1.)
The first term in arithmetic progression is (a)=32
Since the 5th term in the arithmetic progression is 22 so we can write:
t5=22⇒a+(5−1)d=22 [∵ tn=a+(n−1)d ( Where a = first term and d = common difference) ]
⇒a+4d=22⇒32+4d=22⇒4d=22−32
⇒d=−410⇒d=−2.5
Let us assume that number of terms in this progression be n
Now:
Since the last term of this progression is −28 so we can write:
tn=−28⇒a+(n−1)d=−28⇒32+(n−1)(−2.5)=−28 [ ∵a=32andd=−2.5 ]
⇒n−1=−2.5−28−32⇒n−1=24⇒n=25
Hence:
The sum of all terms in this progression is:
Sn=2n(a+l) [ Where a = first term and l = last term ]
=225(32−28)=50
2.)
Since the amount of money each year increased by 2.5% of the amount of money allocated in the previous year so each year added a fixed amount money of the money allocated in the previous year and allocated money from 2005to2014 forms an arithmetic progression.
Now:
Amount of money allocated in 2005 is $2000 which is also a first term (a) in this progression.
Number of terms in this progression is:
n= number of years from 2005to2014=10
Amount of money allocated in 2006 is:
=$(2000+2000×2.5%)=$2050
So, common difference in this progression is:
d=2050−2000=50 [ ∵ Common difference in an arithmetic progression (d) = next term - previous term ]
Hence:
The total amount of money allocated from 2005to2014 inclusive is:
=2n(2a+(n−1)d)
=210(2(2000)+(10−1)50)=$22250
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