Question

The 9th term of an AP is equal to 6 times its second term. If 5th term is 22, find the AP.

Answer 1

9 th term of an AP = 6 * 2nd term

a + ( 9 - 1 ) d = 6 ( a + ( 2 - 1 ) d

a + 8d = 6 ( a + d)

a + 8d = 6a + 6d

8d - 6d = 6a - a

2d = 5a

d = 5/2 a

nw its given tat 5th term is 22

so,.....

22 = a + (n-1)d

22 = a + ( 5 - 1)d

22 = a + 4d

22 = a + 4 (5/2 a ) (the value which we found above for d )

22 = a + 2 * 5 a ( 4 and 2 will get cancelled)

22 = a + 10 a

22 = 11a

a = 2

d =5/2 a

d = 5/2*2

d = 5

nw the AP is: 2,7.12,17,22,27. Hope it helps you.

Answer 2

Let the first term be "a" and common difference "d"

Ninth term = a + (9 - 1)d = a + 8d
Second term = a + (2 - 1)d = a + d

A/Q,
a + 8d = 6(a + d)
a + 8d = 6a + 6d
- 5a + 2d = 0

Fifth term = a + (5 - 1)d = a + 4d = 22

- 5a + 2d = 0
a + 4d = 22

Solving the equations ,
- 11a = - 22
a = 2

Now, 2 + 4d = 22
4d = 20
d = 5

First term = 2
Second term = a + d = 2 + 5 = 7
Third term = a + 2d = 2 + 10 = 12
Fourth term = a + 3d = 2 + 15 = 17
Fifth term = a + 4d = 2 + 20 = 22

The required AP = 2 , 7 , 12 , 17 , 22 ..................

Hope This Helps You!