Find the value of x and y in the arithmetic sequence ( 5,x,13,y....)

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

5, x, 13, y , ....   is an A.P

Then,

a1= 5

a2= 5 + r = x

a3= 5 + 2r = 13

==> 2r = 13-5 = 8

==> r = 8/2 = 4

==> a2= 5 + r = x

==> 5 + 4 = x

==> x =9

==> y = 13 + r

           = 13 + 4 = 17

==> y= 17

pohnpei397's profile pic

pohnpei397 | College Teacher | (Level 3) Distinguished Educator

Posted on

For this sequence, the answer is that x is 9 and y is 17.  The difference between each set of terms in this sequence is 4.

We can find this by the fact that the difference between 5 and 13 is 8.  We know that, in an arithmetic sequence, the difference between each pair of numbers is constant.  Therefore, the difference between 5 and 13 must be split equally into the intervals between 5 and x and x and 13.  8 divided by 2 is 4 and therefore the difference between each pair of terms is 4.

Once we know that the difference between each pair of terms is 4, we simply add 4 to 13 to get the value for y.

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll use another method, to solve the problem. We'll apply the arithmetic mean theorem.

Each term of an arithmetic series, starting with the second one, could be calculated as the semi-sum of it's neighbour terms.

Acoording to this rule, we'll write:

x = (5+13)/2

x = 18/2

x = 9

(x+y)/2 = 13

We'll substitute x by 9:

(9+y)/2 = 13

9+y = 26

We'll subtract 9 both sides:

y = 26-9

y = 17 

The arithmetic sequence is: 5 , 9 , 13 , 17 , ...........

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The  first 4 terms of the given arothmetoc progression are 5,x,a3 and y.

To find x and y..

Solution:

In an arithmetic progression, the rth term ar  is given by:

ar = a1+(r-1)d, where d is the common difference.

So a1 = 5.

a3 = a1+(3-1)d = a1+2d.,

2d = a3-a1 = 13-5 = 8

d = 8/2 =4.

Therefore the common diffrence = 4.

So a2 = x = a1+(2-1)d = 5+1*4 = 9

a4 = y = a1+(4-1)d = 5+3*4 =  17.

So  x= 9 and y = 17.

Tally.

So the the 1st 4 terms of AP = 5, 9, 13 and 17

 

 

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

let:

an = nth term of the sequence.

And

r = difference between any two consecutive terms of sequencee

Then:

a1 = 5

a2 = a1 + r = x

a3 = a1 + 2r = 13

Substituting given value of a1 in the above equation:

5 + 2r = 13

== 2r = 13 - 5 = 8

Therefore:

r = 8/2 = 4

Using values of a1 and r we calculate x any y as follows:

x = a1 + r

= 5 + 5 = 9

y = a1 + 3r

= 5 + 3*4 = 5 + 12 = 17

Answer:

x = 9

y =17

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