Question

write the gp whose 4th term is 54 and the 7th term is 1458​

Answer 1

4th term = 54

7th term = 1458

t4=54

ar3=54⋯(1)

t7=1458

a.r6=1458⋯(2)

12=ar3ar6=541458

r3=27.

r3=33

r=3

Substitute r = 3 in the first equation , we get 

a(3)3=54.

a(27)=54

a=54/27=2.

∴ The G.P form

a,ar,ar2⋯

=2,2(3),2(3)2⋯

=2,6,18⋯

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Answer 2

$\blue\star$$\bold{Question:-}$

Write the gp whose 4th term is 54 and the 7th term is 1458.

$\blue\star$$\bold\red{Given:-}$

$\mathrm{\:\:\:\:4th\: term = t4 = 54}$

$\mathrm{\:\:\:\:7th \:term = t7 = 1458}$

$\blue\star$$\bold{Solution:-}$

$\mathrm{\:\:\:\:\:\:\:t4 = 54}$

$\mathrm{ar³ = 54 }$.......$\fbox{equation 1}$

$\mathrm{\:\:\:\:\:\:\:t7 = 1458}$

$\mathrm{ar⁶ = 1458 }$.......$\fbox{equation\: 2}$

$\mathrm{Now,}$

$\: \: \: \: \: \: \: \: \frac{2}{1} = \frac{ {ar}^{6} }{ {ar}^{3} } = \frac{1458}{54}$

$⇒ {r}^{3} = 27$

$⇒ {r}^{3} = {3}^{3}$

$⇒r = 3$

$\mathrm{Substitute\:r=3\:in\:1st\: equation\:,we\:get}$

$➟a {(3)}^{3} = 54$

$➟a(27) = 54$

$➟a = 54 \div 27 = 2$

$\mathrm{:. the\:gp\:form:-}$

$a,ar, {ar}^{2} ...........$

$⇒2,2(3),2 {(3)}^{2}$

$⇒ 2,6,18....................$

$\mathrm\blue{Ans:-}$$\mathrm\red{2,6,18......}$