Question

the sum of three decreasing number in AP is 27 if - 1 - 1, 3 are added to their respectively the resulting series is GP. find the numbers
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Answer 1

The three numbers are

17,9 and 1

Step-by-step explanation:

Let three number of A.P are

a-d,a and a+d

According to question

$a-d+a+a+d=27$

$3a=27$

$a=\frac{27}{3}=9$

Then, three number after substituting the values of a

9-d,9 and 9+d

$a_1=9-d-1=8-d$

$a_2=9-1=8$

$a_3=9+d+3=12+d$

When three terms are in G.P

Then, by definition of G.P

$\frac{a_2}{a_1}=\frac{a_3}{a_2}$

$\frac{8}{8-d}=\frac{12+d}{8}$

$64=(8-d)(12+d)=96+8d-12d-d^2$

$d^2-8d+12d+64-96=0$

$d^2+4d-32=0$

$d^2+8d-4d-32=0$

$d(d+8)-4(d+8)=0$

$(d+8)(d-4)=0$

$d+8=0$

$d=-8$

$d-4=0$

$d=4$

Substitute the value of d=-8

$a_1=a-d=9-(-8)=17$

$a_2=9$

$a_3=9+(-8)=1$

Substitute the value of d=4

$a_1=9-4=5$

$a_2=9$

$a_3=9+4=13$

But we are given the three decreasing number therefore, the value of d=4 is not possible.

The three numbers are

17,9 and 1

#Learns more:

https://brainly.in/question/1191155:Answered by Jashanjatana

Answer 2

Your Answer:-

let ,

the three number of A.P are a-d , a , a+dtherefore:-

As per given Condition.

If we added -1 , -1 , 3 in them

Therefore:-

$a_{1} = a - d - 1 \\ = 9 - d - 1 \\ = 8 - d$

$a_{2} = a - 1 = 8$

$a_{3} = a + d + 3 = 12 + d$

It is given that this three term are in G.P.

Therefore:-

According to G.P.

"If ratio of two consecutive term are equal then taht is in G.P"

Therfore :-

$\frac{ a_{2}}{ a_{1} } = \frac{ a_{3} }{ a_{2}}$

$\frac{8}{8 - d} = \frac{12 + d}{8}$

$64 = (12 + d)(8 - d)$

$64 = 96 + 8d - 12d - {d}^{2}$

${d}^{2} - 8d + 12d + 64 - 96 = 0$

${d}^{2} - 8d + 12d + 64 - 96 = 0$

${d}^{2} + 8d - 4d - 32 = 0$

$d(d + 8) - 4(d + 8) = 0$

$(d + 8)(d - 4) = 0$

$d = - 8 \: and \: 4$

d = 4 is not possible ( as we have given decreasing no.)

Therefore:-

$a_{1} = 17$

$a_{2} = 9$

$a_{3} = 1$

Therefore:- The three no. are

17 , 9 , 1.

Thank you..