Question

15.
If r and y are inserted between 4 and 16, so that
the resulting sequence becomes an AP, then
(a) x = 8, y = 12
(b) x = 10, y = 12
(c) r=8, y = 10
(d) x =10, y = 14
Full Marks:2
​

Answer 1

Answer:

x=8,y=12 Option(a)

Step-by-step explanation:

If r and y are inserted between 4 and 16

Then 4,x,y, 16 are in an AP

Let d is common difference

then

16 is 4th term

so 16=4+3d

3d=12,d=4

So x=4+4=8

and y=x+4=8+4=12

Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

If x and y are inserted between 4 and 16, so that

the resulting sequence becomes an AP, then

(a) x = 8, y = 12

(b) x = 10, y = 12

(c) x = 8, y = 10

(d) x = 10, y = 14

CONCEPT TO BE IMPLEMENTED

If in an arithmetic progression

First term = a

Common difference = d

Then nth term of the AP

= a + ( n - 1 )d

EVALUATION

Here it is given that x and y are inserted between 4 and 16

So the resulting sequence 4 , x , y , 16 becomes an AP

Let common difference = d

Then ,

x = 4 + d

y = 4 + 2d

16 = 4 + 3d

Now 16 = 4 + 3d gives

4 + 3d = 16

⇒ 3d = 12

⇒ d = 4

Thus we get

x = 4 + d = 4 + 4 = 8

y = 4 + 2d = 4 + 8 = 12

FINAL ANSWER

Hence the correct option is (a) x = 8 , y = 12

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