1.2 Numbers and Algebra

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	M17/5/MATHL/HP1/ENG/TZ1/XX/7
Question Text	An arithmetic sequence $u₁, u₂, u₃$ has $u₁ = -2$ and common difference $d ≠ 0$ . (a) Given that $u₃, u₄$ and $u₆$ are the first three terms of a geometric sequence, find the value of $d$ ;
Total Mark	4
Correct Answer	2
Explanation	n/a
Mark Scheme	Step 1: Consider the given information Given information are $u₁= -2$ , and that $u₃, u₄$ and $u₆$ form a geometric sequence. Step 2: Combine the given information $u₃ =u₁ + 2d, u₄ = u₁+3d, u₆ = u₁ + 5d$ Step 3: We haven’t used the geometric sequence yet so, $u₃, u₄, u₆ = a, ar, ar²$ We can notice that, $(ar)²=a \times ar²\\ u₄²=u₃\times u₆$ Replacing the current variables what we determined in step 2, $(-2+3d)² = (-2+2d)(-2+5d)$ Expand and simplify $d²-2d = d(d-2)=0\\ d=0~or~2$ So, $d=2$
Question Text	(b) Given that $uₙ=18$ , where n is a positive integer, determine the value of $\sum_{r=1}^{n}u_{r}$ .
Total Mark	3
Correct Answer	88
Explanation	n/a
Mark Scheme	Step 1: Point out the given information, $d = 2$ from part (a), $u_{n}=18$ and $u_{1}=-2$ Step 2: Combine the given information to find useful information $u_{n} =u_1+(n-1)d$ $18 = -2 + (n - 1) \times 2,$ so $n = 11$ Step 3: Refer back to the question: determine the value of $\sum_{r=1}^{n}u_{r}$ $\sum_{r=1}^{11}u_{r}=sum$ of the first 11 terms of the arithemtic sequence, given by the formula, $\frac{n(u_{1}+u_{n})}{2}$ $\sum_{r=1}^{11}u_{r}=\frac{11(18-2)}{2}=88$ Answer: 88

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	M17/5/MATHL/HP1/ENG/TZ2/XX/3
Question Text	The 1st, 3rd, and 6th terms of an arithmetic sequence, with common difference $d, d≠0$ , are the first three terms of a geometric sequence, with common ratio r. Given that the 1st term of both sequences is 4. (a) Find the value of $d$ .
Total Mark	4
Correct Answer	1
Explanation	n/a
Mark Scheme	Step 1: Consider the given information. Given information are $u_{1}=4$ and $r_{1}=4$ and that $u_{1}, u_3$ and $u_{6}$ from a geometric sequence. Step 2: Combine the given information. $u_{1}=4, u_3=4+2d, u_6=4=5d$ If we use the information that the 1st, 3rd, and 6th terms of the arithmetic sequence form a geometric sequence, $u_{1}\times d=u_{3}\\ u_{3} \times d=u_{6}$ Then, we can attempt to find the value of $d$ . $\frac{4+2d}{4} = \frac{4+5d}{4+2d}$ Expand and simplify. $(4+2d)(4+2d) = 4(4+5d)\\ d = 1~or~0$ Since $d\neq 0, d=1$ Answer = 1
Question Text	(b) Find the value of $2r$ .
Total Mark	1
Correct Answer	3
Explanation	n/a
Mark Scheme	Step 1: Point out the given information, $d=1 from part (a) and a=4.$ Step 2: Combine the given information. Use the value of d to find the first two terms of the geometric sequence. $a = u₁= 4\\ ar = u₃= 4 + 2(1) = 6$ Hence, we can find $r$ by using ratios. $r=\mathrm{\frac{ar}{a}}=\frac{6}{4}=\frac{3}{2}$ Hence, $2r = 3$

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	N16-TZ0-P1-6(HL)
Question Text	The sum of the first n terms of a sequence $\left\{u_{n} \right\}$ is given by $Sₙ=4n²-5n$ , where $n∈Z^+$ . (a) Write down the value of $u₁$ .
Total Mark	1
Correct Answer	1
Explanation	n/a
Mark Scheme	$u₁= 1$
Question Text	(b) Find the value of $u₇$ .
Total Mark	2
Correct Answer	47
Explanation	n/a
Mark Scheme	$u₇= S₇- S₆= 4(72) - 5(7) - 4(62) + 5(6) = 47$ Answer: 47
Question Text	(c) Given that {uₙ} is an arithmetic sequence, clearly state its common difference.
Total Mark	4
Correct Answer	8
Explanation	n/a
Mark Scheme	Step 1: Consider the given information and identify what we need to prove. If we assume un to be an arithmetic sequence, we can find the expression for the arithmetic sequence by using the following expression. $uₙ= Sₙ - Sₙ₋₁$ Step 2: Combine the given information. Substitute the sum of the arithmetic sequence expression into $Sₙ$ and $Sₙ₋₁$ . $uₙ = (4n²-5n) - (4(n-1)²-5(n-1)) = (4n²-5n) - (4n²-8n+4-5n+5) = 8n+1$ Now, we can find the common difference using the arithmetic sequence. $d = uₙ₊₁ - uₙ = 8(n+1)+1-(8n+1) = 8(constant)$

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	M16-TZ1-P1-1(HL)
Question Text	The seventh term of an arithmetic sequence is equal to 3 and the sum of the first 16 terms is 24. (a) Find the common difference.
Total Mark	3
Correct Answer	-1
Explanation	n/a
Mark Scheme	Step 1: Consider the given information. We know that the 7th term of the arithmetic sequence is 3 and the sum of the first 16 terms is 24. $u₇ = u₁+ 6d = 3 S₁₆= (16/2)(2u₁+(16-1)d) = 8(2u₁+15d) = 24$ Step 2: Combine the given information. We can combine the two equations by expressing u₁ in terms of $d$ . $u₁ = 3 - 6d S₁₆ = 8(2(3-6d) + 15d) = 24 d = -1$ Answer: -1
Question Text	(b) Find the first term.
Total Mark	3
Correct Answer	9
Explanation	n/a
Mark Scheme	Now, we can use the value of $d$ to find the first term. $u₁ = 3 - 6(-1) = 9$ Answer: 9

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	M15-TZ1-P1-12(HL)
Question Text	Let $\left\{u_{n} \right\},n\in Z^{+}$ , be an arithmetic sequence with first term equal to $\alpha$ and common difference of $d$ , where $d\neq 0$ . Let another sequence $\left\{v_{n} \right\},n\in Z^{+}$ , be defined by $v_{n}=3^{u_{n}+2_{n}}$ . (a) (i) Given that $\frac{v_{n+1}}{v_{n}}$ is a constant and can be expressed in terms $3^{d+k}$ , find the value of $k$ . Total Mark: 2 Correct Answer: 2 Explanation: n/a Mark Scheme: $\frac{v_{n+1}}{v_{n}}=\frac{3^{u_{n+1}+2(n+1)}}{3^{u_{n}+2n}}=3^{u_{n+1}+2n+2-u_{n}-2n}=3^{d+2}$ (ii) Write down the first term of the sequence $\left\{ v_{n}\right\}$ . (a) $3^{a}$ (b) $3^{a+2}$ (c) $3^{2a}$ (d) $3^{2a+2}$ Total Mark: 1 Correct Answer: (b) Explanation: n/a Mark Scheme: x (iii) Write down a formula for $vₙ$ in terms of a, d and n. (a) $3^{ad+2n}$ (b) $3^{a+(n-1)d+n}$ (c) $3^{a+nd+2n}$ (d) $3^{a+(n-1)d+2n}$ Total Mark: 1 Correct Answer: (d) Explanation: n/a Mark Scheme: x
Question Text	(b) Let {Sₙ} be the sum of the first n terms of the sequence {vₙ}. (i) Find {Sₙ}, in terms of a, d and n. (a) $\frac{3^{a+3}\left ( (3^{d+3})^{n}-1 \right )}{3^{d+3}-1}$ (b) $\frac{3^{a+2}\left ( (3^{d+2})^{n}-1 \right )}{3^{d+2}-1}$ (c) $\frac{3^{a+2}\left ( (3^{d+1})^{n}-1 \right )}{3^{d+2}-1}$ (d) $\frac{3^{a+2}\left ( (3^{d+2})^{n}-1 \right )}{3^{d+2}-1}$ Total Mark: 1 Correct Answer: (d) Explanation: n/a Mark Scheme: x (ii) Find range of d for which $\sum_{i=1}^{\infty} v_{i}$ exists. (a) $d>2$ (b) $d<2$ (c) $d<-2$ (d) $d<-2$ Total Mark: 3 Correct Answer: (c) Explanation: n/a Mark Scheme: For the sum of infinity to exist, the following condition must be satisfied. $-1<3^{d+2}<1$ $-\frac{1}{9}<3^{d}<\frac{1}{9}$ $3^{d}<3^{-2}$ $d<-2$ (iii) You are now told that $\sum_{i=1}^{\infty }v_{i}$ does exist and is denoted by $S_{\infty }$ . (a) $\frac{3^{a+2}}{3^{d+2}-1}$ (b) $\frac{3^{d+2}}{1-3^{a+2}}$ (c) $\frac{3^{a+2}}{3^{d+2}}$ (d) $\frac{3^{a+2}}{1-3^{d+2}}$ Total Mark: 1 Correct Answer: (d) Explanation: n/a Mark Scheme: Substitute u₁ and r into S∞ equation, as \|r\| is given to be less than 1. (iv) Given that $S_{\infty } =\frac{27}{2}(3^{a})$ find the value of $d$ . Total Mark: 3 Correct Answer: -3 Explanation: n/a Mark Scheme: Utilize your answer from (iii) and substitute it into the equation. $\frac{3^{a+2}}{1-3^{d+2}}=\frac{27}{2}(3^{a})$ $\frac{3^{2}}{1-3^{d+2}}=\frac{27}{2}$ $\frac{1}{1-3^{d+2}}=\frac{3}{2}$ $3-3^{d+3}=2$ $d=-3$

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	M15-TZ1-P1-12(HL)
Question Text	Let $wₙ, n\in Z^{+}$ , be a geometric sequence with first term equal to p and common ratio $q$ , where $p$ and $q$ are both greater than zero. Let another sequence $zₙ$ be defined by $zₙ = log⁡(wₙ)$ . Find $\sum_{\iota =1}^{n}z_{\iota }$ , giving your answer in the form $Ink$ with $k$ in terms of $n, p$ and $q$ . (a) $log(p^n{q^{n}})$ (b) $log(p^n{q^{\frac{(n-1)}{2}}})$ (c) $log(p^n{q^{\frac{n}{2}}})$ (d) $log(p^n{q^{\frac{n(n-1)}{2}}})$
Total Mark	6
Correct Answer	(d)
Explanation	n/a
Mark Scheme	We can express $z_{n}$ in terms of $p$ and $q$ . $w_{n}=pq^{n-1}$ $z_{n}=logpq^{n-1}$ $z_{n}=logp+(n-1)logq$ Using this expression, we can find the first term and common difference of the { $z_{n}$ } sequence. $z_{n+1}-z_{n}=(logp+nlogq)-(logp+(n-1)logq)=logq$ Hence, { $z_{n}$ } is an arithmetic sequence with the first term being $logp$ and the common difference being $logq$ Finally, we can find the sum of the arithmetic sequence as below. $\sum_{i=1}^{n}z_{i}=\frac{n}{2}(2logp+(n-1)logq)=n\left ( logp+logq^{\left ( \frac{n-1}{2} \right )} \right )=nlog\left ( pq^{\left ( \frac{n-1}{2} \right )} \right )=log\left ( p^{n}q^{\frac{n(n-1)}{2}} \right )$ or $\sum_{i=1}^{n}z_{i}=logp+logpq+logpq^{2}+...+logpq^{n-1}=log\left ( p^{n}q^{(1+2+3+...+(n-1))} \right )=log\left ( p^{n}q^{\frac{n(n-1)}{2}} \right )$ Answer: (d)

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	M14/5/MATHL/HP1/ENG/TZ1/X/13
Question Text	A geometric sequence $uₙ$ , with complex terms, is defined by $uₙ₊₁ = (1 - i)uₙ ~and~ u₁=4$ . (a) Find the fourth term of the sequence, giving your answer in the form $x + yi , (x,y ∈ R)$ .
Total Mark	2
Correct Answer	$-8-8i$
Explanation	n/a
Mark Scheme	METHOD 1 Solve the fourth term by following through the sequence. $u₁ = 4\\ u₂ = 4 - 4i\\ u₃ = - 8i\\ u₄ = - 8 - 8$ $i$ METHOD 2 Calculate the fourth term by understanding the pattern. $u₄ = 4(1 - i)³ = - 8 - 8i$ Answer: $-8-8i$
Question Text	(b) Find the sum of the first 20 terms of $\left\{ u_{n }\right\}$ . Giving your answer in the form $a×(1+2ᵐ)$ , find the sum of $a$ and $m$ .
Total Mark	3
Correct Answer	14
Explanation	n/a
Mark Scheme	Use the summation formula for geometric sequences (formula booklet) $S_{20}=\frac{4((1-i)^{20}-1)}{i}$ $=\frac{4((-2i)^{10}-1)}{i}$ $=\frac{4(2^{10}-1)}{i}$ $=4(2^{10}+1)$ Answer: 14
Question Text	(c) Given that $v_n$ is a geometric sequence, its first term can be expressed as $p(1 - i)ᵏ((1 - i)²)ⁿ⁻¹.$ Find the value of $n$ .
Total Mark	2
Correct Answer	16
Explanation	n/a
Mark Scheme	Rearrange the expression for $vₙ$ to find the general formula of the geometric sequence. $vₙ = (4(1 - i)ⁿ⁻¹(4(1-i)ⁿ⁻¹⁺ᵏ)\\ = 16(1-i)ᵏ(1-i)²ⁿ⁻²\\ = 16(1 - i)ᵏ(1 - i)²ⁿ⁻¹(=16(1 - i)ᵏ(-2i)ⁿ⁻¹)$ Hence, the value of p is 16. Answer: 16
Question Text	A third sequence $w_n$ is defined by $wₙ = \| uₙ - uₙ₊₁ \|$ . (d) Given that $wₙ$ forms a geometric sequence with a common ratio of √k, state the value of $k$ . (a) $2(\sqrt{2})^{n}$ (b) $2(\sqrt{2})^{n+1}$ (c) $4(\sqrt{2})^{n}$ (d) $4(\sqrt{2})^{n+1}$
Total Mark	4
Correct Answer	(d)
Explanation	n/a
Mark Scheme	Find the general formula of the geometric sequence. $wₙ = \|4(1 - i)ⁿ⁻¹-4(1 - i)ⁿ\|\\ = 4\|1 - i\|ⁿ⁻¹\|1 - 1 - i\|\\ = 4\|1 - i\|ⁿ⁻¹\\ = 4(√2)ⁿ⁻¹$ Answer: (d)

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	M14/5/MATHL/HP1/ENG/TZ2/XX/9
Question Text	The first three terms of a geometric sequence are $cosx, sin2x$ , and $(-\frac{\pi }{2}<x<\frac{\pi }{2})$ (a) Find the common ratio $r$ .
Total Mark	2
Correct Answer	$2sin(x)$
Explanation	n/a
Mark Scheme	To find the common ratio of the geometric sequence, we can divide the second term by the first term. $r=\frac{sin(2x)}{cos(x)}=\frac{2sin(x)cos(x))}{cos(x)}=2sin(x)$
Question Text	(b) Find the set of values of x for which the geometric series $sinx +sin2x +2 sin²x cosx+…$ converges. (a) $-\frac{\pi}{2}<x<-\frac{\pi }{6} ~or~ \frac{\pi }{6}<x<\frac{\pi }{2}$ (b) $-\frac{\pi}{2}<x<-\frac{\pi }{4} ~or~ \frac{\pi }{4}<x<\frac{\pi }{2}$ (c) $-\frac{\pi}{4}<x<-\frac{\pi }{6} ~or~ \frac{\pi }{6}<x<\frac{\pi }{4}$ (d) $-\frac{\pi}{3}<x<-\frac{\pi }{6} ~or~ \frac{\pi }{6}<x<\frac{\pi }{3}$
Total Mark	4
Correct Answer	(a)
Explanation	n/a
Mark Scheme	For the geometric series to converge, the absolute value of the common ratio should be smaller than 1.\|r\|<1 $\|r\|<1 \\\|2sin~x\|<1$ or $-1 < r < 1 \\-1 < 2sin~x <1$ then $0<sinx<\frac{1}{2}~for-\frac{\pi }{6}<x<\frac{\pi }{6}\\ -\frac{\pi }{2} <x<-\frac{\pi }{6}or\frac{\pi }{6}<x<\frac{\pi }{2}$ Answer: (a)
Question Text	(c) Given that the sum to infinity of this series is $\frac{1}{2(t-\sqrt{15})}$ , find the value of $t$ .
Total Mark	4
Correct Answer	2
Explanation	n/a
Mark Scheme	Use the sum of infinite geometric series formula to find the value of $t$ $S_{\infty }=\frac{cosx}{1-2sinx}=\frac{cos(arcos(\frac{1}{4}))}{1-2sin(arccos(\frac{1}{4}))}=\frac{\frac{1}{4}}{1-\frac{\sqrt{15}}{2}}=\frac{1}{2(2-\sqrt{15})}$ Hence, the value of $t$ is equal to 2. Answer: 2

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	N13-TZ0-P1-7(HL)
Question Text	The sum of the first two terms of a geometric series is 20 and the sum of the first four terms is 40. (a) Find the value of $r²$ .
Total Mark	4
Correct Answer	4
Explanation	n/a
Mark Scheme	Step 1: Consider the given information. The question has provided us the sum of the first two terms and sum of the first tour terms. We can express the given information as an equation. $a+ar=10\\a+ar+ar^2+ar^3=50$ Stop 2: Combine the given information. We can combine the two equation to find the value of $r^2$ . $ar^2+ar^3=40 \\r^2(a+ar)=40\\ r^2=4$ Answer: 4
Question Text	(b) Given $r=2,$ (i) the first term will be in the form $\frac {x}{3}$ . Find $x$ . Total Mark: 1 Correct Answer: 10 Explanation: n/a Mark Scheme: To find the first term, we need to find the value of a using the given $r$ value. $a+2a=10 \\a=\frac{10}{3} \\x=10$ Answer: 10 (ii) find the sum of the first ten terms. Total Mark: 2 Correct Answer: 3410 Explanation: n/a Mark Scheme: We can use the sum of geometric series to find the value of $S_{10}$ . $S_{10}=\frac{10}{3}\left ( \frac{2^{10}-1}{2-1} \right )=3410$ Answer: 3410

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	M13-TZ1-P1-8(HL)
Question Text	The first terms of an arithmetic sequence are $\frac{1}{log_{3}x},\frac{1}{log_{27}x},\frac{1}{log_{243}x},\frac{1}{log_{2187}x},…$ Find $x$ if the sum of the 20 terms of the sequence is equal to 100.
Total Mark	6
Correct Answer	27
Explanation	n/a
Mark Scheme	Tip 1: Find the common difference of the arithmetic sequence. $d=\frac{1}{log_{27}x}-\frac{1}{log_{3}x}=\frac{log_{3}27}{log_{3}x}-\frac{log_{3}3}{log_{3}x}=\frac{2}{log_{3}x}$ Tip 2: Find the value of $x$ by using the given information that the sum of the first 12 terms is equal to 48. The sum of the first 12 terms in the arithmetic sequence can be found by using the following formula. $S_{12}=\frac{12}{2}(2\times \frac{1}{log_{3}x}+11\times \frac{2}{log_{3}x})=48\\ \frac{144}{log_{3}x} =48 \\log_{3}x=3$ Hence, we can find $x$ . $x=3^3=27$ Answer: 27

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	M13-TZ2-P1-6(HL)
Question Text	A geometric sequence has first term $a$ , common ratio r and sum to infinity 36. A second geometric sequence has first term $a$ , common ratio $r³$ and sum to infinity 16. Given that the value of $r$ can be expressed as $r=\frac{\sqrt{x}-1}{2}$ find the value of $x$ .
Total Mark	7
Correct Answer	26
Explanation	n/a
Mark Scheme	Step 1: Consider the given information. We can express the first series as $\frac{a}{1-r}=36$ and the second series as $\frac{a}{1-r^{3}}=16$ Step 2: Combine the given information. To solve for the value of $r$ , we can attempt eliminate $a$ by using the substitution method. $\frac{36(1-r)}{1-r^{3}}=16$ Simplify and obtain $4r^{2}+4r-5=0 \\r=\frac{-1\pm \sqrt{6}}{2}$ Hence, the value of $r$ is equal to $r=\frac{\sqrt{6}-1}{2}$ . Answer: 6

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	M19/5/MATHL/HP1/ENG/TZ2/XX/1
Question Text	In an arithmetic sequence, the sum of the 4th and 6th term is -2. Given that the sum of the first eight terms is 8, determine the sum of the first term and the common difference.
Total Mark	1
Correct Answer	11
Explanation	n/a
Mark Scheme	Attempt to from two equation involving the first term and the common difference to find their values. (1) $\left( u_{1}+3d \right )+(u_{1}+5d)=-2\\ 2u_{1}+8d=-2$ (2) $\frac{8}{2}\left [ 2u_{1}+7d \right ]=8\\ 2u_{1} +7d=2$ Hence, the first term and common difference can be determined as the following $u_{1}=15, d=-4$ The sum of the two terms is equal to 11. Answer: 11

Topic	1.2 Numbers and Algebra
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source	M18/5/MATHL/HP1/ENG/TZ2/XX/5
Question Text	The geometric sequence $u₁, u₂, u₃,…$ has a common ratio $r$ . Consider the sequence $A=\left\{a_{n}=log_{3}\|u_{n}\|:n\in Z^{+} \right\}$ (a) Given that $A$ is an arithmetic sequence, state its common difference $d$ in terms of $r$ . (a) $log_3\|r\|$ (b) $log_3\|r+1\|$ (c) $log_3\|3r\|$ (d) $log_3\|r^2\|$
Total Mark	4
Correct Answer	(a)
Explanation	n/a
Mark Scheme	Tip 1: Find an expression for the common difference $d$ . Using the given information, the arithmetic sequence’s common difference can be expressed as $d=a_{n+1} -a_{n}\\ =log_{3}\left\| u_{n+1}\right\|-log_{3}\left\|u_{n} \right\|\\ =log_{3}\left\| \frac{u_{n+1}}{u_{n}}\right\|$ Tip 2: Use the information that $u_1, u_2, u_3$ is a geometric sequence. Since we know that $u_n$ is a geometric sequence, we can express the common difference in terms of $r$ . $d=log_{3}\|r\|$
Question Text	(b) Find the value of $d$ .
Total Mark	3
Correct Answer	-1
Explanation	n/a
Mark Scheme	We are given the information that the sum to infinity of the geometric sequence is equal to 4. $\frac{8}{1-r}=9\\ r=\frac{1}{3}\\ d=log_{3}\|\frac{1}{3}\|=-1$ Answer: -1 $\underline{..................................................................}$

1.2 Numbers and Algebra

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