What is the 6th term of a geometric sequence if the difference between its 3rd and 1st term is 9 and that between its 4th and 2nd term is 18?
This is an easy to moderate level difficulty SAT Math Question. The formula to find the nth term of a GP is given in the Hint / Formula tab. Use the formula and the steps outlined to solve the question. If you need further assistance, step wise explanation is provided in the subsequent tabs.
The nth term of a geometric sequence a_{n} = \a r^{n-1} \\)
Where a is the first term, 'n' is the number of terms and 'r' is the common ratio.
To find the 6th term of the geometric sequence, you need the first term 'a' and the common ratio 'r'.
Data given: Difference between the 3rd and the 1st term; the difference between the 4th and the 2nd term.
Difference between the 3rd and the 1st term is 9.
a_{3} - a = 9
The 3rd term of the sequence a_{3} = a * r^{3 - 1} = a * r^{2}
Therefore, a_{3} - a = a * r^{2} - a = 9 or a(r^{2} - 1) = 9 .... eqn (1)
Difference between the 4th and the 2nd term is 18.
a_{4} - a_{4} = 9
The 4th term of the sequence a_{4} = a * r^{4 - 1} = a * r^{3}
And the 2nd term of the sequence a_{2} = a * r^{2 - 1} = a * r
Therefore, a_{4} - a_{2} = a * r^{3} - ar = 18 or ar(r^{2} - 1) = 18 .... eqn (2)
Divide eqn(2) by eqn (1) to get r, the common ratio
\ar \lbrack r^2 - 1\rbrack \over {a \lbrack r^2 - 1\rbrack} \\) = \18 \over 9\\)
or the common ratio r = 2
Substitute 'r' in eqn (1) to determine the first term 'a'
a(r^{2} - 1) = 9
a(2^{2} - 1) = 9
or 3a = 9 or a = 3.
Last step: Compute the 6th term using a and r
a_{6} = a * r^{6 - 1} = a * r^{5}
a_{6} = 3 * 2^{5} = 3 * 32 = 96
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