Question

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

Answer 1

Answer:

Parameters of AP

First term = 4; Common difference = 6

Explanation:

Let

a = first term of AP

d = common difference

$a_{n}$ = nth term of AP = a + (n - 1)d

Given:

Third term = 16

=> a + (3 - 1)d = 16

=> a + 2d = 16 .......(i)

Given:

7th term exceeds 5th term by 12

=> a + (7 - 1)d = a + (5 - 1)d + 12

=> a + 6d = a + 4d + 12

=> 2d = 12

=> d  = 6 .........(ii)

Substituting for "d" in (i), we get:

a + 2(6) = 16

=> a + 12 = 16

=> a = 4

The AP will be determined as:

a = first term = 4

d = common difference = 6

Verification:

The AP is: 4, 10, 16, 22, 28, 34, 40, 46, ....

We can see that:

third term = 16 √

7th term (40) exceeds 5th term (28) by 12 √

Answer 2

Hope it helps ✌️✌️......

Plz Plz mark me as brainliest