Question

Find the smallest of the three numbers in arithmetic progression, if the product
of the first and the third numbers is 252 and the sum of the three numbers is 48.
​

Answer 1

GIVEN :–

• The product of the first and the third numbers of A.P. is 252.

• And the sum of the three numbers is 48.

TO FIND :–

• The smallest of the three numbers in A.P. = ?

SOLUTION :–

• Let the three numbers of A.P. is a , a+d , a+2d.

• According to the first condition –

$\\\bf \implies (a)(a + 2d) = 252$

$\\\bf \implies {a}^{2} + 2ad= 252\: \: \: \: \: \: - - - eq.(1)$

• According to the second condition –

$\\\bf \implies (a) + (a + d) + (a + 2d)=48$

$\\\bf \implies a+a+d+a+2d=48$

$\\\bf \implies 3a+3d=48$

$\\\bf \implies 3(a+d)=48$

$\\\bf \implies a+d= \dfrac{48}{3}$

$\\\bf \implies a+d=16$

$\\\bf \implies d=16 - a\: \: \: \: \: \: - - - eq.(2)$

• Put the value of 'd' in eq.(1) –

$\\\bf \implies {a}^{2} + 2a(16 - a)= 252$

$\\\bf \implies {a}^{2} +32a - 2 {a}^{2}= 252$

$\\\bf \implies 32a -{a}^{2}= 252$

$\\\bf \implies {a}^{2} - 32a + 252 = 0$

$\\\bf \implies {a}^{2} -18a - 14a+ 252 = 0$

$\\\bf \implies a(a - 18) - 14(a - 18)=0$

$\\\bf \implies (a -1 4)(a - 18) = 0$

$\\\large \implies { \boxed{ \bf a = 14,18}}$

• According to eq.(2) –

$\\\bf \implies d=16 - a$

$\\\bf \implies d=16 - 14,16- 18$

$\\\large \implies { \boxed{ \bf d = 2, - 2}}$

• When a = 14 , d = 2 :–

$\\\bf \implies \: 14,16,18 \: \: are \: \: in \: \:A.P.$

• When a = 18 , d = -2 :–

$\\\bf \implies \: 18,16,14 \: \: are \: \: in \: \:A.P.$

• Hence , The smallest term of A.P. is 14 .

Answer 2

${ \huge ❥{\underline{\underline {\rm{Given:}}}}}$

The product of the first and the third numbers of Arithmetic progress is 252. And the sum the three numbers is 48.

We have to find the smallest of the 3 numbers in Arithmetic progress;

${ \huge❥{\underline{\underline{ \rm{Solution:}}}}}$

Let the three numbers of A.P. be x, x+y, x+2y

• According to the first condition

(x)(x+2y)= 252

$\rightarrow \sf {x}^{2} + 2xy = 252 - - - - (i)$

• According to the second condition

(x)+(x+y)+(x+2y)=48

=x+x+y+x+2y=48

= 3x+3y=48

= 3(x+y)=48

$= \sf x + y = \frac{48}{3}$

= x+y=16

= y= 16-x - - - -(ii)

Put value of y in (i)

$\sf {x}^{2} + 2y(1 - x) = 252$

$\rightarrow \sf {x}^{2} + 32x - {2x}^{2} = 252$

$\sf \rightarrow {x}^{2} - 32x + 252 = 0 \\ \\ \sf \rightarrow {x}^{2} - 18x - 14x + 252 = 0$

= x(x-18)-14(x-18)=0

= (x-14)(x-18)=0

= x=14,18

•According to (ii)

y=16-x

y= 16-14,16-18

y= 2, -12

When x= 14, y=2

14,16,18 are in A.P.

When x=18, y= -2

18,16,14 are in A.P.

Therefore, the smallest term of A.P. is 14.

${ \large{ \underline{ \sf{Hope \: it \: helps \: you...}}}}$