Given that the first term of the arithmetic sequence an is a, the tolerance is B, {BN} is the arithmetic sequence with the first term B and the common ratio a, if A1 = B1, A2 = B2, find the general formula of an and BN Given that the first term of the arithmetic sequence {an} is a, the tolerance is B, {BN} is an arithmetic sequence with the first term of B and the common ratio of A. if A1 = B1, A2 = B2, find the general formula of {an}, {BN}

Given that the first term of the arithmetic sequence an is a, the tolerance is B, {BN} is the arithmetic sequence with the first term B and the common ratio a, if A1 = B1, A2 = B2, find the general formula of an and BN Given that the first term of the arithmetic sequence {an} is a, the tolerance is B, {BN} is an arithmetic sequence with the first term of B and the common ratio of A. if A1 = B1, A2 = B2, find the general formula of {an}, {BN}

A1 = a, B1 = B, A1 = B1, then B = a
a2=a+b b2=ab
A2 = B2, then a + B = AB, B = a is substituted
2a=a²
a(a-2)=0
A = 0 (equal ratio sequence, common ratio not equal to 0, rounding off) or a = 2
b=a=2
an=a+b(n-1)=2+2(n-1)=2n
bn=ba^(n-1)=2×2^(n-1)=2ⁿ
The general term formula of sequence {an} is an = 2n; the general term formula of sequence {BN} is BN = 2 & # 8319
A1 = B1, i.e. a = b
And A2 = B2, that is, a + B = b * a, 2A = a ^ 2, a = 2 = B
Therefore:
an=2n
bn=2^n
A = B, a + B = Ba is solved
A1 = a, B1 = B, A2 = a + B, B2 = AB, then:
A = B and a + B = ab
That is: 2A = AB, a = 0 [rounding off] or B = 2, thus a = b = 2
an=2n，bn=2^n
a1=a=b1=b
a2=a+b=ab=b2
So a = b = 2 or a = b = 0
So an = 2n, BN = 2 Λ n or an = BN = 0
Because A1 = B1, a = B
an=a+(n-1)a=an
bn=a*a^（n-1）=a^n
And A2 = B2
2a=a^2
A=2
an=2n bn=2^n
In the arithmetic sequence {an} with tolerance D (D ≠ 0) and the arithmetic sequence {BN} with common ratio Q, we know that A1 = B1 = 1, A2 = B2, a8 = B3, and find the value of D and P
The formula of sequence of numbers with equal difference and equal ratio
a1=b1=1
a2=1+d
a8=1+7d
b2=q
b3=q*q
therefore
1+7d=q*Q
1+d=q
have to
D=5
Q=6
As shown in the figure, the center angle of the circle is 30 degrees, and the radii are 1, 3, 5, 7 The area of the shadow of the sector is S1, and the area of the shadow of the sector is S2
So what is S50
I found the answer, but I don't understand,
Please give me a solution,
S1=[π*(3^2)/12]-[π*(1^2)/12]=2*π/3S2=[π*(7^2)/12]-[π*(5^2)/12]=2*πS3=[π*(11^2)/12]-[π*(9^2)/12]=10*π/3…… The general formula is Sn = {π * [(4 * n-1) ^ 2),] / 12} - {π * [(4 * n-3) ^ 2),] / 12}
Let the point (m, n) move on the image of the line x + y = 1 in the first quadrant, then the maximum value of log2m + log2n is______ .
∵ point (m, n) moves on the image of line x + y = 1 in the first quadrant ∵ m + n = 1, M > 0, n > 0, ∵ log2m + log2n = log2 (MN) ≤ log2 (M + N2) 2 = log22-2 = - 2, if and only if m = n = 12 "=" holds. So the answer is: - 2
As shown in the figure, the center angle of the circle is 30 degrees, and the radii are 1, 3, 5, 7 The area of the shadow part is recorded as S1,
As shown in the figure, the center angle of the circle is 30 degrees, and the radii are
1、3、5、7…… The fan-shaped shape, shadow
The areas of the parts are S1, S2, S3, S4, respectively
Then S50 = (the result remains Π) graph
T
Subtract the area of triangle o1ab from the area of sector o1ao2b and multiply by 2
Where ∠ ao1b = 120 degree
Can log (4) 6 be reduced to log (2) 3? And how much is log (3) x = 1 / 3
log(4)6
=log4(2*3)
=log4(2)+log4(3)
=1/2+1/2*log2(3)
Log (3) x = 1 / 3 x = 3 times root (3)
As shown in the figure, the semicircle areas on the three sides of a right angle triangle from small to large are recorded as S1, S2 and S3 in turn, then the relationship among S1, S2 and S3 is______ .
Let the diameters of the three semicircles be: D1, D2, D3, S1 = 12 × π × (D12) 2 = d128 π, S2 = 12 × π × (D22) 2 = d228 π, S3 = 12 × π × (D32) 2 = d328 π. From the Pythagorean theorem, it can be concluded that: D12 + D22 = D32, ｜ S1 + S2 = π 8 (D12 + D22) = d328 π = S3
Log (a) - x =? Can it be simplified?
The beauty of Mathematics
What does your log (a) - x mean?
Is a the logarithm of base-x?
This formula can't be simplified any more
It is known that the area of triangle ABC, triangle DCE and triangle CEF are S1, S2 and S3 respectively. When S1 = 4 and S2 = 5, S3=
These three triangles are regular triangles, and B, C, e, f are on the same line, a, D, G are on the same line
In the title "known triangle ABC, triangle DCE, triangle CEF" should be "known triangle ABC, triangle DCE, triangle EFG". According to the meaning of the title, the triangle GED is similar to the triangle DCA, and Ge: de = DC: AC. because the triangle ABC, triangle DCE, and triangle EFG are regular triangles, they are both similar
Simplify log one third bottom 36
process
Original formula = log 3 ^ (- 1) bottom 36
=- log3 bottom (9 * 4)
=- log3 bottom 9-log3 bottom 4
=- log3 bottom 3 ^ 2 - log3 bottom 2 ^ 2
=- 2 * log3 bottom 3-2 * log3 bottom 2
=- 2 - 2 * log3 bottom 2