Math, asked by rakesh29308, 9 months ago

If the numbers x+3, 2x+1 and x-7 are
in A P , find the numbers​

Answers

Answered by AlluringNightingale
2

Answer :

0 , -5 , -10

Note :

★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.

★ If a1 , a2 , a3 , . . . , an are in AP , then

a2 - a1 = a3 - a2 = a4 - a3 = . . .

★ The common difference of an AP is given by ; d = a(n) - a(n-1) .

★ The nth term of an AP is given by ;

a(n) = a1 + (n - 1)d .

★ If a , b , c are in AP , then 2b = a + c .

★ The sum of nth terms of an AP is given by ; S(n) = (n/2)×[ 2a + (n - 1)d ] .

or S(n) = (n/2)×(a + l) , l is the last term .

★ The nth term of an AP can be also given by ; a(n) = S(n) - S(n-1) .

Solution :

• Given : (x + 3) , (2x + 1) , (x - 7) are in AP .

• To find : The numbers

Since ,

(x + 3) , (2x + 1) , (x - 7) are in AP thus ,

=> 2(2x + 1) = (x + 3) + (x - 7)

=> 4x + 2 = 2x - 4

=> 4x - 2x = -4 - 2

=> 2x = -6

=> x = -6/2

=> x = -3

Now ,

• 1st no. = x + 3 = -3 + 3 = 0

• 2nd no. = 2x + 1 = 2•(-3) + 1 = -6 + 1 = -5

• 3rd no. = x - 7 = -3 - 7 = -10

Hence ,

Required numbers are : 0 , -5 , -10

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