the third term of an A.P is 1 and the 6th term is-11 . determine its 15th term and the term
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a+2d=1-------(I)
a+5d=-11-------(II)
Subtracting equation (II) from equation (I)
=a+2d-1-(a+5d+11)=0
=a+2d-1-a-5d-11=0
=-3d-12=0
=-3d=12
d=-4
put d=-4 in equation (I)
a+2×(-4)=1
a-8=1
a=9
therefore a=9 & d=-4
a15=a+(n-1)d
a15=9+(15-1)×(-4)
a15=9+14×(-4)
=9-56
=-47
So,15th term is -47
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a+5d=-11-------(II)
Subtracting equation (II) from equation (I)
=a+2d-1-(a+5d+11)=0
=a+2d-1-a-5d-11=0
=-3d-12=0
=-3d=12
d=-4
put d=-4 in equation (I)
a+2×(-4)=1
a-8=1
a=9
therefore a=9 & d=-4
a15=a+(n-1)d
a15=9+(15-1)×(-4)
a15=9+14×(-4)
=9-56
=-47
So,15th term is -47
Hope it is helpful.....
If you like my answer mark me as BRAINLIST and follow me .
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