Question

find five terms in GP such that their product is 1024 and fifth term is square of third term​

Answer 1

Answer:

The required terms are

1, 2, 4, 8, 16

1,-2, 4, -8, 16

Step-by-step explanation:

Concept used:

The n th term of G.P is $t_n=ar^{n-1}$

Let the G.P be $\bold{a,ar,ar^2,ar^3,ar^4.........}$

Given:

Product of first five terms =1024

$\implies\:a.ar.ar^2.ar^3.ar^4=1024$

$\implies\:a^5r^{10}=1024$

$\implies\:(ar^2)^5=2^{10}$

$\implies\:(ar^2)^5=(2^2)^5$

$\implies\:ar^2=2^2$

$\implies\:ar^2=4$........(1)

Also

$t_5=t_3^2$

$\implies\:ar^4=(ar^2)^2$

$\implies\:ar^4=a^2r^4$

$\implies\:a=1$

Put a=1 in (1) we get

$\implies\:r^2=4$

tex]\implies\:r=2,-2[/tex]

When r=2, The 5 terms are

[tex]a=1[\tex]

[tex]ar=1(2)=2[\tex]

[tex]ar^2=1(4)=4[\tex]

[tex]ar^3=1(8)=8[\tex]

[tex]ar^4=1(16)=16[\tex]

When r= - 2, The 5 terms are

[tex]a=1[\tex]

[tex]ar=1(-2)=-2[\tex]

[tex]ar^2=1(4)=4[\tex]

[tex]ar^3=1(-8)=-8[\tex]

[tex]ar^4=1(16)=16[\tex]

Answer 2

Answer:

1   2  4  8  16

1  -2  4  -8  16

Step-by-step explanation:

Find five numbers in G. P. such that their

product is 1024 and fifth term is square of the

third term.​

Let say 5 numbers in GP are

a , ar , ar² , ar³  ar⁴

fifth term is square of the  third term.​

=> ar⁴ = (ar²)²

=> ar⁴ = a²r⁴

=> a = a²

=> 1 = a

five numbers are

1 , r , r² , r³  r⁴

Products of  five number of GP

= 1 * r * r² * r³ *  r⁴

= r¹⁰

product is 1024

=> r¹⁰  = 1024

=> r = ±2

GP

1   2  4  8  16

1  -2  4  -8  16