arithmetic progression...

Asked by deepitapai | 8th Mar, 2009, 04:05: PM

Expert Answer:

nth term of an A.P. with first term as a and common diff as d is given by,

nth term=a+(n-1)d

So acc to qn,

a+2d=15

So,

 a=15-2d

sum of n terms of an A.P. with first terms as a and common diff as d is given by,

sum=n/2[2a+(n-1)d]

So,

 acc to qn,

10/2[2a+9d]=125

2a+9d=25

2(15-2d)+9d=25

30-4d+9d=25

5d=-5

d= -1

so

a=15-2d=17

so, tenth term=

a+9d

=17+9(-1)

=17-9

=8

 

Answered by  | 8th Mar, 2009, 08:15: PM

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