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]
Answer: 1.25
Step-by-step explanation:
Given: A college-entrance exam is designed so that scores are normally distributed with a mean
= 500 and a standard deviation
= 100.
A z-score measures how many standard deviations a given measurement deviates from the mean.
Let Y be a random variable that denotes the scores in the exam.
Formula for z-score = 
Z-score = 
⇒ Z-score = 
⇒Z-score =1.25
Therefore , the required z-score = 1.25
y = -
x - 34
the equation of a line in slope-intercept form is
y = mx + c ( m is the slope and c the y-intercept )
to calculate m use the gradient formula
m = ( y₂ - y₁ ) / (x₂ - x₁ )
with (x₁, y₁ ) = (- 68, 0 ) and (x₂, y₂ ) = (0, - 34 )
m =
=
= - 
the y-intercept = (0, - 34 ) ⇒ c = - 34
y= -
x - 34
2a. speed= distance
---------------
time
60÷69 = 52.1739 miles per hour.
so for the next one with the airplane flies 296 km in 276 min is the same formula.
296/276 = 39.9839 mph.
I couldn't see the first question clear...sorry but apply same formula.
Hoping this will help you