Question

Three numbers are in AP and their sum is 15. If 1, 3, 9 are added to them respectively, the resulting numbers are in GP. Find the numbers.

Answer 1

Given

⇒Three number are AP and their Sum is 15

We can write

⇒a₁ + a₂ + a₃ = 15

Formula

⇒Aₙ = a + (n-1)d

Now

⇒a + a + d + a + 2d = 15

⇒3a + 3d = 15

⇒3(a + d) = 15

⇒a + d = 5      (i)

Now it's given

⇒a+1 , a+d+3 and a + 2d +9

Now this number are in GP, we use this formula

⇒b² = ac

⇒( a+d+3)² = (a+1)(a+2d+9)       (ii)

Now Take (i)st equation

⇒a + d = 5  

⇒a=5-d   put the value on (ii)eq

⇒( a+d+3)² = (a+1)(a+2d+9)       (ii)

⇒( 5-d+d+3)² = (5-d+1)(5-d+2d+9)

⇒(5+3)² = (6-d)(14+d)

⇒(8)² = 84+ 6d-14d-d²

⇒64 = 84-8d-d²

⇒d²+8d-20 = 0

⇒d²+10d-2d-20=0

⇒d(d+10)-2(d+10)=0

⇒(d+10)(d-2)=0

⇒d=-10 and d = 2

Now put the value of d on first (i)eq , we get

⇒a=5-d       (i)

When d = -10

⇒a = 5-(-10)

⇒a = 5+10

⇒a = 15

We get sequences

15 , 5 , -5 , -15

When d = 2

⇒a = 5-2

⇒a= 3

We get sequences

3 , 5 ,7,9

Answer 2

Question:-

Three numbers are in AP and their sum is 15. If 1, 3, 9 are added to them respectively, the resulting numbers are in GP. Find the numbers.

Given:-

• Three numbers are in A.P and their sum is 15.
• Is 1 , 3 , 9 are added to then the riddles results are in G.P .

To Find:-

• Find the numbers.

Solution:-

Here ,

Formula to be used:-

$\tt\implies \: An = a + ( n - 1 )d$

Now ,

$\tt\implies \: a + a + d + a + 2d = 15$

$\tt\implies \: 3a + 3d = 15$

$\tt\implies \: 3( a + d ) = 15$

$\tt\implies \: a + d = \dfrac { 15 } { 3 }$

$\tt\implies \: a + d = 5$

$\tt\implies \: a = 5 - d . . . . . . . ( 1 )$

Here ,

a + 1 , a + d + 3 and a + 2d + 9

These numbers are in G.P:-

$\tt\implies \: { b }^{ 2 } = ac$

$\tt\implies \: { ( a + d + 3 ) }^{ 2 } = ( a + 1 )( a + 2d + 9 ) . . . . . . . ( 2 )$

Substitute a + d = 5 in equation ( 2 ):-

( 5 - d + d + 3 )² = ( 5 - d + 1 )( 5 - d + 2d + 9 )

$\tt\implies \: {( 5 + 3 )}^{ 2 } = ( 6 - d )( 14 + d )$

$\tt\implies \: {( 8 )}^{ 2 } = 84 + 6d - 14d - {d }^{ 2 }$

$\tt\implies \: { d }^{ 2 } + 8d - 20 = 0$

$\tt\implies \: { d }^{ 2 } + 10d - 2d - 20 = 0$

$\tt\implies \: d( d + 10 ) - 2( d + 10 ) = 0$

$\tt\implies \: ( d + 10 )( d - 2 ) = 0$

$\tt\implies \: d = 2 , -10$

So ,

d = - 10 substitute in equation 1

$\tt\implies \: a = 5 - ( - 10 )$

$\tt\implies \: a = 15$