what is the seventh term of a geometric sequence whose fourth term is 128 and common ratio equal to 4
Answer (1)
The seventh term of a geometric sequence whose fourth term is 128 and common ratio equal to 4 is 8192Let's answer this the easiest way and by using a formulaThe easy way is to get to the 7th term simply multiply until you get there[tex]a_4[/tex] = 128[tex]a_5[/tex] = 128 (4)=512[tex]a_6[/tex] = 512 (4)=2048[tex]a_7[/tex] = 2048(4)=8192Therefore, the 7th term is 8192Now, by using the formula[tex]a_n = a_1 (r)^n^-^1[/tex]Since we're only looking for [tex]a_4[/tex], we can pretend that [tex]a_4[/tex] is the first term and [tex]a_7[/tex] is the fourth term because 4 , 5 , 6 , 7 can be considered as 1 , 2, 3, 4 in their positions and solve[tex]a_4 = 128 (4)^4^-^1[/tex]= 128 (4)³= 128 (64)=8192Therefore, the 7th term is 8192We arrived at the same answer using different ways. You can even solve for a sub 1 first and then use that to solve for a sub 7.
