First term ,a=4 , common difference =4-7=-3, n =50
sum of first 50terms= (50/2)[2×4+(50-1)(-3)]
=25×[8+49]×-3
=25×57×-3
=25× -171
= -42925
derivation of the formula for the sum of n terms
Progression, S
S=a1+a2+a3+a4+...+an
S=a1+(a1+d)+(a1+2d)+(a1+3d)+...+[a1+(n−1)d] → Equation (1)
S=an+an−1+an−2+an−3+...+a1
S=an+(an−d)+(an−2d)+(an−3d)+...+[an−(n−1)d] → Equation (2)
Add Equations (1) and (2)
2S=(a1+an)+(a1+an)+(a1+an)+(a1+an)+...+(a1+an)
2S=n(a1+an)
S=n/2(a1+an)
Substitute an = a1 + (n - 1)d to the above equation, we have
S=n/2{a1+[a1+(n−1)d]}
S=n/2[2a1+(n−1)d]
Secant is the same as 1/cos so you can just write it as 1/2pi then you can put that into your calculator as 1/cos(2pi) and it equals 1.
Answer:
If you're talking about things like Algebra and Geometry, its Geometry
If you're not, can you be more specific ? :D
Answer:
<em>The voltage at the middle source is</em> 
Step-by-step explanation:
<u>Voltage Sources in Series</u>
When two or more voltage sources are connected in series, the total voltage is the sum of the individual voltages of each source.
The figure shown has three voltage sources of values:
The sum of these voltages is:
Operating:
We know the total voltage is 
, thus:
Equating the real parts and the imaginary parts independently:
4+a=6
1+b=-3
Solving each equation:
a = 2
b = -4
The voltage at the middle source is
