Question

The sum of three terms in A.P. is 125 and their product is 45.Find the numbers.​

Answer 1

Correct Question: The sum of three terms in A.P. is 15 and their product is 45.Find the numbers.

GIVEN:

Sum of three terms in AP = 15

Product of those three terms = 45

TO FIND:

Those three terms of AP

SOLUTION:

Let the numbers be (a - d), a and (a + d)

$= > (a - d) + a + (a + d) = 15 \\ \\ = > a - d + a + a + d = 15 \\ \\ = > 3a = 15 \\ \\ = > a = \frac{15}{3} \\ \\ = > a = 5$

Now,

$= > (a - d) a (a + d) = 45 \\ \\ = > a( {a}^{2} - {d}^{2} ) = 45 \\ \\ = > 5( {5}^{2} - {d}^{2} ) = 45 \\ \\ = > (25 - {d}^{2} ) = \frac{45}{5} \\ \\ = > 25 - {d}^{2} = 9 \\ \\ = > - {d}^{2} = 9 - 25 \\ \\ = > - {d}^{2} = - 16 \\ \\ = > d = \sqrt{16} \\ \\ = > d = 4$

Substitute (a) and (d) to find the numbers.

$= > a - d = 5 - 4 = 1 \\ \\ = > a = 5 \\ \\ = > a + d = 5 + 4 = 9$

Therefore, the numbers are 1, 5 and 9.

Answer 2

$\large{\underline{\sf{Correct\:Question-}}}$

The sum of three terms in A.P. is 15 and their product is 45.Find the numbers.

→$\sf{\red{Given}}$

Sum of three terms in A.P = 15

Product = 45

→$\sf{\red{To\: find}}$ -

The numbers

→$\sf{\red{Solution}}$ -

Let the three number in A.P be ( a - d ) , ( a ) , ( a + d )

♦Now it is given that when these terms sum up they equals to 125.

→( a - d ) + ( a ) + ( a + d ) = 15

→ a - d + a + a + d = 15

→ 3a = 15

→ a = $\dfrac{15}{3}$

→a = 5

♦ Now it is given that product = 45

→ ( a - d ) ( a ) ( a + d ) = 45

→a ( a - d )( a + d ) = 45

→a $\sf({a}^{2}-{d}^{2})$ = 45

→5 $\sf({5}^{2}-{d}^{2})$ = 45

→$\sf({5}^{2}-{d}^{2})$ = $\dfrac{45}{5}$

→$\sf(25-{d}^{2})$ = 9

→$\sf{d}^{2}$ = 25-9

→ $\sf{d}^{2}$ = 16

→ d = $\sf\sqrt16$

→ d = 4

♦ Now taking d = 4

Given numbers →

→( a - d ) = 5 - 4 = 1

→ a = 5

→( a + d ) = 5 + 4 = 9

$\rule{200}2$

Numbers are = 1, 5 and 9

$\rule{200}2$

♦Verification -

~ Sum

1 + 5 + 9 = 15

~ Product

(1)(5)(9) = 45

$\rule{200}2$