Question

diameter the AP whose third term is 16 and the 7th term exceeds the 5th term by 12​

Answer 1

Hey User !!!

Let a be the First term, a3 be the third term, a5 be the 5th term and a7 be the 7th term... (1)

Let the common difference be "d"

Common difference is equal in AP  

So,

a7 = a5 + d + d = a5 + 2d ...(2)

From Equation (1) & (2)

a5 + 12 = a5 + 2d  

2d = 12

d = 6

From Given, we get that

a3 = 16

a3 = a + 2d = 16

a + ( 2 × 6 ) = 16              [ We know that d = 6 ]

a + 12 = 16

a = 4

So first term is 4 ... We can find AP by adding d continuously

So, AP is 4, 10, 16, 22, 28..

Hope it helps !!!

Answer 2

Answer:

a+2d=16

a+6d-(a+4d)=12

2d=12

d=6

Put d =6 in eq 1

a+2×6=16

a=16-12=4

So ap is :-

4,10,16,22,28,34,40...