If there are two terms am and an such that √ (am * an) = 2 √ 2A1, then the minimum value of 1 / M + 4 / N?

If there are two terms am and an such that √ (am * an) = 2 √ 2A1, then the minimum value of 1 / M + 4 / N?

Let the common ratio of the sequence of equal ratio be q, then it is obtained from A7 = A6 + 2a5
a6*q=a6+2a6/q
Since an > 0, q = 1 + 2 / Q is obtained by dividing both sides of the above formula by A6
The solution is q = 2 or q = - 1
Because all the items are positive, q = 2
There are two terms am, an, such that √ (am * an) = 2 √ 2A1, so am * an = 8A1 ^ 2
That is, a1q ^ (m-1) * A1 * q ^ (n-1) = 8A1 ^ 2
SO 2 ^ (M + n-2) = 8
So m + n-2 = 3, so m + n = 5
Therefore, 1 / M + 4 / N = 1 / 5 * (M + n) * (1 / M + 4 / N) = 1 / 5 * (5 + 4m / N + n / M) > = 1 / 5 * (5 + 4) = 9 / 5
If and only if M = 5 / 3, n = 10 / 3, the equal sign holds
The above solution is wrong, because m, n are positive integers!!
A number minus two fifths of it is equal to seven fifths of it. What's the number
Let this number be X
X-2X/5=7/50
3X/5=7/50
X=7/50*5/3
X=7/30
Let this number be X
Then X-5 of 2x = 50 of 7
3x out of 5 = 7 out of 50
X = 7 / 50 △ 3 / 5 = 7 / 30
A: that's 7 out of 30
X-2X/5=7/50
How much is minus ninety-nine and fifteen sixteenth times minus eight?
=-99*(-8)+15/16*(-8)
=792-8.5
=783.5
1585/2
If the middle term of any two terms am, an is 4A1, then the minimum value of 1 / M + 4 / n
How much? How much
A7 = A6 + 2a5, A4 * q ^ 3 = A4 * q ^ 2 + 2a4q, divide both sides by a4q, get Q ^ 2-q-2 = 0, q = 2A1 * 2 ^ (m-1) * A1 * 2 ^ (n-1) = (4A1) ^ 2, divide both sides by (A1) ^ 2, get m + n = 6, so m = 6-n substitute 1 / M + 4 / N to get 1 / (6-n) + 4 / N, let this formula be t, that is t = 1 / (6-n) + 4 / N = (24-3n) / N * (6-n), get TN ^ 2
Six times the difference between a number minus 7 is equal to the sum of 2.4 times of this number plus 8.4
Let this number be X
Then 6 (X-7) = 2.4x + 8.4
6X-42=2.4X+8.4
3.6X=50.4
X=14
This number is 14
6〔x-7〕＝2.4x+8.4,x＝14
Let this number be X
(x-7)×6=2.4x+8.4
x=14
Let this number be X
(x-7)*6=2.4x+8.4
x=14
Let this number be x 6 * (X-6) = 2.4x + 8 * and a multiplier sign
Let's say this number is y
Then (Y-7) * 6 = 2.4 * y + 8.4
We get y = 14
Simple calculation of 2 / 98 * 99
98 * 2 / 99
= 98 x 2 / 99
= (99-1) x 2 / 99
= 99x2/ 99 - 2/99
= 2- 2/99
=1 and 97 out of 99
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If there are am and an, then am * an under the root sign is equal to twice the root
2A1, find the minimum value of 2 / M + 8 / n
Let the common ratio qa5q ^ 2 = q5q + 2q5, that is, Q ^ 2-q-2 = 0, the solution is q = 2 or q = - 1 (incompatible), and am * an is equal to 2 times the root 2A1 under the root sign, that is, am * an is equal to (root 8) A1 under the root sign, then am * an = 8a1a1 * 2 ^ (m-1) * A1 * 2 ^ (n-1) = 8A1 ^ 22 ^ (M + n-2) = 82 ^ (M + n) = 32, M + n = 5, because Mn is a positive integer, so when m = 2, n = 3
6 times of a number plus 2 equals 8 times of it minus 14 to find the number
10 points in one minute!
Let this number be X
6X+2=8X-14
8X-6X=14+2
2X=16
X=8
X=8
Let a number be X
6x+2=8x-14
2x=16
X=8
Calculation: 100-99 + 98-97 + 96-95 + +2-1．
The original formula = (100-99) + (98-97) + (96-95) + +（2-1）=1+1+… +1=50．
In {an}, a1 + A2 = 30, A3 + A4 = 60, then A7 + A8=
Two hundred and forty
A3 = A1 * q ^ 2, A4 = A2 * q ^ 2, so A3 + A4 = (a1 + A2) * q ^ 2 = 30 * q ^ 2 = 60
So Q ^ 2 = 2
Similarly, a7 = A1 * (Q ^ 2) ^ 3, a8 = A2 * q ^ 2) ^ 3
So A7 + A8 = (a1 + A2) * (Q ^ 2) ^ 3 = 30 * 2 ^ 3 = 30 * 8 = 240
(a7+a8)/(a3+a4)=((a3+a4)/(a1+a2))^2
a7+a8=240
If the sequence {an} is an equal ratio sequence, {an + a (n + 1)} is also an equal ratio sequence;
prove:
（an+a(n+1))/(a(n-10+an) =q(a(n-1)+an)/(a(n-1)+an)=q
So the sequence {an + a (n + 1)} is also an equal ratio sequence, and the common ratio is also the original common ratio;
A4 + A5 = 120, the sequence {an + a (n + 1)} is an equal ratio sequence with 30 as the first term and 2 as the common ratio,
a7+a8=30*2^(7-1)=1920