If in a geometric sequence, sum of #4^(th)# and #6^(th)# is #80# and sum of #7^(th)# and #9^(th)# is #640#, what is the first term and common ratio?

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Answer 1 · 2017-11-27T04:45:34

Common ratio is #2# and first term too is #2#.

If in a geometric progression, #a_1# is the first term and #r# is common ratio, then #n^(th)# term #a_n=a_1r^(n-1)#

Now sum of #4^(th)# and #6^(th)# is #80#, therefore

#a_4+a_6=80# or #a_1r^3+a_1r^5=80# i.e. #a_1r^3(1+r^2)=80#

Similarly as sum of #7^(th)# and #9^(th)# is #640#, therefore

#a_7+a_9=640# or #a_1r^6+a_1r^8=80# i.e. #a_1r^6(1+r^2)=640#

Dividing latter by former, we get

#(a_1r^6(1+r^2))/(a_1r^3(1+r^2))=640/80#

or #r^3=8# i.e. #r=2#

Putting this in #a_1r^3(1+r^2)=80#, we get

#a_1xx8xx5=80# or #a_1=2#