It is known that the tolerance of the arithmetic sequence is equal to the common ratio of the arithmetic sequence, and they are all equal to D. It is also known that A1 = B1, A3 = 3B3, A5 = 5b5, find an, BN

It is known that the tolerance of the arithmetic sequence is equal to the common ratio of the arithmetic sequence, and they are all equal to D. It is also known that A1 = B1, A3 = 3B3, A5 = 5b5, find an, BN

a1=b1=1,d=±1.
An = ± n, BN = 1 or the nth power of - 1
In other words, the sequence an is a positive and negative natural number, and BN is a constant sequence of 1 or an equal ratio sequence of positive and negative 1 alternately. Tolerance = common ratio = ± 1. In this way, both sequences can meet the requirements of the topic
From an = a1 + (n-1) d; BN = B1 * d ^ (n-1); A1 = B1, A3 = 3B3, A5 = 5b5;
The results show that A1 = B1; a1 + 2 * d = 3 * B1 * (d ^ 2); a1 + 4 * d = 5 * B1 * (d ^ 4); a1 + 4 * d = 5 * B1 * (d ^ 4);
The solution is A1 = B1 = 2 / (5 (d ^ 3) - 3D);
Substituting in the general formula, we can get an, BN.
The tolerance of the sequence is not equal to 3, and the ratio of a to B is not equal to 5
Finding an and BN
a1=1 .1
a3=3b3,a1+2d=3*b1*d^2.2
a5=5b5,a1+4d=5*b1*d^4.3
From 1,2,3, we get
D ^ 2 = 1 / 5 or D ^ 2 = 1,
Because D is greater than 0 and not equal to 1
So d = √ 5 / 5
a1=b1=-√5
an=n*√5/5-6√5/5
bn= -√5*(√5/5)^(n-1)
If the tolerance D of the arithmetic sequence {an} is not equal to 0, and a1.a3.a5. Constitutes the arithmetic sequence, what is the common ratio
Let an = a1 + (n-1) d
A1.a3.a5
a3*a3=a1*a5
(a1+2d)^2=a1*(a1+4d）
The solution is d = 0, please check the question
Arithmetic sequence {a}_ In n}, if the tolerance D ≠ 0, A1, A3 and A9 are in equal proportion sequence, then (a1 + a3 + A9) / (A2 + A4 + A10)=_________
(a1 + 2D) / a = (a + 8D) / (a + 2D) gives A1 = D, an = n * D
（a1+a3+a9)/(a2+a4+a10)=13/16
It's very simple, isn't it? But I can't help it when I'm in bed. Just follow the routine steps^
50 Mixed fractions here?
Don't be too hard, it's better to be in the fifth grade
1、 2/3÷1/2－1/4×2/5 2、 2－6/13÷9/26－2/33、 2/9＋1/2÷4/5＋3/8 4、 10÷5/9＋1/6×45、 1/2×2/5＋9/10÷9/20 6、 5/9×3/10＋2/7÷2/57、 1/2＋1/4×4/5－1/8 8、 3/4×5/7×4/3－1/29、 23－8/9×1/27÷1...
What is rational number in Mathematics
In mathematics, all numbers are divided into rational numbers and irrational numbers
That is to say, any number is either rational or irrational
Irrational number refers to the infinite non cyclic decimal
Rational numbers are numbers other than irrational numbers, that is, numbers other than infinite acyclic decimals
Integers and fractions are called rational numbers
Generally speaking, rational numbers include integers, finite decimals and infinite cyclic decimals. Infinite non cyclic decimals are not rational numbers,
Root 2 is not
Integers and fractions are called rational numbers
The power supply voltage is constant, R1 = R2 = 6 Ω, when the switches S1 and S2 are open, the ammeter indication is 0.5 A; when the switches S1 and S2 are closed, the ammeter indication is 1.5 A, the power supply voltage = (), the resistance of resistance R3 = (); when the switches S1 and S2 are closed, the total resistance of the circuit = ()
I estimate three resistors in parallel
S1, S1 closed, current flow through R1, R2 is 1.5-0.5 = 1a, then 1A * 3 Ω = 3V
R3 = 3V / 0.5A = 6ohm
Total resistance = 6 / 3 = 2 ohm
69+69×99 540÷36
69+69×99
=69×（1+99）
=6900
540÷36
=540÷9÷4
=60÷4
=15
1: Original formula = 69 * (99 + 1) = 69002: original formula = 60 / 4 = 30 / 2 = 15~~~
69+69×99=69*（1+99）=69*100=6900
540/36=15
Hope it can help you, please accept
69+69x99
=69x(99+1)
=69x100
=6900
540÷36
=540÷4÷9
=60÷4
=15
69+69×99=69×(99+1)=69×100=6900
540÷36=540÷(6×6)=540÷6÷6=90÷6=15
I hope I can help you. O(∩_ ∩)O
The oral calculation of decimal addition and subtraction
1.5+6.8= 2.6-1.3= 45.6-2.6= 29.8+3.2= 56.3-22.5= 48.2-5.69= 23.6+6.6= 5.5+2.3= 23.6+9.5= 5.3+10.2= 21-2.3= 23.2-21.6= 12.3+56.2=21.9+14= 6.14+3.83= 7.93-1.64= 5.23-2.36= 3.44+3.36= 7.74-5.85= 9.35-4.7...
The rule of adding bracket in mathematics of grade two
(a)=a
-(a)=-a
-(-a)=a
When the - sign is in front of the bracket, the + sign inside the bracket changes to the - sign, and the - sign changes to the + sign. (when encountering the X sign, △ sign does not change)
When the + sign is in front of the bracket, the + sign and - sign in the bracket remain unchanged. (when encountering the X sign, △ sign does not change)
Hope to help you.