What are 180 ° - 126 ° 43 ′ 12 ″ 23 ° 25 ″ * 4 quarter * 33 ° 22 ′ 36 ° 35 ″ + 48 ° 56 ′ 23 ″ - 23 ° 59 ′ 48 ″

What are 180 ° - 126 ° 43 ′ 12 ″ 23 ° 25 ″ * 4 quarter * 33 ° 22 ′ 36 ° 35 ″ + 48 ° 56 ′ 23 ″ - 23 ° 59 ′ 48 ″

180°-126°43′12′′=53°16′48′′
23°25〃*4 =92°1′40′′
Quarter * 33 ° 22 ′ = 8 ° 20 ′ 30 ′
36°35〃+48°56′23〃-23°59′48〃 =60°57′10′′

It is known that in the equal ratio sequence {an}, a1 + an = 66, A2 times a (n-1) = 128, and the first n terms and Sn = 126, then the values of N and Q are?

A2 * a (n-1) = A1 * an = 128 and a1 + an = 66
The solution is A1 = 2 an = 64 or A1 = 64 An = 2
When A1 = 2 and an = 64
The sum formula is 126 = (2-64q) / (1-Q)
We get q = 2, n = 6
When A1 = 64 and an = 2
There are 126 = (64-2q) / (1-Q)
Q = 1 / 2, n = 6
☆⌒_ - [hope to help you~

In the equal ratio sequence {an}, a1 + an = 66, a2an-1 = 128, Sn = 126, then the value of n is ()
A. 5B. 6C. 7D. 8

According to Weida's theorem, we can deduce that A1 and an are two of the equations x2-66x + 128 = 0, ∧ A1 = 2 & nbsp; and an = 64, so QN-1 = 32; or A1 = 64 and an = 2, so QN-1 = 132

22 times 33 divided by 66

22=2*11
33=3*11
22*33=2*11*3*11=6*11*11=66*11
So: 22 * 33 / 66 = 66 * 11 / 66 = 11