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.
Answers
➝ Given :-
- Ap : 13, 19, 25,....
- 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)].
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.