A2=12 a+d=12 eq............... 1

a7-a4=15

a+6d -( a+3d)/15

a+6d-a-3d=15

3d=15 (a-a=0)

d=15/3

d= 5

so put in eq 1

a+d=12

a+5=12

a=12-5

a=7

so ap is

7,12,17,...... so on

Given:
- The second term of the arithmetic progression (AP) is 12.
- The seventh term of the AP exceeds the fourth term by 15.

To Find:
- The AP.

Solution:
Let's assume that the first term of the AP is 'a' and the common difference is 'd'.

Using the given information:
1. The second term of the AP is 12:
The second term of an AP can be calculated using the formula:
T2 = a + (2-1)d
Substituting the given value, we have:
12 = a + d

2. The seventh term of the AP exceeds the fourth term by 15:
The seventh term of an AP can be calculated using the formula:
T7 = a + (7-1)d
The fourth term of an AP can be calculated using the formula:
T4 = a + (4-1)d
According to the given condition:
T7 - T4 = 15
(a + 6d) - (a + 3d) = 15
3d = 15
d = 5

Substituting the value of 'd' in the first equation:
12 = a + 5
a = 7

The AP:
Using the values of 'a' and 'd' obtained above, the AP can be written as:
a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, ...

Substituting the values of 'a' and 'd' in the above expression:
7, 7 + 5, 7 + 2(5), 7 + 3(5), 7 + 4(5), 7 + 5(5), ...
7, 12, 17, 22, 27, 32, ...

Therefore, the arithmetic progression (AP) with a second term of 12 and the seventh term exceeding the fourth term by 15 is:
7, 12, 17, 22, 27, 32, ...

Note:
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