Question

, determine the AP whose third term is 16 and the 7 term exceeds the 15 by 12

Answer 1

Given;
Third Term of A. P = a+2d .
As per the question,
a + 2d = 16 .
a = 16 - 2d .

15 term of A. P = a+14d
7th term of A. P = a+6d.

As per the question,
a+6d - 12 = a+ 14d

6d - 14d = 12

-8d = 12 .

d = -12/8 = -3/2 .

The first term of A. P = 16 - 2(-3/2) = 16+3 = 19 .

The A. P is 19 , 17 1/2 , 16 ..............

Answer 2

$a_{3} = 16$
a + 2d = 16
a = 16 - 2d

$a_{7} = a_{15} + 12$
a + 6d = a + 14d + 12
14d - 6d = - 12
8d = -12
d = -12/8
d = -3/2

a = 16 - 2(-3/2)
= 16 + 3
= 19
$a_{1} = 19$
$a_{2} = a_{1} + d = 19 + (-3/2) = (38-3)/2 = 35/2 = 17.5$
$a_{3} = a_{2} + d = 35/2 + (-3/2) = 32/2 = 16$

Required AP :
19, 17.5 , 16, .....

Hope it helps