Question

For what value of n are the nth terms of the two AP equal
63,65,67.....
3,10,17

Answer 1

63,65,67..

a=63

d=2

let no. of terms be n

tn=a+(n-1)d

=63+(n-1)2

3,10,17...

a=3

d=7

tn=a+(n-1)d

3+(n-1)7=63+(n-1)2

3+7n-7=63+2n-2

7n-4=61+2n

5n=57

n=57/5

now,

3+(n-1)7

Answer 2

Solution:-

The given two AP are

Let the one AP written as AP :- 63,65,67,...

And, the second AP as A'P':-3,10,17

Now, solving first AP:- 63,65,67

a=63 ,d=65-63=2 n=?

an=a+(n-1)d

an=63+(n-1)2 ----------------(1)

Now, solving second A'P':-3,10,17 ,.....

a=3 ,d=10-3=7 , n=?

an=a+(n-1)d

an=3+(n-1)7 -------------(2)

ATQ,

AP=A'P'

So,

63+(n-1)2=3+(n-1)7

63+2n-2=3+7n-7

61+2n= -4+7n

61+4=7n-2n

65=5n

n=65/5

n=13

Therefore, the two ap have 13th term equal