Assume the total cost of a tertiary education will be R75 000 when your child enters university in 18 years. You presently have
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Well... we know that... there are 365 days in a year, unless is a leap-year, but we'll use 365 anyway
and each day has 24hrs, each hr has 60 minutes
so. let us use those ratios
so.. multiply and simplify, cancelling out any like-units atop and bottom
notice, all we do, is use the ratios, in a way, that if we need one unit to be changed, we flip the ratio
for example, to toss away "year" unit, since in the first fraction is at the bottom, then we put it on the top on the ratio, year/year = 1, effectively cancelling the unit
and each day has 24hrs, each hr has 60 minutes
so. let us use those ratios
so.. multiply and simplify, cancelling out any like-units atop and bottom
notice, all we do, is use the ratios, in a way, that if we need one unit to be changed, we flip the ratio
for example, to toss away "year" unit, since in the first fraction is at the bottom, then we put it on the top on the ratio, year/year = 1, effectively cancelling the unit
The given sequence is
a₁, a₂, ...,
Because the given sequence is an arithmetic progression (AP), the equation satisfied is
where
d = the common difference.
The common difference may be determined as
d = a₂ - a₁
The common difference is the difference between successive terms, therefore
d = a₃ - a₂ = a₄ - a₃, and so on..
The sum of the first n terms is
Example:
For the arithmetic sequence
1,3,5, ...,
the common difference is d= 3 - 1 = 2.
The n-th term is
For example, the 10-term is
a₁₀ = 1 + (10-1)*2 = 19
Th sum of th first 10 terms is
S₁₀ = (10/2)*(1 + 19) = 100
a₁, a₂, ...,
Because the given sequence is an arithmetic progression (AP), the equation satisfied is
where
d = the common difference.
The common difference may be determined as
d = a₂ - a₁
The common difference is the difference between successive terms, therefore
d = a₃ - a₂ = a₄ - a₃, and so on..
The sum of the first n terms is
Example:
For the arithmetic sequence
1,3,5, ...,
the common difference is d= 3 - 1 = 2.
The n-th term is
For example, the 10-term is
a₁₀ = 1 + (10-1)*2 = 19
Th sum of th first 10 terms is
S₁₀ = (10/2)*(1 + 19) = 100