How do you find #S_n# for the geometric series #a_1=343#, #a_n=-1#, r=-1/7?

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Answer 1 · 2017-07-10T12:34:24

#S_n = 300#

#a_1=343 , a_n= = -1 , r= -1/7#

Given that

#a_n = "last term"#

#a_1 = "first term"#

#r = -1/7#

#T_n = a_1 cdot r^(n-1)#

Recall that #T_n = a_n#

Hence #-># #a_n = a_1 cdot r^(n-1)#

#a_n= a_1*r^(n-1) or 343 * (-1/7)^(n-1) = -1 or (-1/7)^(n-1) = -1/343 # or

#(-1/7)^(n-1) = (-1/7)^3 :. n-1 = 3 or n =4 #

#S_n = a_1 * ( r^n -1)/(r-1) = 343 * ((-1/7)^4-1 ) /(-1/7-1)#

# =343*(1/2401-1)/(-8/7) = 343 * (-2400/2401)* (-7/8) #

#= cancel343* 300/cancel343 =300#

#S_n = 300# [Ans]