If sum of first 6 terms of an ap is 36 and that of the first 16 terms is 256, find the sum of the first 10 terms.
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S6=6/2(2a+6-1d)
36=3(2a+5d)
2a+5d=12 eqn (1)
Again s16=16/2[2a+15d]
256=8(2a+15d)
2a+15d=32 eqn2
From 1&2 10d=20,d=2
Or a=1
Now s10=10/2(2×1+10-1×d)
=5(2+9×2)
=5×20
=100
36=3(2a+5d)
2a+5d=12 eqn (1)
Again s16=16/2[2a+15d]
256=8(2a+15d)
2a+15d=32 eqn2
From 1&2 10d=20,d=2
Or a=1
Now s10=10/2(2×1+10-1×d)
=5(2+9×2)
=5×20
=100
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