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:
A = 360
B= 180
C=270
Step-by-step explanation:
Sometimes the outlier, if it's too large, can throw off the mean, making it larger and smaller, so it isn't as accurate
sorry if this doesnt make sense if you need me to explain it more I will
A and D
should be the answers.
