Question

50 points_____________________

Insert four no btw 4 & 9 such that the resulting sequence is in AP.

Answer 1

Insert four no btw 4 & 9 such that the resulting sequence is in AP.

Let AP be 4,x1,x2,x3,x4,9 :

Here first term(a) = 4,

Last no = 9,

No of terms = 6.

Let common difference be d.

=) an = a + (n-1)d

=) 9 = 4 + (n-1)d

=) 9- 4 = (n-1)d

=) 5 = (6-1)d

=) 5 = 5d

=) 1 = d

Hence x1 = a+d

= 4 + 1 = 5.

=) x2 = a + 2d

= 4 + 2(1)

= 4 + 2 = 6.

=) x3 = a + 3d

= 4 + 3(1)

= 7.

=) x4 = a + 4d

= 4 + 4(1)

= 4 + 4 = 8.

Hence four no between 4 & 9 are 5,6,7 and 8.

Answer 2

Answer :-

→ 5, 6, 7, 8 .

Step-by-step explanation :-

Let the required numbers be x₁ , x₂ , x₃ and x₄ .

In this AP, we have

→ First term ( a₁ ) = 4 .

→ Last term = 9 .

→ Number of terms = 6 .

Let the common difference be d . Then,

°•° 6th term = 9 .

⇒4 + ( 6 - 1 )d = 9 .

⇒ 5d = 9 - 4 .

⇒ 5d = 5 .

⇒d = 5/5 .

•°• d = 1 .

•°• x₁ = ( 4 + d ) = 4 + 1 = 5 .

•°• x₂ = ( 4 + 2d ) = 4 + 2 × 1 = 6 .

•°• x₃ = ( 4 + 3d ) = 4 + 3 × 1 = 7 .

•°• x₄ = ( 4 + 4d ) = 4 + 3 × 1 = 8 .

Hence, the required numbers are 5, 6, 7, 8 .