The sum of first n terms of the sequence (6, 36, 216, ..., 6n, ...) is 55,986. Under these conditions, whereas log 2 and log 3 = 0.30 = 0.48, the value of log n is:

a) 0,78
b) 1,26
c) 1,56
d) 1,08

Help me guys! Do I have to use the sum formula of arithmetical progression ?

Question

The sum of first n terms of the sequence (6, 36, 216, ..., 6n, ...) is 55,986. Under these conditions, whereas log 2 and log 3 = 0.30 = 0.48, the value of log n is:

a) 0,78 
b) 1,26
c) 1,56
d) 1,08

Help me guys! Do I have to use the sum formula of arithmetical progression ?

anonymous · Answer

i guess we need n right?

anonymous · Answer

yep, i guess we need n to find log n.

anonymous · Answer

$$55968=\frac{6^n-1}{6-1}$$
$$5\times 55968=6^n-1$$
$$ 299839=6^n$$

anonymous · Answer

i made a mistake somewhere, let me think

anonymous · Answer

I did this: (6 + an)n / 2 = 55986  and then: (6 + an)n = 111972

anonymous · Answer

but it just doesnt make sense. how do i find an and n at the same time?

anonymous · Answer

n = 6

anonymous · Answer

so 
$$\ln(6)=\ln(3)+\ln(2)$$ and we are done

anonymous · Answer

How did you find n = 6??

anonymous · Answer

i found the 6 by brute force but we can also use the formula

anonymous · Answer

$$6\frac{6^n-1}{6-1}=55986$$
$$6^n-1=\frac{5}{6}\times 55986$$
$$6^n=46656$$

anonymous · Answer

since n is an integer, we can guess and check. if that is too much work then a calculator will find it via 
$$n=\frac{\ln(46656)}{\ln(6)}$$

anonymous · Answer

which formula did you use? i dont think ive learnt it