Question

the value of x for which 2x,(x+10) and (3 x+2) are the three consecutive terms of an AP ,is​

Answer 1

(x+10)−2x=(3x+2)−(x+10)

2(x+10)=2x+3x+2

4x+20=5x+2

x=18

Answer 2

Answer:

Analysis→

Here the question conveys that we've to find a suitable value of x for which 2x, (x+10) and (3x+2) are three consecutive terms of an AP. And we know that when three consecutive terms are in AP, the sum of first and last terms is equal to the twice of second term.

Given→

• $\rm{1^{st}\:term(a_1)= 2x}$
• $\rm{2^{nd}\:term(a_2)= (x+10)}$
• $\rm{3^{rd}\:term(a_3)= (3x+2)}$

To Find→

The value of x.

Answer→

$\large{\underline{\boxed{\leadsto{\rm{a_1+a_3=2\times a_2}}}}}$

$\implies\rm{a_1+a_3=2\times a_2}$

$\implies\rm{2x+(3x+2)=2(x+10)}$

$\implies\rm{5x+2=2x+20}$

$\implies\rm{3x+2=20}$

$\implies\rm{3x=18}$

$\implies\rm{x=\dfrac{18}{3}}$

$\implies\rm{x=\dfrac{\cancel{18}}{\cancel{3}}}$

$\implies\rm{x=6}$

${\boxed{\boxed{\implies{\bold{x=6\checkmark}}}}}$

☞ Hence the value of x is 6 for which 2x, (x+10) and (3x+2) are three consecutive terms in an AP.

Know More→

1st Term:-

$\implies\rm{2x}$

$\implies\rm{2(6)}$

$\implies\rm{12}$

${\boxed{\boxed{\implies{\bold{12\checkmark}}}}}$

2nd Term:-

$\implies\rm{x+10}$

$\implies\rm{(6)+10}$

$\implies\rm{16}$

${\boxed{\boxed{\implies{\bold{16\checkmark}}}}}$

3rd Term:-

$\implies\rm{3x+2}$

$\implies\rm{3(6)+2}$

$\implies\rm{18+2}$

$\implies\rm{20}$

${\boxed{\boxed{\implies{\bold{20\checkmark}}}}}$

Hence AP:-

${\boxed{\boxed{\implies{\bold{12,16,20,. . . . . . . . .\infty\:\checkmark}}}}}$

HOPE IT HELPS.