How do you find #a_1# for each geometric series #S_n=33#, #a_n=48#, r=-2?

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Answer 1 · 2017-12-03T02:20:56

#a_1=3#

For geometric series, #S_n=a_1*(1-r^n)/(1-r)# and #a_n=a_1*r^(n-1)#

Hence #a_1*[1-(-2)^n]/[1-(-2)]=33# or #a_1*(1-r^n)=99# and #a_1*(-2)^(n-1)=48#

After dividing first equation to second one,

#[a_1*(1-(-2)^n)]/[a_1*(-2)^(n-1)]=99/48#

#(1-(-2)^n)/(-2)^(n-1)=33/16#

#16*(1-(-2)^n)=33*(-2)^(n-1)#

#16-16*(-2)^n=-33/2*(-2)^n#

#-1/2*(-2)^n=16#

#(-2)^n=-32#, so #n=5#

Thus,

#a_1*(-2)^(5-1)=48#

#16a_1=48#, so #a_1=3#