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Let $g_1, g_2, g_3,\dots$ be a geometric sequence. If $g_{23} = 16$ and $g_{28} = 24$, what is $g_{43}$?

 May 21, 2023
 #1
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Let r be the common ratio of the geometric sequence. Then, we have g23​=g1​r23−1=16 and g28​=g1​r28−1=24. Dividing these two equations, we get r5=2, so r=21/5. Therefore, g43​ = g1​r43−1 = 16r42 = 16(21/5)42 = 16*(2^4) = 128​.

 May 21, 2023
 #3
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wrong

Guest May 21, 2023
 #2
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16 ==F *  R^22,             

24 ==F *  R^27, solve for F, R

 

F==2.687314174158 - this is the first term

R ==1.0844717711977 - this is the common ratio.

 

The 43rd term ==2.687314174158  x  1.0844717711977^(43 - 1)==81 - the 43rd term

 

Your sequence looks like this:

 

2.687314174 , 2.914316362 , 3.160493827 , 3.427466339 , 3.716990491 , 4.030971261 , 4.371474543 , 4.740740741 , 5.141199508 , 5.575485736 , 6.046456892 , 6.557211815 , 7.111111111 , 7.711799262 , 8.363228605 , 9.069685338 , 9.835817722 , 10.66666667 , 11.56769889 , 12.54484291 , 13.60452801 , 14.75372658 , 16 , 17.35154834 , 18.81726436 , 20.40679201 , 22.13058988 , 24 , 26.02732251 , 28.22589654 , 30.61018802 , 33.19588481 , 36 , 39.04098376 , 42.33884481 , 45.91528202 , 49.79382722 , 54 , 58.56147564 , 63.50826722 , 68.87292303 , 74.69074083 , 81 , >>Number of terms = 43>>Total Sum == 1008.087058

 May 21, 2023
 #4
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right!

Guest May 21, 2023

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