Question

Find the arithmetic progression whose third term is 16 and the seventh term exceeds its fifth term by 12.

Answer 1

Here, let us take the first term of the A.P as a and the common difference of the A.P as d

Now, as we know,

`a_n = a + (n - 1)d`

So for the 3rd term (n = 3)

`a_3 = a + (3 - 1)d`

16 = a + 2d

a = 16 - 2d ......(1)

Also for 5th term (n = 5)

`a_5 = a + (5 - 1)d`

= a + 4d

For 7th term (n = 7)

`a_7 = a + (7 - 1)d`

= a + 6d

Now we are given

`a_7 = 12 + a_5`

a + 6d = 12 + a + 4d

6d - 4d = 12

2d = 12

d = 6

Substituting the value of d in (1), we get,

`a = 16 - 2(6)`

= 16 - 12

= 4

So the first term is 4 and the common difference is 6

Therefore the A.P is 4, 10, 16, 22,....