Question

For what value of n, are the nth terms of two APs;$63,65,67$.....and$3,10,17$.....equal?
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Answer 1

(A)

$AP:3,10,\mathrm{17.........}\text{AP:69,65,67}$

$\text{d = 7}{\text{d}}^1=2$

$\text{a = 3}{\text{a}}^1\text{= 63}$

${\text{a}}_n=3+\left(n-1\right)7{\text{a}}_n=63+\left(n-1\right)2$

$\text{= 7n - 4}\text{2n + 61}$

$\text{7n - 4 = 2n + 61}$

$\text{5n = 65}$

$\text{n = 13}$

(B)

$a_3=16\text{a + 6a - a - 4d = 12}$

$a+\left(3-1\right)d=16\text{2d = 15}$

$a+2d=16\to \left(1\right)\text{d = 6}$

$\text{put d = 6}$

$a+2\left(6\right)=16$

$a=16-12$

$a=4$

$\therefore \text{a = 4}$

$\text{a + d = 4 + 6 = 10}$

$\text{a + 2d = 4 + 12 = 16}$

$\therefore AP=4,10,\mathrm{16.......}$