Question

The sum of 3 numbers in AP is 18. If the product of first and third number is 5 times the common difference, find the numbers.​

Answer 1

$\huge\underline\mathrm{SOLUTION:-}$

AnswEr:

• The three numbers are 2, 6, 10.

Given That:

• The sum of 3 numbers in AP is 18. The product of the first and third number is 5 times the common difference.

Need To Find:

• The three numbers = ?

ExPlanation:

Let the first three numbers of A.P are (a-d), a and (a+d) where d is the common difference.

• Sum of the first three numbers = (a-d) + a + (a+d) = 18 (Given)

➠ (a-d) + a + (a + d) = 18

➠ a - d + a + a + d = 18

➠ 3a = 18

a = 6..................................(1)

Also, product of first and third term = 5 times common difference.

So the equation becomes

➠(a - d)(a + d) = 5d

➠a² - d² = 5d

Substituting a = 6 which we obtained from equation (1)

36 - d² = 5d

= d² + 5d - 36 = 0

= d² + 9d - 4d - 36 = 0

= d(d + 9) + 9(d - 4) = 0

= (d + 9)(d - 4) = 0

d = -9 and d = 4

d = -9 is neglected

• Hence d = 4

ThereFore:

• Three numbers are (a - d) + a + (a + d) = 2, 6, 10.

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Answer 2

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

The sum of 3 numbers in an A.P. is 18. If the product of first and third number is 5 times the common difference.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The numbers.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Let the three numbers in Arithmetic progression :

• (a-d)
• a
• (a+d)

A/q

$\longrightarrow\sf{(a-d)+a+(a+d)=18}\\\\\longrightarrow\sf{a\cancel{-d} +a+a \cancel{+d}=18}\\\\\longrightarrow\sf{3a=18}\\\\\longrightarrow\sf{a=\cancel{\dfrac{18}{3} }}\\\\\\\longrightarrow\sf{\green{a=6}}$

&

$\longrightarrow\sf{(a-d)(a+d)=5d}\\\\\longrightarrow\sf{a^{2} \cancel{+ad-ad} -d^{2} =5d}\\\\\longrightarrow\sf{a^{2} -d^{2} =5d}\\\\\longrightarrow\sf{(6)^{2} -d^{2} =5d\:\:\:\:[\therefore a=6]}\\\\\longrightarrow\sf{36-d^{2} =5d}\\\\\longrightarrow\sf{d^{2} +5d-36=0}\\\\\longrightarrow\sf{d^{2} +9d-4d-36=0\:\:\:\:[factorise]}\\\\\longrightarrow\sf{d(d+9)-4(d+9)=0}\\\\\longrightarrow\sf{(d+9)(d-4)=0}\\\\\longrightarrow\sf{d+9=0\:\:Or\:\:d-4=0}$

$\longrightarrow\sf{\green{d\neq -9\:\:Or\:\:d=4}}$

∴ We know that negative value isn't acceptable .

$\star$Common difference (d) = 4

Now;

$\dag\:{\underline{\underline{\bf{The\:numbers\::}}}}}$

$\bullet\sf{(a-d)=6-4=\boxed{2}}}\\\\\bullet\sf{(a)=\boxed{6}}}\\\\\bullet\sf{(a+d)=6+4=\boxed{10}}}$