Asked by | 22nd Jun, 2009, 11:09: PM
Let the middle term be a.So it's given that a=20
Since out of 11 terms, the middle term has to be the 6th term, so it means that the 6th term is a=20.
Let the comon difference be d.
So the 5 terms before the 6th tem are
the 5terms after the 6th term will be
a+d, a+2d, a+3d, a+4d, a+5d
Since the sum of the 11 terms is
So the required sum is 11a=220.
Answered by | 23rd Jun, 2009, 01:11: AM
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