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Find the sum of the series 1 + 7/5 + 13/25 + 19/125 ... to infinite terms.
a) 25/8
b) 2
c) 15/2
d) 12/5
e) 4
a) 25/8
b) 2
c) 15/2
d) 12/5
e) 4
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17 Sep 2013, 06:28
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shameekv
Sum of infinite GP = a/(1-r) where a is the first term, and r, the common ratio and |r|<1.
S = 1+7/5+13/25+19/125...
S/5 = 1/5+7/25+13/125...
Subtract : \(4S/5 = 1+6/5+6/25+6/125+.... = 1+6(1/5+1/25+1/125+...) = 1+[6*(1/5)]/[1-1/5] = 1+3/2 = 5/2\)
Thus, S = 5*5/4*2 = 25/8
A.
16 Apr 2023, 19:50
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