Isochromatic In the known equal ratio sequence {an}, A3 = 16, common ratio q = 1 / 2 Q: if the sum of the first n terms of the sequence {an} is Sn = 124, find the value of n The first n terms of {an} and Sn = - 2n ^ - n are known Ask: seek a1 + a3 + A5 + +The value of a25

Isochromatic In the known equal ratio sequence {an}, A3 = 16, common ratio q = 1 / 2 Q: if the sum of the first n terms of the sequence {an} is Sn = 124, find the value of n The first n terms of {an} and Sn = - 2n ^ - n are known Ask: seek a1 + a3 + A5 + +The value of a25

A3 = 16, so A2 = 32, A1 = 64
sn=a1+a2+a3+a3*[1-(1/2)^(n-3)]
64+32+16+16*[1-(1/2)^(n-3)]=124
(1/2)^(n-3)=1/4
n-3=2
n=5

Two mathematical problems of equal difference and equal ratio
In the arithmetic sequence, S4 = 21, the last four terms of S = 67, Sn = 286, find n?
In the equal ratio sequence, A1 times A2 times A3 = 2, a (n-2) a (n-1) an = 4, A1 times A2 times A3 times A4 Multiply an = 64, find n?

A: n = 26
It is known that the sum of the first four terms is 21, the sum of the last four terms is 67, and the sum of the first n terms is 286
A1+A2+A3+A4=21.(1)
A(n-3)+A(n-2)+A(n-1)+An=67.(2)
(1) + (2)
(A1+A2+A3+A4)+[A(n-3)+A(n-2)+A(n-1)+An]=88
(A1+An)+[A2+A(n-1)]+[A3+A(n-2)]+[A4+A(n-3)]=88
In arithmetic sequence
∵(A1+An)=[A2+A(n-1)]=[A3+A(n-2)]=[A4+A(n-3)]
∴4*(A1+An)=88
(A1+An)/2=11
Sn=[(A1+An)/2]*n
It is known that Sn = 286
286=11n
n=26

In the 220 V voltage and 220 V capacitor, use resistance to limit the charging voltage to only 10 V?

It is only a matter of time. The length of time is related to the value of resistance. It is not the voltage of only 10V

a/xy .b/yz
c/a-b .1/(b-a)^2
1/2a .1/5a^2b
b/a(b+1) a/b(b+1)
2a/2a+1 4(2a-1)/4a^2-4a+1
a-1/(a+1)^2-4 1-a/2-4a+2a^2

a/xy .b/yz =ab/zxy^2
c/a-b .1/(b-a)^2=-c/(b-a)^3
1/2a .1/5a^2b=1/10a^3b
b/a(b+1) a/b(b+1)=1/(b+1)^2
2A / 2A + 1 4 (2a-1) / 4A ^ 2-4a + 1 = wrong

How many currents can 35 square soft copper wires admit per square when the voltage is 660 / 1140 v
660 V and 1140 V, 55 kW motor, find 35 square soft copper wire per square can withstand how much current, maximum can drive how many kilowatts

The current multiplied by the voltage is equal to the electric power. The standard copper core wire of one square meter can bear three 55kW motors when the voltage is 668v, or the starting current is large
Thank you!

How to calculate the distance on the graph when the actual distance is known, and how to calculate the actual distance when the distance on the graph is known?

This requires the scale of this image
If the scale is 1: X
The actual distance is 1 △ 1: x = x times of the distance on the graph
And the distance on the graph is 1 / X of the actual distance

As shown in the figure, the power supply voltage is 10V, and the resistance value of resistance R is 10 Ω. After the circuit is connected, it will be within 100s
This is different from other questions on the Internet. Don't copy it. The picture shows a resistor R connected in series with a lamp L and a power supply

It should be 9V-------

Among these perfect square numbers, the last two digits are 25

Only the last two digits are 25
So 35, 45, 55, 65, 75, 85, 95 satisfy the condition, if you want the formula,

When the load resistance of a power supply is 4 Ω and 9 Ω, the output power of the external circuit of the power supply is equal, what is the internal resistance of the power supply?

[E/(r+4)]^2*4=[E/(r+9)]^2*9
To calculate,

1 234 56789 10 11 12 13 15 16 what is the law? What is the sum of the numbers in line 15?

1 234 56789 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 …… Obviously, the last number in each row is a complete square number, and its arithmetic square root is the number of rows. Therefore: the first number in the nth row = (n-1) ^ 2 + 1 = n ^ 2 - 2n + 2 the last number in the nth row = n ^ 2 how many numbers in the nth row = n ^ 2 - (n-1) ^ 2 = 2n - 1 the sum of the numbers in the nth row = (n ^ 2 - 2n + 2 + n ^ 2) * (2n - 1) / 2 = (n ^ 2 - N + 1) * (2n - 1) = 2n ^ 3 - 3N ^ 2 + 3N - 1 = n ^ 3 + (n - 1) ^ 3