Question

For what value of n are the nth term of two APs 63,65,67........... and 3,10,17........equal?

Answer 1

Answer:

Required numeric value of n is 13.

Step-by-step explanation:

Let the sequence 63, 65, 67..... be AP${}_{1}$ and sequence 3, 10, 17.... be AP$_{2}$

Here,

First term of AP$_{1}$ = 63

Common Difference between the terms of AP$_{1}$ = 65 - 63 = 2

First term of AP$_{2}$ = 3

Common Difference between the terms of AP$_{1}$ = 10 - 3 = 7

From the properties of arithmetic progressions :

• nth term = a + ( n - 1 )d, where a is the first term, d is the common difference between the terms and n is the total number of terms.

According to the question :

= > nth term of AP$_{1}$ = nth term of AP$_{2}$

= > first term of AP$_{1}$ + { ( n - 1 ) common difference between the terms of AP$_{1}$ } = first term of AP$_{2}$ + { ( n - 1 )common difference between the terms of AP$_{2}$ }

Substituting the values from above :

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

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

= > 60 = 5( n - 1 )

= > 60 / 5 = n - 1

= > 12 = n - 1

= > 12 + 1 = n

= > 13 = n

Hence the 13th terms of both the APs are equal.

Answer 2

Answer:

13

Step-by-step explanation:

nth term = a+(n-1)d

for 1 St sequence ,here a=63 & d=2

nth term is 63+(n-1)2

for 2nd sequence,here a=3 & d= 7

nth term is 3+(n-1)7

as given both are equal

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

60=(n-1)5

12=n-1

n=13