Answer:
900,000
Step-by-step explanation:
I'm not sure how to explain rounding, but this should be the answer if ur asking for rounding to the nearest hundred thousand and not the hundredth thousand
The answer is No it is not
Answer:
100
Step-by-step explanation:
We have the sum of first n terms of an AP,
Sn = n/2 [2a+(n−1)d]
Given,
36= 6/2 [2a+(6−1)d]
12=2a+5d ---------(1)
256= 16/2 [2a+(16−1)d]
32=2a+15d ---------(2)
Subtracting, (1) from (2)
32−12=2a+15d−(2a+5d)
20=10d ⟹d=2
Substituting for d in (1),
12=2a+5(2)=2(a+5)
6=a+5 ⟹a=1
∴ The sum of first 10 terms of an AP,
S10 = 10/2 [2(1)+(10−1)2]
S10 =5[2+18]
S10 =100
This is the sum of the first 10 terms.
Hope it will help.
5 is the difference.
Note: Minuend - Subtrahend = Difference
For t=0
<span>Allan borrows--------------------------- > 1870 dollars
for t=6 years
</span>F1 = P*(1 +(r/m))^n
i=r/m
n=m*t---------- >1*6=6
we have
P1=1870
r=8%
m=1
t=6 years
F1 = 1870*(1 +(0.08/1))^6------------------ >2967.45 dollars
for t=2
Allan borrows--------------------------- > 1240 dollars
for t=6 years
F2 = P2*(1 +(r/m))^n
i=r/m
n=m*t---------- >1*4=4
we have
P2=1240
r=8%
m=1
t=4 years------------> (6-2)=4 years
F2 = 1240*(1 +(0.08/1))^4------------------ >1687 dollars
F1+F2=2967.45+1687=4654.45 dollars
the answer is 4654.45 dollars