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Q33. Let a₁, a₂, a₃, ... be the sequence of all positive integers that are relatively prime to 75, where a₁ < a₂ < a₃ < ⋯.
(The first five terms of the sequence are: a₁ = 1, a₂ = 2, a₃ = 4, a₄ = 7, a₅ = 8.)

Then the value of a₂₀₀₈ is defined as p10³ + q10² + r10 + s. Find (p + q + r + s)

Solution:
Let's see how many numbers are in the given sequence out of first 'n' positive integers:
In first 15 integers, we need to discard all the multiples of 3 and 5.
1,2,4,7,8,11,13,14 = 8 such integers.
So:
n - n/3 - n/5 + n/15 = 8n/15
We need to find the 2008th term in the sequence.
8n/15 = 2008 => n = 3765.
Since this is not in sequence, go one step back and check if that's valid:
3764 is not a multiple of of 3 or 5. So valid.
Answer: 3 + 7 + 6 + 4 = 20.

Q34. Given that

$x + (1 + x)^2 + (1 + x)^3 + \ldots + (1 + x)^n = a_0 + a_1x + a_2x^2 + \ldots + a_nx^n,$
where each $a_r$ is an integer, $r = 0, 1, 2, \ldots, n$.

Find the value of $n$ such that
$a_0 + a_2 + a_3 + a_4 + \ldots + a_{n-2} + a_{n-1} = 60 - \frac{n(n+1)}{2}.$

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Solution:

 Let $x = 1$, we have
$a_0 + a_1 + a_2 + \ldots + a_n = 1 + 2^2 + 2^3 + \ldots + 2^n = 2^{n+1} - 3.$

Also, Using Binominal theorem:
$a_1 = 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}, \quad a_n = 1,$

So
$60 - \frac{n(n+1)}{2} + \frac{n(n+1)}{2} + 1 = 2^{n+1} - 3 \Rightarrow n = 5.$

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Q35.

Given that 

$a_{n+1} = \dfrac{a_{n-1}}{1 + n a_{n-1} a_n}$, where $n = 1, 2, 3, \ldots$ and $a_0 = a_1 = 1$, find the value of

$$\left( 2 \times 10^4 - \dfrac{1}{a_{199} a_{200}} \right).$$

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Solution:
Multiply both sides by a_n, invert and you will get:

$$\frac{1}{a_{n+1}{a}_n}-\frac{1}{a_n{a}_{n-1}}=\frac{1+n{a}_{n-1}{a}_n}{a_{n-1}{a}_n}-\frac{1}{a_n{a}_{n-1}}=n\mathrm{.}$$

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Therefore

$$\sum_{n=1}^{199} \left( \frac{1}{a_{n+1} a_n} - \frac{1}{a_n a_{n-1}} \right) = \frac{1}{a_{200} a_{199}} - \frac{1}{a_0 a_1} \Rightarrow \sum_{n=1}^{199} n = \frac{199 \times 200}{2} = 19900.$$

Hence

$$\frac{1}{a_{200} a_{199}} = 1 + 19900 = 19901. \text{ So, value of } \left(2 \times 10^4 - \frac{1}{a_{199} a_{200}}\right) = 99.$$

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Q36.

Suppose that 

$(u_n)$ is a sequence of real numbers satisfying

$$u_{n+2} = 2u_{n+1} + u_n,$$

and that $u_3 = 9$ and $u_6 = 128$. What is $u_5$?

Solution:

Plugging in 

$n = 4$, we get

$$128 = 2u_5 + u_4.$$

Plugging in $n = 3$, we get

$$u_5 = 2u_4 + 9.$$

This is simply a system of two equations with two unknowns. Substituting gives

$$128 = 5u_4 + 18 \Rightarrow u_4 = 22, \text{ and } u_5 = \frac{128 - 22}{2} = 53$$

Q37.

For each positive integer 

$n$, the mean of the first $n$ terms of a sequence is $n$. What is the square root of the 2008th term of the sequence to the nearest integer?

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Solution:

Since the mean of the first n terms is n, the sum of the first n terms is n². Thus, the sum of the first 2007 terms is 2007² and the sum of the first 2008 terms is 2008². Hence, the 2008ᵗʰ term is 2008² − 2007² = (2008 + 2007)(2008 − 2007) = 4015

To find the nearest integer to sqrt(4015):

1. Find the perfect squares around 4015:
   $63^2=3969$ and $64^2=4096$. So $\sqrt{4015}\in(63,64)$.

2. Compare 4015 to the midpoint square $(63.5)^2$:
   $(63.5)^2=63^2+63+\tfrac14=3969+63+0.25=4032.25$.

Since $4015<4032.25$, we have $\sqrt{4015}<63.5$.
Therefore $\sqrt{4015}$ is closer to 63 than to 64.
If we take the other approach, i.e. take difference  4096 - 3969 = 127, divide by 2 = 63.5 and then see where 4015 falls, it will work here but it may fail sometimes.
To see why, note that (63.5)^2 = 4032.25.
To figure out if sqrt(4032.3) is closer to 63 or 64, if we compare with 3969 + 63.5 = 4032.5 we will concluded that it's closer to 63 but in reality it's closer to 64 since sqrt(4032.3) > 63.5.

Q38.

A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with terms 247, 475, and 756 and end with the term 824. Let S be the sum of all the terms in the sequence. What is the largest prime number that always divides S?

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Solution:
Let's try to build the smallest such sequence:
111 (one number is enough as it satisfies all the given constraints).

With 3 terms:
First term: 247
Second term: 472
Last term: 724

So each digit appears at each place once.
If the digits are a,b,c then sum is:
abc + bca + cab
= (a + b + c)(100 + 10 +1) = 111.(a+b+c)
111 = 37*3
So 37 is a potential answer.

To reason about it formally:
If I start with abc, sequence will be like:
abc
bc_
c__
..
..
__a
_ab
All the blanks can have any digit.
Again each original digit appears at each place once for sure.
Similarly, even if you add any digit which was not there originally, it will have to appear at least once at every place so that 'ab' can go from left to right.
So the earlier logic of 111 and hence 37 holds.
So answer = 37.

Q39. How many non-similar triangle have angles whose degree measures are distinct positive integers in arithmetic progression?
Solution:
a +  (a+d) + (a+2d) = 180
=> 3a + 3d = 180
=> a + d = 60
So middle term is 60.
Example sequence:
59,60,61
58,60,62
....
So total 59 such sequences are possible.
Answer = 59.

Q40.

 Let $a_1, a_2, \ldots$ be a sequence for which
$a_1 = 2$, $a_2 = 3$ and $a_n = \frac{a_{n-1}}{a_{n-2}}$ for each positive integer $n \geq 3$.

What is $a_{2006}$?

Solution:

 Looking at the first few terms of the sequence

$2, 3, \frac{3}{2}, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, 2, 3, \frac{3}{2} \ldots$

Clearly, the sequence repeats every 6 terms. Since $2006 \equiv 2 \pmod{6}$,
$a_{2006} = a_2 = 3$.

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