What is the least value of n for which the series 1+3+3²+3³.....to n terms is greater than 7000
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Answered by
12
1+3+3^2......n terms
= 3^n -1)/2 >7000
3^n -1>14000
3^n >14001
take log
n log 3> log 14001
n> log (14001)/log 3
n> 4.14/0.477
n>8.6
so take it 9
its 9
Mirmita:
thank you so much for the amswer
ur welcome
would you mind helping me with another math problem?
here it goes
if the first term of an AP is 4 and the sum of first 4 terms is equal to 1/3 of sum of next 4 terms,then what will be the sum of first 8 terms?
where to give ans
ohhh
I have posted the question look if you can find it
Answered by
6
Answer:
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1+3+3^2......n terms
= 3^n -1)/2 >7000
3^n -1>14000
3^n >14001
take log
n log 3> log 14001
n> log (14001)/log 3
n> 4.14/0.477
n>8.6
so take it 9
its 9
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