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What is the sum of a 6-term geometric series if the first term is 21 and the last term is 352,947?
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so it means a_1=21 and a_6=352947 you have: \[\LARGE Sn=a_1\cdot {1-r^n \over 1-r }\] but you first need to find r so here we go: \[\LARGE a_n=a_1\cdot r^{n-1}\] \[\LARGE a_6=21\cdot r^{6-1}\] \[\LARGE 352947=21\cdot r^5\] \[\LARGE {352947\over 21 }= r^5\] \[\LARGE 16807= r^5\] now find here r ... I can't do it without calculator :( as soon as you do it. substitute here: \[\LARGE Sn=a_1\cdot {1-r^n \over 1-r }\] \[\LARGE S_6=a_1\cdot {1-r^6 \over 1-r }\] Good Luck! :)
so is 16807 r?
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