Question

the sum of 3 numbers in an AP is 12 and their product is 48. find those numbers
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Answer 1

Given :

• The sum of 3 numbers in an AP is 12.
• The product of the numbers is 48.

To Find :

• The terms in AP.

Solution :

Let the first term be(a-d)

Let the second term be a

Let the third term be (a+d)

Case 1 :

The three terms adds up to 12.

Equation :

$\longrightarrow$ $\sf{(a-d)+(a)+(a+d)\:=\:12}$

$\longrightarrow$ $\sf{a-d+a+a+d =12}$

$\longrightarrow$ $\sf{3a=12}$

$\longrightarrow$ $\sf{a=\dfrac{12}{3}}$

$\longrightarrow$ $\sf{a=4\:\:\:(1)}$

Case 2 :

Product of these 3 terms is 48.

Equation :

$\longrightarrow$ $\sf{(a-d)(a)(a+d)=48}$

$\longrightarrow$ $\sf{(4-d)(4)(4+d)=48}$

$\bold{\big[From\:equation\:(1)\:a\:=\:4\big]}$

$\longrightarrow$ $\sf{4(4-d)(4+d)=48}$

$\longrightarrow$ $\sf{(16-4d)(4+d)=48}$

$\longrightarrow$ $\sf{16(4+d)-4d(4+d)=48}$

$\longrightarrow$ $\sf{64+16d-16d-4d^2=48}$

$\longrightarrow$ $\sf{64-4d^2=48}$

$\longrightarrow$ $\sf{-4d^2=48-64}$

$\longrightarrow$ $\sf{-4d^2=-16}$

$\longrightarrow$ $\sf{d^2=\dfrac{-16}{-4}}$

$\longrightarrow$ $\sf{d^2=\dfrac{16}{4}}$

$\longrightarrow$ $\sf{d^2=4}$

$\longrightarrow$ $\sf{d=\sqrt{4}}$

$\longrightarrow$ $\sf{d=\:\pm\:2}$

Now, when d = 2, let's form the AP.

$\large{\boxed{\sf{First\:Term\:=\:(a-d)\:=\:4-2\:=\:2}}}$

$\large{\boxed{\sf{Second\:Term\:=\:t_2\:=\:a\:+\:d\:=\:2\:+2\:=\:4}}}$

$\large{\boxed{\sf{Third\:Term\:=\:t_3\:=\:t_2\:+\:d\:=\:4\:+\:2\:=\:6}}}$

When, d = - 2,

$\large{\boxed{\sf{First\:Term\:=\:(a-d)\:=\:4-(-2)\:=\:4+2=6}}}$

$\large{\boxed{\sf{Second\:Term\:=\:t_2\:=\:a\:+\:d\:=\:2\:+2\:=\:4}}}$

$\large{\boxed{\sf{Third\:Term\:=\:t_3\:=\:t_2\:+\:d\:=\:4\:+\:(-2)\:=\:4-2=2}}}$