Question

Determine the A.P. whose $3^{rd}$ term is 16 and $7^{th}$ term exceeds the $5^{th}$ term by 12.

Answer 1

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"

The 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.......