Algebra

239 soru

Soru 41Soru

Let A=(2113)A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} and B=(x21y)B = \begin{pmatrix} x & 2 \\ 1 & y \end{pmatrix} be 2×22 \times 2 matrices with integer entries xx and yy. If the matrix product ABAB is singular and det(A+B)=0\det(A + B) = 0, what is the value of x+yx + y?

Cevabı ve açıklamayı göster

Cevap: 3-3

Cevap

The value of x+yx + y is 3-3.
Since matrix ABAB is singular, det(AB)=det(A)det(B)=0\det(AB) = \det(A)\det(B) = 0. Evaluating det(A)=70\det(A) = 7 \neq 0, we find det(B)=xy2=0\det(B) = xy - 2 = 0, giving xy=2xy = 2. Computing A+B=(2+x303+y)A+B = \begin{pmatrix} 2+x & 3 \\ 0 & 3+y \end{pmatrix}, its determinant is (2+x)(3+y)=6+3x+2y+xy=0(2+x)(3+y) = 6 + 3x + 2y + xy = 0. Substituting xy=2xy = 2 yields 3x+2y=83x + 2y = -8. The integer pair satisfying both xy=2xy = 2 and 3x+2y=83x + 2y = -8 is x=2x = -2 and y=1y = -1. Therefore, x+y=3x + y = -3.

Adım Adım Çözüm

1
Calculate the determinant of matrix AA.
det(A)=(2)(3)(1)(1)=6+1=7\det(A) = (2)(3) - (1)(-1) = 6 + 1 = 7.
Knowing det(A)\det(A) helps simplify the condition det(AB)=0\det(AB) = 0 using determinant properties.
2
Use the singularity of ABAB to find a relationship between xx and yy.
det(AB)=det(A)det(B)=7(xy2)=0    xy=2\det(AB) = \det(A)\det(B) = 7(xy - 2) = 0 \implies xy = 2.
The determinant of a product of matrices equals the product of their determinants.
3
Form the matrix A+BA + B and compute its determinant.
A+B=(2+x303+y)    det(A+B)=(2+x)(3+y)(3)(0)=(2+x)(3+y)=6+3x+2y+xyA + B = \begin{pmatrix} 2+x & 3 \\ 0 & 3+y \end{pmatrix} \implies \det(A+B) = (2+x)(3+y) - (3)(0) = (2+x)(3+y) = 6 + 3x + 2y + xy.
Setting this determinant to zero gives a second equation involving xx and yy.
4
Substitute xy=2xy = 2 into det(A+B)=0\det(A+B) = 0 and solve for integer values of xx and yy.
6+3x+2y+2=0    3x+2y=86 + 3x + 2y + 2 = 0 \implies 3x + 2y = -8. Testing integer pairs (x,y)(x,y) for xy=2xy = 2 gives x=2x = -2 and y=1y = -1.
3(2)+2(1)=83(-2) + 2(-1) = -8 is satisfied only by x=2,y=1x = -2, y = -1 among integer pairs.
5
Compute x+yx + y.
x+y=2+(1)=3x + y = -2 + (-1) = -3.
This answers the question directly.

Anahtar Kavram

Determinants of matrix products and sums, matrix singularity, and integer solutions to matrix equations.
Soru 42Soru

If the matrix P=(x386)P = \begin{pmatrix} x & 3 \\ 8 & 6 \end{pmatrix} is a singular matrix, what is the value of xx?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of xx is 44.
For a 2×22 \times 2 matrix to be singular, its determinant must be zero. For matrix PP, the determinant is (x6)(38)=6x24(x \cdot 6) - (3 \cdot 8) = 6x - 24. Setting 6x24=06x - 24 = 0 gives 6x=246x = 24, which simplifies to x=4x = 4.

Adım Adım Çözüm

1
State the condition for a matrix to be singular.
A matrix is singular if its determinant is equal to zero: det(P)=0\det(P) = 0.
By definition, a singular matrix has a determinant equal to zero.
2
Calculate the determinant of matrix PP.
\det(P) = (x \times 6) - (3 \times 8) = 6x - 24.
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is calculated using the formula adbcad - bc.
3
Solve the linear equation 6x24=06x - 24 = 0 for xx.
6x = 24 \implies x = 4.
Adding 2424 to both sides and dividing by 66 isolates xx.

Anahtar Kavram

Singular matrix condition and 2×22 \times 2 determinant calculation
Soru 43Soru

Find the maximum integer value of mm for which the quadratic inequality x2mx+9>0x^2 - mx + 9 > 0 holds for all real values of xx.

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The maximum integer value of mm is 55.
For the quadratic expression x2mx+9x^2 - mx + 9 to remain strictly positive for all real values of xx, the quadratic curve must lie completely above the x-axis. Because the coefficient of x2x^2 is positive (1>01 > 0), this requires the discriminant to be strictly negative (D<0D < 0). Evaluating b24ac<0b^2 - 4ac < 0 gives m236<0m^2 - 36 < 0, which simplifies to 6<m<6-6 < m < 6. The largest integer strictly less than 66 is 55.

Adım Adım Çözüm

1
Determine the condition for the quadratic expression to be positive for all real values of xx.
Since the leading coefficient is 1>01 > 0, the condition is that the discriminant D=b24ac<0D = b^2 - 4ac < 0.
A parabola opening upward lies entirely above the horizontal axis when it has no real roots.
2
Calculate the discriminant using the coefficients of the quadratic expression.
D=(m)24(1)(9)=m236<0D = (-m)^2 - 4(1)(9) = m^2 - 36 < 0.
Here a=1a = 1, b=mb = -m, and c=9c = 9.
3
Solve the quadratic inequality for mm.
m236<0    6<m<6m^2 - 36 < 0 \implies -6 < m < 6.
The roots of m236=0m^2 - 36 = 0 are m=6m = -6 and m=6m = 6, and the expression is negative strictly between these boundary values.
4
Determine the maximum integer value within the open interval (6,6)(-6, 6).
The maximum integer value is 55.
The boundary value 66 is excluded by the strict inequality m<6m < 6.

Anahtar Kavram

Quadratic Inequalities and Discriminant Conditions for Positive Definiteness
Soru 44Soru

The sum of the first three terms of an arithmetic progression (AP) with a positive common difference dd is 2121. If 22 is added to the first term, 33 is added to the second term, and 99 is added to the third term, the resulting three numbers form consecutive terms of a geometric progression (GP). What is the sum of the first 1010 terms of this arithmetic progression?

Cevabı ve açıklamayı göster

Cevap: 210

Cevap

The sum of the first 10 terms of the arithmetic progression is 210.
Representing the AP terms as 7d,7,7+d7-d, 7, 7+d and adding the specified values produces GP terms 9d,10,16+d9-d, 10, 16+d. Solving 102=(9d)(16+d)10^2 = (9-d)(16+d) yields d=4d=4. Consequently, the first term of the AP is 33. Using S10=102[2(3)+9(4)]S_{10} = \frac{10}{2}[2(3) + 9(4)] gives the final answer 210.

Adım Adım Çözüm

1
Express AP terms symmetrically and solve for the middle term.
The middle term is a=7a = 7, making the terms 7d7-d, 77, and 7+d7+d.
Choosing terms ad,a,a+da-d, a, a+d allows the sum equation 3a=213a = 21 to directly isolate the middle term.
2
Set up the geometric progression relation to determine common difference dd.
The GP terms are 9d9-d, 1010, and 16+d16+d. Solving 102=(9d)(16+d)10^2 = (9-d)(16+d) gives d2+7d44=0d^2 + 7d - 44 = 0, yielding d=4d = 4.
In any geometric progression, the square of the middle term equals the product of the first and third terms.
3
Determine the first term a1a_1 and calculate S10S_{10}.
The first term is a1=74=3a_1 = 7 - 4 = 3, and the sum S10=102[2(3)+(101)(4)]=210S_{10} = \frac{10}{2}[2(3) + (10-1)(4)] = 210.
Applying the AP sum formula Sn=n2[2a1+(n1)d]S_n = \frac{n}{2}[2a_1 + (n-1)d] with n=10n=10, a1=3a_1=3, and d=4d=4.

Anahtar Kavram

Integrating AP and GP structural relationships to solve for sequence parameters and evaluate finite sums
Soru 45Soru

A binary operation \ast on the set of real numbers is defined by ab=a+b+5a \ast b = a + b + 5. What is the identity element of this operation?

Cevabı ve açıklamayı göster

Cevap: -5

Cevap

The identity element of the binary operation is 5-5.
By the definition of an identity element ee, the relation ae=aa \ast e = a must hold for all real numbers aa. Applying the rule ab=a+b+5a \ast b = a + b + 5 gives a+e+5=aa + e + 5 = a. Subtracting aa from both sides leads to e+5=0e + 5 = 0, which gives e=5e = -5.

Adım Adım Çözüm

1
Set up the identity element equation
ae=a    a+e+5=aa \ast e = a \implies a + e + 5 = a
By definition of an identity element, operating any element aa with ee yields aa.
2
Solve the equation for ee
e=5e = -5
Subtracting aa from both sides gives e+5=0e + 5 = 0, which yields e=5e = -5.

Anahtar Kavram

Identity Element of a Binary Operation
Soru 46Soru

Given the matrices A=(k231)A = \begin{pmatrix} k & 2 \\ 3 & 1 \end{pmatrix} and B=(1120)B = \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix}, what is the smaller positive integer value of kk for which det(A2+B)=4\det(A^2 + B) = 4?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The smaller positive integer value of kk is 5.
Evaluating A2A^2 via matrix multiplication gives (k2+62k+23k+37)\begin{pmatrix} k^2 + 6 & 2k + 2 \\ 3k + 3 & 7 \end{pmatrix}. Adding BB gives A2+B=(k2+72k+13k+57)A^2 + B = \begin{pmatrix} k^2 + 7 & 2k + 1 \\ 3k + 5 & 7 \end{pmatrix}. Computing det(A2+B)=(k2+7)(7)(2k+1)(3k+5)=k213k+44\det(A^2 + B) = (k^2 + 7)(7) - (2k + 1)(3k + 5) = k^2 - 13k + 44. Setting this to 4 results in k213k+40=0k^2 - 13k + 40 = 0, which yields roots k=5k = 5 and k=8k = 8. The smaller value is 5.

Adım Adım Çözüm

1
Compute matrix A2A^2 using matrix multiplication (row by column).
A2=(k231)(k231)=(k2+62k+23k+37)A^2 = \begin{pmatrix} k & 2 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} k & 2 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} k^2 + 6 & 2k + 2 \\ 3k + 3 & 7 \end{pmatrix}
Matrix multiplication requires taking the dot product of rows of the first matrix with columns of the second matrix.
2
Add matrix BB to A2A^2.
A2+B=(k2+62k+23k+37)+(1120)=(k2+72k+13k+57)A^2 + B = \begin{pmatrix} k^2 + 6 & 2k + 2 \\ 3k + 3 & 7 \end{pmatrix} + \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} k^2 + 7 & 2k + 1 \\ 3k + 5 & 7 \end{pmatrix}
Matrix addition is performed by adding corresponding entries.
3
Calculate the determinant of A2+BA^2 + B and set it equal to 4.
det(A2+B)=(k2+7)(7)(2k+1)(3k+5)=7k2+49(6k2+13k+5)=k213k+44=4\det(A^2 + B) = (k^2 + 7)(7) - (2k + 1)(3k + 5) = 7k^2 + 49 - (6k^2 + 13k + 5) = k^2 - 13k + 44 = 4
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is adbcad - bc.
4
Solve the quadratic equation for kk.
k213k+40=0    (k5)(k8)=0    k=5 or k=8k^2 - 13k + 40 = 0 \implies (k - 5)(k - 8) = 0 \implies k = 5 \text{ or } k = 8
Factoring the quadratic equation gives the two possible integer roots.

Anahtar Kavram

Matrix multiplication, matrix addition, determinant of a 2x2 matrix, and quadratic equation solving.
Soru 47Soru

The 2nd2^{\text{nd}}, 5th5^{\text{th}}, and 14th14^{\text{th}} terms of a non-constant arithmetic progression (A.P.) form three consecutive terms of a geometric progression (G.P.). If the first term of the A.P. is 33, what is the sum of the first 66 terms of the A.P.?

Cevabı ve açıklamayı göster

Cevap: 108

Cevap

108
With first term a=3a=3, the terms T2=3+dT_2 = 3+d, T5=3+4dT_5 = 3+4d, and T14=3+13dT_{14} = 3+13d form a geometric progression. Therefore, (3+4d)2=(3+d)(3+13d)(3+4d)^2 = (3+d)(3+13d). Expanding gives 9+24d+16d2=9+42d+13d29 + 24d + 16d^2 = 9 + 42d + 13d^2, which simplifies to 3d218d=03d^2 - 18d = 0. Since the sequence is non-constant (d0d \neq 0), d=6d = 6. Using the A.P. sum formula Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a + (n-1)d], S6=3[2(3)+5(6)]=3[6+30]=108S_6 = 3[2(3) + 5(6)] = 3[6 + 30] = 108.

Adım Adım Çözüm

1
Express the 2nd2^{\text{nd}}, 5th5^{\text{th}}, and 14th14^{\text{th}} terms of the A.P. in terms of the first term a=3a=3 and common difference dd.
T2=3+dT_2 = 3 + d, T5=3+4dT_5 = 3 + 4d, and T14=3+13dT_{14} = 3 + 13d.
The nthn^{\text{th}} term of an A.P. is given by Tn=a+(n1)dT_n = a + (n-1)d.
2
Set up the condition for consecutive terms of a G.P. and solve for dd.
(3+4d)2=(3+d)(3+13d)    9+24d+16d2=9+42d+13d2    3d218d=0    d=6(3 + 4d)^2 = (3 + d)(3 + 13d) \implies 9 + 24d + 16d^2 = 9 + 42d + 13d^2 \implies 3d^2 - 18d = 0 \implies d = 6.
For consecutive terms in a G.P., the middle term squared equals the product of the adjacent terms (T52=T2×T14T_5^2 = T_2 \times T_{14}).
3
Calculate the sum of the first 66 terms of the A.P. using a=3a=3 and d=6d=6.
S6=62[2(3)+(61)(6)]=3[6+30]=3×36=108S_6 = \frac{6}{2}[2(3) + (6-1)(6)] = 3[6 + 30] = 3 \times 36 = 108.
The sum formula for an A.P. is Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a + (n-1)d].

Anahtar Kavram

Simultaneous conditions connecting A.P. and G.P. terms combined with sequence sum formulas.
Tahmini Süre:2m 0s
Soru 48Soru

A binary operation \ast on the set of real numbers is defined by ab=a+b2aba \ast b = a + b - 2ab. If (x3)2=38(x \ast 3) \ast 2 = 38, find the value of xx.

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The value of xx is 3.
Applying the binary operation definition sequentially yields x3=35xx \ast 3 = 3 - 5x for the inner expression, and (35x)2=15x7(3 - 5x) \ast 2 = 15x - 7 for the composite expression. Equating 15x7=3815x - 7 = 38 leads to 15x=4515x = 45, giving x=3x = 3.

Adım Adım Çözüm

1
Evaluate the inner binary operation expression x3x \ast 3
x3=35xx \ast 3 = 3 - 5x
Apply the definition ab=a+b2aba \ast b = a + b - 2ab with a=xa = x and b=3b = 3.
2
Evaluate the outer binary operation (35x)2(3 - 5x) \ast 2
(35x)2=15x7(3 - 5x) \ast 2 = 15x - 7
Substitute the result from step 1 into the outer operation definition with a=35xa = 3 - 5x and b=2b = 2.
3
Set the resulting expression equal to 38 and solve the linear equation
x=3x = 3
Solve 15x7=3815x - 7 = 38 by adding 7 to both sides to get 15x=4515x = 45, then dividing by 15.

Anahtar Kavram

Nested composition of defined binary operations
Soru 49Soru

Find the positive value of xx that satisfies the simultaneous equations y=x+2y = x + 2 and y=x24y = x^2 - 4.

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The positive value of xx is 3.
Equating the linear equation y=x+2y = x + 2 and the quadratic equation y=x24y = x^2 - 4 yields x2x6=0x^2 - x - 6 = 0. Factorizing this quadratic equation gives (x3)(x+2)=0(x - 3)(x + 2) = 0, which yields roots x=3x = 3 and x=2x = -2. Selecting the positive value gives 3.

Adım Adım Çözüm

1
Equate the linear and quadratic equations
x+2=x24x + 2 = x^2 - 4
Since both expressions are equal to yy, set them equal to each other to solve for xx.
2
Rearrange into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0
x2x6=0x^2 - x - 6 = 0
Subtract xx and 22 from both sides of the equation.
3
Factorize the quadratic expression
(x3)(x+2)=0(x - 3)(x + 2) = 0
Find two factors of 6-6 that add up to 1-1, which are 3-3 and 22.
4
Determine the roots and select the positive value
x=3x = 3
Setting each factor to zero gives x=3x = 3 or x=2x = -2. Selecting the positive root yields 33.

Anahtar Kavram

Solving simultaneous linear and quadratic equations by substitution
Soru 50Soru

A binary operation \circ defined on the set of real numbers R{1}\mathbb{R} \setminus \{1\} is given by ab=a+baba \circ b = a + b - ab. What is the inverse of 33 under this operation?

Cevabı ve açıklamayı göster

Cevap: 32\frac{3}{2}

Cevap

The inverse of 33 under the given operation is 32\frac{3}{2}.
To find the inverse of 33 under the operation ab=a+baba \circ b = a + b - ab, we must first determine the identity element ee. Setting ae=aa \circ e = a yields a+eae=aa + e - ae = a, which simplifies to e(1a)=0e(1 - a) = 0. For all a1a \neq 1, the identity element is e=0e = 0. Next, using the definition of inverse 3x=03 \circ x = 0, we substitute into the operational formula to obtain 3+x3x=0    32x=0    x=323 + x - 3x = 0 \implies 3 - 2x = 0 \implies x = \frac{3}{2}. Thus, the option specifying 32\frac{3}{2} is correct.

Adım Adım Çözüm

1
Find the identity element ee of the operation \circ.
e=0e = 0
By definition of identity, ae=aa \circ e = a. Substituting into the operational formula gives a+eae=a    e(1a)=0a + e - ae = a \implies e(1 - a) = 0. Since a1a \neq 1, e=0e = 0.
2
Set up the inverse equation for 33, letting xx be the inverse of 33.
3x=03 \circ x = 0
By definition of an inverse element, aa1=ea \circ a^{-1} = e.
3
Expand 3x3 \circ x using the operation rule and solve for xx.
x=32x = \frac{3}{2}
3+x3x=0    32x=0    2x=3    x=323 + x - 3x = 0 \implies 3 - 2x = 0 \implies 2x = 3 \implies x = \frac{3}{2}.

Anahtar Kavram

Identity and Inverse Elements in Binary Operations
Soru 51Soru

The sum of the first nn terms of an arithmetic progression is given by Sn=2n2+3nS_n = 2n^2 + 3n. What is the value of the 7th7^{\text{th}} term of the progression?

Cevabı ve açıklamayı göster

Cevap: 29

Cevap

The 7th7^{\text{th}} term of the arithmetic progression is 2929.
For any sequence, the nn-th term is found using Tn=SnSn1T_n = S_n - S_{n-1}. Substituting n=7n = 7 gives S7=2(7)2+3(7)=119S_7 = 2(7)^2 + 3(7) = 119 and S6=2(6)2+3(6)=90S_6 = 2(6)^2 + 3(6) = 90. Thus, T7=11990=29T_7 = 119 - 90 = 29.

Adım Adım Çözüm

1
State the relationship between the nn-th term TnT_n and the sum of first nn terms SnS_n
Tn=SnSn1T_n = S_n - S_{n-1}
The sum of the first nn terms minus the sum of the first n1n-1 terms equals the nn-th term.
2
Calculate the sum of the first 7 terms (S7S_7)
S7=2(7)2+3(7)=119S_7 = 2(7)^2 + 3(7) = 119
Substitute n=7n = 7 into the sum formula Sn=2n2+3nS_n = 2n^2 + 3n.
3
Calculate the sum of the first 6 terms (S6S_6)
S6=2(6)2+3(6)=90S_6 = 2(6)^2 + 3(6) = 90
Substitute n=6n = 6 into the sum formula Sn=2n2+3nS_n = 2n^2 + 3n.
4
Compute the 7th7^{\text{th}} term (T7T_7)
T7=11990=29T_7 = 119 - 90 = 29
Subtract S6S_6 from S7S_7.

Anahtar Kavram

Relationship between the nth term and the sum of first n terms of an AP
Soru 52Soru

The sum of the first nn terms of an arithmetic progression (A.P.) is 210210. If the first term is 33 and the last term is 3939, what is the value of nn?

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The number of terms nn is 1010.
Using the standard sum formula Sn=n2(a+l)S_n = \frac{n}{2}(a + l) for an A.P. with known first term a=3a = 3 and last term l=39l = 39, we set 210=n2(3+39)=21n210 = \frac{n}{2}(3 + 39) = 21n. Solving for nn gives n=10n = 10.

Adım Adım Çözüm

1
Identify the given parameters of the arithmetic progression.
First term a=3a = 3, last term l=39l = 39, and sum Sn=210S_n = 210.
These values are required to apply the sum formula for an A.P.
2
Apply the sum formula Sn=n2(a+l)S_n = \frac{n}{2}(a + l).
210=n2(3+39)=n2(42)=21n210 = \frac{n}{2}(3 + 39) = \frac{n}{2}(42) = 21n.
The sum of nn terms in an A.P. with a known first and last term is given by n2(a+l)\frac{n}{2}(a + l).
3
Solve for nn.
n=21021=10n = \frac{210}{21} = 10.
Dividing the total sum by 2121 gives the exact number of terms.

Anahtar Kavram

Sum of an Arithmetic Progression using first and last terms
Tahmini Süre:1m 30s
Soru 53Soru

Given the simultaneous equations x+2y=7x + 2y = 7 and x2+3xy+y2=19x^2 + 3xy + y^2 = 19, find the product of all possible values of yy that satisfy the system.

Cevabı ve açıklamayı göster

Cevap: -30

Cevap

The product of all possible values of y that satisfy the system is -30.
Rearranging the linear equation gives x = 7 - 2y. Substituting this into the quadratic equation x^2 + 3xy + y^2 = 19 produces (7 - 2y)^2 + 3(7 - 2y)y + y^2 = 19. Expanding and combining like terms yields y^2 + 7y - 30 = 0. Solving for y gives y = 3 and y = -10. Multiplying these values together gives a product of -30.

Adım Adım Çözüm

1
Isolate x in the linear equation
x = 7 - 2y
Expressing one variable in terms of the other enables substitution into the quadratic equation.
2
Substitute x into the quadratic equation and expand
(7 - 2y)^2 + 3(7 - 2y)y + y^2 = 19
This reduces the system to a single quadratic equation in terms of y.
3
Simplify the resulting expression into standard quadratic form
y^2 + 7y - 30 = 0
Expanding yields (49 - 28y + 4y^2) + (21y - 6y^2) + y^2 = 19, which reduces to y^2 + 7y - 30 = 0.
4
Calculate the product of the roots of y
y_1 * y_2 = -30
By Vieta's formulas, the product of roots for y^2 + ay + b = 0 is b/1 = -30 (or factoring gives y = 3 and y = -10, with product 3 * (-10) = -30).

Anahtar Kavram

Solving simultaneous linear and quadratic equations via substitution and applying quadratic root properties
Tahmini Süre:2m 0s
Soru 54Soru

The sum of the first four terms of an arithmetic progression (A.P.) is 3232, and the sum of the next four terms is 9696. What is the common difference of the progression?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The common difference of the arithmetic progression is 44.
The sum of the first four terms yields 2a+3d=162a + 3d = 16. The sum of the first eight terms is 32+96=12832 + 96 = 128, which gives 2a+7d=322a + 7d = 32. Subtracting these two linear equations gives 4d=164d = 16, leading to d=4d = 4.

Adım Adım Çözüm

1
Formulate an equation for the sum of the first 4 terms.
2a+3d=162a + 3d = 16
The sum of the first nn terms of an A.P. is Sn=n2(2a+(n1)d)S_n = \frac{n}{2}(2a + (n-1)d). Substituting n=4n = 4 and S4=32S_4 = 32 gives 2(2a+3d)=322(2a + 3d) = 32, which simplifies to 2a+3d=162a + 3d = 16.
2
Formulate an equation for the sum of the first 8 terms.
2a+7d=322a + 7d = 32
The total sum of the first 8 terms is the sum of the first 4 terms plus the sum of the next 4 terms (S8=32+96=128S_8 = 32 + 96 = 128). Substituting n=8n = 8 gives 4(2a+7d)=1284(2a + 7d) = 128, which simplifies to 2a+7d=322a + 7d = 32.
3
Solve the system of simultaneous linear equations for dd.
d=4d = 4
Subtracting (2a+3d=16)(2a + 3d = 16) from (2a+7d=32)(2a + 7d = 32) eliminates 2a2a, resulting in 4d=164d = 16, which yields d=4d = 4.

Anahtar Kavram

Sum of an Arithmetic Progression
Soru 55Soru

If 2x+y=322^{x+y} = 32 and log3x+log3y=1+log32\log_3 x + \log_3 y = 1 + \log_3 2, what is the value of x2+y2x^2 + y^2?

Cevabı ve açıklamayı göster

Cevap: 1313

Cevap

The value of x2+y2x^2 + y^2 is 1313.
Expressing 3232 as 252^5 gives x+y=5x + y = 5. Converting 11 to log33\log_3 3 allows combining the right-hand side to log3(3×2)=log36\log_3(3 \times 2) = \log_3 6, so log3(xy)=log36    xy=6\log_3(xy) = \log_3 6 \implies xy = 6. Evaluating x2+y2=(x+y)22xyx^2 + y^2 = (x+y)^2 - 2xy yields 522(6)=2512=135^2 - 2(6) = 25 - 12 = 13.

Adım Adım Çözüm

1
Simplify the exponential equation using index laws.
x+y=5x + y = 5
Since 32=2532 = 2^5, 2x+y=25    x+y=52^{x+y} = 2^5 \implies x + y = 5.
2
Simplify the logarithmic equation using logarithm laws.
xy=6xy = 6
log3x+log3y=log3(xy)\log_3 x + \log_3 y = \log_3(xy) and 1+log32=log33+log32=log3(3×2)=log361 + \log_3 2 = \log_3 3 + \log_3 2 = \log_3(3 \times 2) = \log_3 6. Therefore, xy=6xy = 6.
3
Use the algebraic identity to find x2+y2x^2 + y^2.
x2+y2=13x^2 + y^2 = 13
Using (x+y)2=x2+y2+2xy(x+y)^2 = x^2 + y^2 + 2xy, we have 52=x2+y2+2(6)    25=x2+y2+12    x2+y2=135^2 = x^2 + y^2 + 2(6) \implies 25 = x^2 + y^2 + 12 \implies x^2 + y^2 = 13.

Anahtar Kavram

Solving simultaneous equations involving indices and logarithms using index laws and logarithmic identities.
Soru 56Soru

The 3rd3^{\text{rd}} and 6th6^{\text{th}} terms of an Arithmetic Progression (A.P.) are 1313 and 2828 respectively. What is the sum of the first 1010 terms of the progression?

Cevabı ve açıklamayı göster

Cevap: 255255

Cevap

The sum of the first 1010 terms of the progression is 255255.
The 3rd3^{\text{rd}} term is a+2d=13a + 2d = 13 and the 6th6^{\text{th}} term is a+5d=28a + 5d = 28. Subtracting these equations gives 3d=153d = 15, so d=5d = 5, which leads to a=3a = 3. Using the sum formula S10=102[2(3)+(101)5]S_{10} = \frac{10}{2}[2(3) + (10 - 1)5], we obtain 5(6+45)=2555(6 + 45) = 255.

Adım Adım Çözüm

1
Set up equations for the given terms using the nth term formula Tn=a+(n1)dT_n = a + (n - 1)d.
a+2d=13a + 2d = 13 and a+5d=28a + 5d = 28.
The 3rd3^{\text{rd}} term corresponds to n=3n=3 and the 6th6^{\text{th}} term corresponds to n=6n=6.
2
Solve the simultaneous equations for aa (first term) and dd (common difference).
Subtracting the first equation from the second gives 3d=15d=53d = 15 \Rightarrow d = 5. Substituting d=5d = 5 into the first equation yields a+2(5)=13a=3a + 2(5) = 13 \Rightarrow a = 3.
To find any property of an A.P., the first term aa and common difference dd must be determined.
3
Calculate the sum of the first 1010 terms using Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a + (n - 1)d].
S10=102[2(3)+(101)5]=5[6+45]=5(51)=255S_{10} = \frac{10}{2}[2(3) + (10 - 1)5] = 5[6 + 45] = 5(51) = 255.
Applying the formula for the sum of the first nn terms with n=10n = 10, a=3a = 3, and d=5d = 5.

Anahtar Kavram

Finding the sum of the first nn terms of an Arithmetic Progression given two specific terms.
Soru 57Soru

If log10(x)+log10(5)=2\log_{10}(x) + \log_{10}(5) = 2, what is the value of xx?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

20
Using the logarithmic product identity logb(A)+logb(B)=logb(AB)\log_b(A) + \log_b(B) = \log_b(AB), the left side simplifies to log10(5x)=2\log_{10}(5x) = 2. Converting to exponential form gives 5x=102=1005x = 10^2 = 100, which simplifies directly to x=20x = 20.

Adım Adım Çözüm

1
Apply the product law of logarithms
log10(5x)=2\log_{10}(5x) = 2
The sum of logarithms with the same base equals the logarithm of their product: logb(A)+logb(B)=logb(AB)\log_b(A) + \log_b(B) = \log_b(AB).
2
Convert the logarithmic equation to exponential form
5x=102=1005x = 10^2 = 100
By definition of logarithms, if logb(y)=c\log_b(y) = c, then y=bcy = b^c.
3
Solve for xx
x=1005=20x = \frac{100}{5} = 20
Divide both sides of the linear equation by 55.

Anahtar Kavram

Logarithmic Product Law and Exponential Conversion
Tahmini Süre:45s
Soru 58Soru

If 75+1227=k3\sqrt{75} + \sqrt{12} - \sqrt{27} = k\sqrt{3}, what is the value of kk?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of kk is 4.
Simplifying each surd into its basic form yields 75=53\sqrt{75} = 5\sqrt{3}, 12=23\sqrt{12} = 2\sqrt{3}, and 27=33\sqrt{27} = 3\sqrt{3}. Combining the coefficients gives (5+23)3=43(5 + 2 - 3)\sqrt{3} = 4\sqrt{3}. Equating 434\sqrt{3} to k3k\sqrt{3} shows that k=4k = 4.

Adım Adım Çözüm

1
Simplify each individual surd by factoring out perfect squares
75=53\sqrt{75} = 5\sqrt{3}, 12=23\sqrt{12} = 2\sqrt{3}, 27=33\sqrt{27} = 3\sqrt{3}
To combine surds through addition or subtraction, they must be converted to similar surds.
2
Combine the coefficients of the like surds
53+2333=(5+23)3=435\sqrt{3} + 2\sqrt{3} - 3\sqrt{3} = (5 + 2 - 3)\sqrt{3} = 4\sqrt{3}
Like terms with the same radical factor 3\sqrt{3} can be added and subtracted directly.
3
Compare the resulting coefficient with k3k\sqrt{3}
k=4k = 4
By direct comparison of coefficients of 3\sqrt{3}, kk equals 4.

Anahtar Kavram

Simplification and combining of similar surds
Soru 59Soru

Two 2×22 \times 2 matrices are given as A=(2143)A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix} and B=(102k)B = \begin{pmatrix} 1 & 0 \\ 2 & k \end{pmatrix}. If the determinant of the product matrix ABAB is equal to 3030, what is the value of kk?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The value of k is 3.
The determinant of a product of square matrices equals the product of their individual determinants: det(AB)=det(A)det(B)\det(AB) = \det(A) \cdot \det(B). Evaluating det(A)\det(A) yields (2)(3)(1)(4)=10(2)(3) - (-1)(4) = 10, and det(B)\det(B) yields (1)(k)(0)(2)=k(1)(k) - (0)(2) = k. Substituting these into det(AB)=30\det(AB) = 30 gives 10k=3010k = 30, leading directly to k=3k = 3.

Adım Adım Çözüm

1
Evaluate the determinant of matrix A
\det(A) = 10
Using the 2×22 \times 2 determinant formula det(abcd)=adbc\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc, we calculate det(A)=(2)(3)(1)(4)=6+4=10\det(A) = (2)(3) - (-1)(4) = 6 + 4 = 10.
2
Evaluate the determinant of matrix B
\det(B) = k
Calculating the determinant of matrix BB yields det(B)=(1)(k)(0)(2)=k\det(B) = (1)(k) - (0)(2) = k.
3
Use the product property of determinants to solve for k
k = 3
Since det(AB)=det(A)det(B)\det(AB) = \det(A) \cdot \det(B), we have 10k=3010k = 30. Dividing both sides by 10 yields k=3k = 3.

Anahtar Kavram

Determinant of Matrix Product
Tahmini Süre:1m 30s
Soru 60Soru

Given the matrices A=(231m)A = \begin{pmatrix} 2 & 3 \\ 1 & m \end{pmatrix} and B=(42)B = \begin{pmatrix} 4 \\ -2 \end{pmatrix}, if the matrix product AB=(26)AB = \begin{pmatrix} 2 \\ 6 \end{pmatrix}, what is the value of mm?

Cevabı ve açıklamayı göster

Cevap: 1-1

Cevap

The value of mm is 1-1.
Multiplying the 2×22 \times 2 matrix by the 2×12 \times 1 column vector yields a 2×12 \times 1 result where the second row entry is 1(4)+m(2)=42m1(4) + m(-2) = 4 - 2m. Setting this equal to the second row element 66 gives 42m=64 - 2m = 6, which simplifies to m=1m = -1.

Adım Adım Çözüm

1
Compute the matrix product ABAB
AB=(2(4)+3(2)1(4)+m(2))=(8642m)=(242m)AB = \begin{pmatrix} 2(4) + 3(-2) \\ 1(4) + m(-2) \end{pmatrix} = \begin{pmatrix} 8 - 6 \\ 4 - 2m \end{pmatrix} = \begin{pmatrix} 2 \\ 4 - 2m \end{pmatrix}
Matrix multiplication requires taking the dot product of each row of the first matrix with the column vector.
2
Equate the resulting vector to the given vector (26)\begin{pmatrix} 2 \\ 6 \end{pmatrix}
42m=64 - 2m = 6
Corresponding elements of equal matrices must be equal.
3
Solve the linear equation for mm
2m=642m=2m=1-2m = 6 - 4 \Rightarrow -2m = 2 \Rightarrow m = -1
Isolating the variable mm by subtracting 44 from both sides and dividing by 2-2.

Anahtar Kavram

Matrix Multiplication and Equality of Matrices
ÖncekiSayfa 3 / 12Sonraki
Algebra Alıştırma Soruları — JAMB UTME — Sayfa 3 | Examkin