Question

if the no. (3x+2),(2x+3),and (2x-5) are in A.P then value of x is
a. 5
b. 7
c. 9
d. 11​

Answer 1

GiveN:

(3x +2), (2x + 3) and (2x -5) are in AP

✏ To finD:

Find the value of x ......?

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✝️ How to solve?

In an AP, the consecutive terms have an common difference. It is a successive additive series in which common difference is always same for consecutive terms.

Let a, b and c are in AP

❇ Then, b - a = c - b

➙ b + b = a + c

➙ 2b = a + c

So, here we can say that, twice the second term = sum of first and third term in the series. Or the second term is the average of the first and third term.

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✝️ Solution:

Let,

a = 3x + 2

b = 2x + 3

c = 2x - 5

Then, On the basis of above relation,

➙ 2(2x + 3) = 3x + 2 + 2x - 5

➙ 4x + 6 = 5x - 3

➙ 5x - 4x = 6 + 3

➙ x = 9

Verification:

If x = 9, then

a = 3(9) + 2 = 29

b = 2(9) + 3 = 21

c = 2(9) - 5 = 13

Here, b - a = c - b = - 8, common difference.

So, a, b and c are in AP. (Verified!)

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Answer 2

Answer:

GIVEN,

A.P-------> (3x+2),(2x+3),(2x-5)

d= a2-a1=a3-a2

a1= 3x+2

a2= 2x+3

a3=2x-5

$(2x + 3) - (3x + 2) = (2x - 5) - (2x + 3)$

$2x + 3 - 3x - 2 = 2x - 5 - 2x - 3$

$- x + 1 = - 5 - 3$

$1 - x = - 8$

$x = 1 + 8$

$x = 9$

Step-by-step explanation:

HOPE IT HELPS YOU OUT.......