Find the sum to infinity of the following arithmetico-geometrico sequence
1) 1, 2/4, 3/16, 4/64,...
2)3, 6/5, 9/25, 12/125, 15/625,...
3) 1, -4/3, 7/9, -10/27,...
PLEASE TELL ME THE ANSWER...
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Answers
Answer:
1. let S=1+2/4+3/16....... be equation no.1
this is in AGP,
r be ratio of this AGP
r=1/4, according to question
now multiply S by r that gives you
1/4S=1/4+2/16+3/64....... be equation no. 2
subtract 1 and 2 according to their denominators
so that
S=1+2/4+3/16.....
1/4S=. 1/4+2/16+......
3/4S= 1+1/4+1/16...... be equation no.3
now equation 3 is in GP with common ratio 1/4 and first term is 1
3/4S= 1/(1-1/4) {sum of infinite
3/4S=1/(3/4). terms in GP=
3/4S=4/3. a/1-r}
S=4/3x4/3=16/9
similarly you can do second and third questions of your own
hope you are happy with the solution..