in an AP a3 =21,a10=63 find a100
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Answered by
1
a10=63
10×10=100
a3=21
a=21/3
a=7
a100=7×100
=700
10×10=100
a3=21
a=21/3
a=7
a100=7×100
=700
Answered by
1
a3=3 and a10=63
let a3=a+2d=21 eq first
and
a10=a+9d=63 eq second
by elimination method
we get d=6
from eqn first
a+2×6=21
a+12=21
a=21-12
a=9
now a100=a+99d
=9+99×6
=9+594
=603
let a3=a+2d=21 eq first
and
a10=a+9d=63 eq second
by elimination method
we get d=6
from eqn first
a+2×6=21
a+12=21
a=21-12
a=9
now a100=a+99d
=9+99×6
=9+594
=603
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