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I think its impossible

Me amd a frind have spent an hour on this as question and im about to gube up because we hsve gotten NOWHERE please can someone explain 2(ii) :smile:received_1008883735863805.jpeg
Reply 1
Original post by Saffronwillow
Me amd a frind have spent an hour on this as question and im about to gube up because we hsve gotten NOWHERE please can someone explain 2(ii) :smile:received_1008883735863805.jpeg


Note that what you have is an arithmetic sequence with common difference -3 and first term 102.

Now you know the formula for the sum of the first n terms of an arithmetic sequence, don't you?
Original post by Zacken
Note that what you have is an arithmetic sequence with common difference -3 and first term 102.

Now you know the formula for the sum of the first n terms of an arithmetic sequence, don't you?


Yeah its 102+99+96=297
Reply 3
Original post by Saffronwillow
Yeah its 102+99+96=297


Huh? The sum to n terms is given by n2(2a+(n1)d)\frac{n}{2}(2a + (n-1)d). Just plug in a=102a = 102 and d=3d=-3.

For the value for which the sum if zero, just set the equation to 0 and solve for nn, ignoring non-positive solutions.
Original post by Zacken
Huh? The sum to n terms is given by n2(2a+(n1)d)\frac{n}{2}(2a + (n-1)d). Just plug in a=102a = 102 and d=3d=-3.

For the value for which the sum if zero, just set the equation to 0 and solve for nn, ignoring non-positive solutions.


Oooh, THANK YOU

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