高数问题~帮我解答下~~又加了两题~

LATER

find the value of a and d(arithmetic progression)
T2+T3=29
S6-S4=71

更新~
1）given that 3,m,n,192 are the first four consecutive terms of a geometric progression.find the three consecutive terms which added up to 16128～～
   2）the price of a new car is RM80000.It is given that the price of the car depreciates at a constant rate of 5% yearly.Calculate the minimum number of years required for      the price of the car to drop to less than RM45000.

caviiy

问题就这样？？？

LATER

回复 2# caviiy


    又编辑过了~~

xlwenlx

Tn=a+(n-1)d           Sn=n/2 [2a+(n-1)d]
T2=a+d                    S6=6a+15d
T3=a+2d                  S4=4a+6d
T2+T3=2a+3d         S6-S4=2a+9d
T2+T3=29                S6-S4=71
2a+3d=29                2a+9d=71
2a=29-3d--------1    2a=71-9d-----------2

2-1 :          subs d=7 into 1:
42-6d=0   2a=29-3(7)
6d=42       2a=8   
d=7              a=4

对吗？

LATER

回复 4# xlwenlx


    答案对了~谢谢~
还有一题~能不能帮忙解解看~
the sum of the first terms of an a.p is given by Sn=3n2-5n.find the first term

suki89

a=4, d=7.

xlwenlx

回复 5# LATER

没有a或d吗？

LATER

回复 7# xlwenlx


    没有哦~~

LATER

回复 6# suki89


    对了~

南湘楚

find the value of a and d(arithmetic progression)
T2+T3=29
S6-S4=71
LATER 发表于 2012-3-22 05:07 PM

Tn=[a+(n-1)d]
T2=[a+(2-1)d]
     =a+d
T3=[a+(3-1)d]
     =a+2d

(a+d)+(a+2d)=29
2a+3d=29----------1st equation

Sn=n/2[2a+(n-1)d]
S6=6/2[2a+(6-1)d]
     =6a+15d
S4=4/2[2a+(4-1)d]
     =4a+6d

6a+15d-(4a+6d)=71
                  2a+9d=71-----------2nd equation

2nd equation-1st equation:
9d-3d=71-29
      6d=42
        d=7
suds d=7 into 1st equation:
2a+3(7)=29
   a=4