Question

for what value of n , are the nth terms of two aps 63,65,67........and 3,10,17.........equal​

Answer 1

Answer:

n=13

Step-by-step explanation:

AP₁= 63, 65, 67 ...  [ a= 63 d= 2 ]

AP₂ = 3,10, 67  ..... [ A= 3 D=7 ]

an = An

a+(n-1) d = A + (n-1) D

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

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

60 = 5n - 5= 5(n-1)

60/5 = n-1

12 = n-1

n= 13

hope it helps

Answer 2

Answer:-

First term of first AP = A = 63 and common difference = D = 65-63 = 2

$\large\sf{Therefore,}$

$\small\sf{An=A+(n-1)D}$

$\small\sf{An=63+(n-1)2}$

First term of second AP = a=3 and common difference, d = 10-3 = 7.

$\large\sf{Therefore,}$

$\small\sf{an=a+(n-1)d}$

$\small\sf{an=3+(n-1)7}$

$\large\sf{A/C,}$

$\small\sf{An=an}$

$\small\sf{63+(n-1)2=3+(n-1)7}$

$\small\sf{63+2n-2=3+7n-7}$

$\small\sf{63=5n}$

$\small\sf{n=13}$

Hence, the 13th term of both the APs are equal.