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, b and c
Step-by-step explanation:
in my opinion, I think its a, b and c...
Answer: 10
Step-by-step explanation:
Answer:
B. The solution is valid because all steps to solve the inequality for F are correct.
Step-by-step explanation:
F - 32 ≤ 0
Add 32 to both sides of the equation to have;
F -32 + 32 ≤ 0 + 32
F ≤ 0 + 32
F ≤ 32
It can be observed that to solve for F, the steps are correct. Thus the solution is valid. Therefore, the correct choice in the given question is the solution is valid because all steps to solve the inequality for F are correct.
Answer:
8
Step-by-step explanation:
