Question

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

Answer 1

$\LARGE{ \underline{\underline{ \sf{Required \: answer:}}}}$

GiveN:

• Product of five terms in GP = 1024
• The Fifth term is square of 3rd term.

To FinD:

• Five no. of the GP?

Step-by-Step Explanation:

Let the terms of the GP be:

$\rm{ \dfrac{a}{ {r}^{2} }, \: \: \dfrac{a}{r}, \:\: a, \:\: ar, \:\: a {r}^{2} }$

According to question,

⇒ Product of all five terms = 1024

⇒ a/r² × a/r × a × ar × ar² = 1024

⇒ a⁵ = 1024

⇒ a⁵ = 4⁵

⇒ a = 4

And,

⇒ Fifth term = (Third term)²

According to the terms we have chosen,

⇒ ar² = a²

⇒ r² = a

⇒ r = $\pm$√a

Putting the value of r,

⇒ r = $\pm$2

Finding the terms:

When a = 4 and r = 2

• a/r² = 1
• a/r = 2
• a = 4
• ar = 8
• ar² = 16

When a = 4 and r = -2

• a/r² = 1
• a/r = -2
• a = 4
• ar = -8
• ar² = 16

Answer 2

So, 5 GP are

1, 2, 4, 8, 16

or

1, -2, 4, -8, -16

${\underline{\underline{\large{\sf{\pink{\bigstar{ \: \:Given :-}}}}}}}$

• Product of GP is 1024.
• Fifth term is square of the third term.

${\underline{\underline{\large{\sf{\pink{\bigstar{ \: \: To \: \: find :-}}}}}}}$

• Five numbers.

${\underline{\underline{\Large{\sf{\red{\bigstar{ \: \:Solution :-}}}}}}}$

We know that GP is Geometric Progression in which we have asked to find Five numbers in GP such that their product is 1024 and also fifth term is square of them.

Let GP be,

$\\ \: \: \: \: \: \: \: \: \: \: \: \bull{ \sf{ \: \: \: \: \frac{a}{ {r}^{2} } \: \:, \: \: \frac{a}{r} , \: a \: , \: ar \: , \: a {r}^{2} }}$

${\underline{\underline{{\sf{\pink{\bigstar{ \: \: Condition \: \: 1 :-}}}}}}}$

Product of GP is 1024 :-

$\\ \implies{ \sf{ \: \: \: \: \frac{a}{ {r}^{2} } \times \frac{a}{r} \times \: a \: \times \: ar \: \times \: a {r}^{2} = 1024 }}$

${\sf{\implies{a × a × a × a × a = 1024}}}$

${\sf{\implies{a⁵ = 4⁵}}}$

${\underline{\boxed{\large{\green{\sf{\implies{a = 4 }}}}}}}$

${\underline{\underline{{\sf{\pink{\bigstar{ \: \: Condition \: \: 2 :-}}}}}}}$

Fifth term is square of the third term :-

${\sf{\implies{(a)² = a(r)²}}}$

${\sf{\implies{a \times a \: = a \times {r}^{2} }}}$

${\sf{\implies{a = {r}^{2} }}}$

Putting the value of a,

${\sf{\implies{ {r}^{2} = 4}}}$

${\sf{\implies{r = \sqrt{4} }}}$

${\underline{\boxed{\large{\green{\sf{\implies{r =\pm \: \: 2 }}}}}}}$

${\underline{\underline{ \large{\sf{\red{\bigstar{ \: \:Therefore \: \: 5 \: \: G.P \: \: \: \: are :-}}}}}}}$

By putting the positive value of r

$\\ \large \implies{ \sf{ \: \: \: \: \frac{4}{ {2}^{2} } \: , \: \frac{4}{2} \: , \: 4 \: , \: 4(2) \: , \: 4 {(2)}^{2} }}$

$\\ \large \implies{ \sf{ \: \: \: \: \frac{4}{ 4 } \: , \: \frac{4}{2} \: , \: 4 \: , \: 8 \: , \: 4 (4) }}$

$\color{blue}{ \underline{ \boxed{\large{\sf{\implies{1 , 2, 4, 8, 16}}}}}}$

By putting the negative value of r,

$\\ \large \implies{ \sf{ \: \: \: \: \frac{4}{ { - 2}^{2} } \: , \: \frac{4}{ - 2} \: , \: 4 \: , \: 4( - 2) \: , \: 4 {( - 2)}^{2} }}$

$\\ \large \implies{ \sf{ \: \: \: \: \frac{4}{ 4 } \: , \: \frac{4}{ - 2} \: , \: 4 \: , \: - 8 \: , \: 4 ( - 4) }}$

$\color{blue}{ \underline{ \boxed{\large{\sf{\implies{1 , \: - 2 \: , \: 4 \: , \: - 8 \: , - 16}}}}}}$

So, 5 GP are

1, 2, 4, 8, 16

or

1, -2, 4, -8, -16