Question

For what value of n are the nth terms of the following two APs the same 13, 19, 25, ... and 69, 68, 67, ...? Also, find this term.​

Answer 1

➝ Given :-

1. Ap : 13, 19, 25,....
2. Ap : 69, 68, 67,....

➝ To find :-

• term of the Ap

➝ Solution :-

Let nth terms of the given progressions be t and T respectively.

➛ The first AP is 13, 19, 25, ....

Let its first term be a and common difference be d. Then,

→ a = 13

→ d = (19-13) = 6.

So, its nth term is given by

↪ t= a+(n-1) d

↪ t = 13+ (n - 1) × 6

↪ t = 6n +7

_____________

➛ The second AP is 69,68, 67, ....

Let its first term be A and common difference be D. Then,

→ A = 69

→ D = (68-69) = -1.

So, it nth term is given by

↪T = A + (n - 1) ×D

↪T = 69+ (n - 1) ×(-1)

↪ T = 70 - n

Now, t = T = 6n + 7 = 70 – n

7n = 63 → n = 9.

Hence, the 9th term of each AP is the same.

This term = 70 - 9 = 61 [ T = (70 - n)].

Answer 2

Given,

a 1

=13

d 1

=19−13=6

T 1

(n)=13+(n−1)6........(1)

a 2

=69

d 2

=68−69=−1

T 2

(n)=69+(n−1)(−1)........(2)

equating (1) and (2), we get,

13+(n−1)6=69+(n−1)(−1)

7n=70−7=63

∴n=9

T 1

(9)=13+(9−1)6=61...........A

T 2

(9)=69+(9−1)(−1)=61........B

both the values of A and B are same.