CBSE Class 10 Answered
arithmetic progression...
Asked by deepitapai | 08 Mar, 2009, 04:05: PM
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 | 08 Mar, 2009, 08:15: PM
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