Question

Find arithmetic progression whose third term is 16 and the seventh 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

a3 = 16

a7 = a5 + 12 ............ (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

given

a3 = 16

a + 2d = 16

a7 - a5 =12

As a7 = a+6d

a5 = a+4d

So

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

a+6d - a-4d = 12

2d=12

d = 6

Substitute d in a+2d = 16

a +2(6)=16

a + 12 = 16

a = 16 - 12

a = 4