How many terms of the progression 3,6,9,12,...must be taken at least to have a sum not less than 2000?
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We know that
An = a+(n-1)d
So here,
An=2000
a=3,
d=3
2000=3+(n-1)*3
=1997=(n-1)*3
1997/3=(n-1)
665=n-1
n=666
Therefore, there should be less than 666 terms to get the ans which is near to 2000
An = a+(n-1)d
So here,
An=2000
a=3,
d=3
2000=3+(n-1)*3
=1997=(n-1)*3
1997/3=(n-1)
665=n-1
n=666
Therefore, there should be less than 666 terms to get the ans which is near to 2000
Gobind11:
wrong
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