Question

find an ap whose second term is 10 and the 6th term exceeds the 4th term by 12

Answer 1

10=a +(2-1)d..(i)
12=a+(6-1)d-{a+(4-1)d}=a+5d-a-3d=2d
12=2d
6=d(put this in eq i
10=a+6
4=a
Acc to question AP is:-
4,4+6=10,10+6=16...and so on
4,10,16,22.....

Answer 2

$\huge\sf{\underline{\underline{Solution:}}}$

Given :-

• 2nd term is 10 and the 6th term exceeds the 4th term by 12.

To Find :-

• A.P

_______________________________________

⇒ a + 5d = a + 3d  +12

⇒ 5d - 3d = 12

⇒ 2d = 12

⇒ d = 12/2

⇒ d = 6

⇒ a + d = 10

⇒ a + 6 = 10  

⇒ a = 10 - 6

⇒ a = 4

Now Finding 6th and 4th term

6th Term

⇒a + 5d

⇒ 4 + 5(6)

⇒4 + 30

⇒30

4th Term

⇒a + 3d

⇒4 + 3(6)

⇒4 + 18

⇒22

Hence, the terms of A.P are 4, 10, 16, 22, 28 etc