If third and fourth terms of an infinite geometric sequence are $27$ $27$ and $20 \frac{1}{4}$ $20 1/4$ , find its tenth term and sum up to infinity?

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If third and fourth terms of an infinite geometric sequence are $27$ $27$ and $20 \frac{1}{4}$ $20 1/4$ , find its tenth term and sum up to infinity?

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Answer 1 · 2018-01-14T07:56:14

Tenth term is $\frac{59049}{16384}$ $59049/16384$ sum of infinite series is $192$ $192$

Explanation:

Let the first term of geometric sequence be $a$ $a$ and common ratio be $r$ $r$ . Then third term is $a r^{2} = 27$ $ar^2=27$ and fourth term is $a r^{3} = 20 \frac{1}{4} = \frac{81}{4}$ $ar^3=20 1/4=81/4$ . Dividing latter by former, we get

$\frac{a r^{3}}{a r^{2}} = \frac{\frac{81}{4}}{27}$ $(ar^3)/(ar^2)=(81/4)/27$ or $r = \frac{81}{4} \times \frac{1}{27} = \frac{3}{4}$ $r=81/4xx1/27=3/4$

and hence $a \times {(\frac{3}{4})}^{2} = 27$ $axx(3/4)^2=27$ or $a = 27 \times \frac{4}{3} \times \frac{4}{3} = 48$ $a=27xx4/3xx4/3=48$

and tenth term is $48 \times {(\frac{3}{4})}^{9} = 3 \times 4^{2} \times \frac{3^{9}}{4^{9}} = \frac{3^{10}}{4^{7}}$ $48xx(3/4)^9=3xx4^2xx3^9/4^9=3^10/4^7$

= $\frac{59049}{16384}$ $59049/16384$

As $| r | < 1$ $|r|<1$ , sum of infinite series is $\frac{a}{1 - r} = \frac{48}{\frac{1}{4}} = 192$ $a/(1-r)=48/(1/4)=192$

Answer 2 · 2018-01-14T08:02:45

Tenth term $a_{10} = 3.6041$ $a_10 = color (purple)(3.6041)$

$S_{\infty} = \frac{a}{1 - r} = 192$ $S_oo = a / (1- r) = color(green)(192)$

Explanation:

Nth term of a G. S $a_{n} = a \cdot r^{n - 1}$ $a_n = a * r^(n-1)$

Third term $a_{3} = a \cdot r^{2} = 27$ $a_3 = a * r^2 = 27$ Eqn (1)

Fourth term $a_{4} = a \cdot r^{3} = 20 (\frac{1}{4})$ $a_4 = a * r^(3) = 20(1/4)$ Eqn (2)

Dividing Eqn (2) by (1),

$\frac{a r^{3}}{a r^{2}} = r = \frac{20 (\frac{1}{4})}{27} = (\frac{3}{4})$ $(a r^3) / (a r^2) = r = (20(1/4)) / 27 =( 3/4)$

Substituting the value of ‘r’ in Eqn (1),

$a r^{2} = 27 = a \cdot {(\frac{3}{4})}^{2}$ $a r^2 = 27 = a * (3/4)^2$

$a = \frac{27}{{(\frac{3}{4})}^{2}} = 48$ $a = 27 / (3/4)^2 = 48$

Tenth term $a_{10} = a \cdot r^{9} = 48 \cdot {(\frac{3}{4})}^{9} = 3.6041$ $a_10 = a * r^9 = 48 * (3/4)^9 = color(purple)(3.6041)$

Sum of geometric sequence of infinite series with ‘r’ less than 1 is given by the formula,

$S_{\infty} = \frac{a}{1 - r} = \frac{48}{1 - \frac{3}{4}} = 192$ $S_oo = a / (1- r) = 48 / (1 - 3/4) = color(green)(192)$

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If third and fourth terms of an infinite geometric sequence are $27$ $27$ and $20 \frac{1}{4}$ $20 1/4$ , find its tenth term and sum up to infinity?

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If third and fourth terms of an infinite geometric sequence are $27$ $27$ and $20 \frac{1}{4}$ $20 1/4$ , find its tenth term and sum up to infinity?