On a geometry sequence, total the 4th and the 6th terms equals 30. While the 3rd term equals 3. The 8th term equals?

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On a geometry sequence, total the 4th and the 6th terms equals 30. While the 3rd term equals 3. The 8th term equals?

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Answer 1 · 2018-01-21T05:33:16

$8^{t h}$ $8^(th)$ term is $96$ $96$

Explanation:

Let the first term of geometric sequence be $a_{1}$ $a_1$ ( $a_{n}$ $a_n$ indicating $n^{t h}$ $n^(th)$ term) and common ratio be $r$ $r$ . Then $a_{3} = a_{1} r^{2}$ $a_3=a_1r^2$ , $a_{4} = a_{1} r^{3}$ $a_4=a_1r^3$ , $a_{6} = a_{1} r^{5}$ $a_6=a_1r^5$ and $a_{8} = a_{1} r^{7}$ $a_8=a_1r^7$ .

As $a_{3} = 3$ $a_3=3$ , we have $a_{1} r^{2} = 3$ $a_1r^2=3$ or $a_{1} = \frac{3}{r^{2}}$ $a_1=3/r^2$

andas total of $4^{t h}$ $4^(th)$ and the $6^{t h}$ $6^(th)$ terms equals $30$ $30$ , we have

$a_{1} r^{3} + a_{1} r^{5} = 30$ $a_1r^3+a_1r^5=30$

i.e. $a_{1} r^{2} (r + r^{3}) = 30$ $a_1r^2(r+r^3)=30$

or $3 (r + r^{3}) = 30$ $3(r+r^3)=30$

or $r^{3} + r = 10$ $r^3+r=10$ or $r^{3} + r - 10 = 0$ $r^3+r-10=0$

i.e. $(r - 2) (r^{2} - 2 r + 5) = 0$ $(r-2)(r^2-2r+5)=0$

or $(r - 2) ({(r - 1)}^{2} + 4) = 0$ $(r-2)((r-1)^2+4)=0$

as ${(r - 1)}^{2} + 4$ $(r-1)^2+4$ is always positive, we have $r - 2 = 0$ $r-2=0$ or $r = 2$ $r=2$

Hence $a_{1} = \frac{3}{4}$ $a_1=3/4$ and $a_{8} = a_{1} r^{7} = \frac{3}{4} \times 2^{7} = 96$ $a_8=a_1r^7=3/4xx2^7=96$

i.e. $8^{t h}$ $8^(th)$ term is $96$ $96$ .

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On a geometry sequence, total the 4th and the 6th terms equals 30. While the 3rd term equals 3. The 8th term equals?

1 Answer

Explanation:

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