A1=5,
d=12-5=7,
n=50,
a50=a+(n-1)d,
=5+(49)7=5+343=348,
a15=5+14(7)=5+98=103,
s15=15/2(5+103)
=15/2(108),
=15×54=810

Given:
An arithmetic progression (AP) with the first term (a) as 5, the common difference (d) as 12 - 5 = 7, and the number of terms (n) as 50.

To find:
The last term of the AP and the sum of the last 15 terms.

Solution:

Finding the last term:
The formula to find the nth term (Tn) of an AP is given by:
Tn = a + (n - 1)d

Substituting the given values, we have:
T50 = 5 + (50 - 1)7
T50 = 5 + 49 * 7
T50 = 5 + 343
T50 = 348

Therefore, the last term of the AP is 348.

Finding the sum of the last 15 terms:
The formula to find the sum of the first n terms (Sn) of an AP is given by:
Sn = (n/2)(2a + (n - 1)d)

We need to find the sum of the last 15 terms, so we will use the formula with n = 15.
Substituting the given values, we have:
S15 = (15/2)(2 * 5 + (15 - 1)7)
S15 = (15/2)(10 + 14 * 7)
S15 = (15/2)(10 + 98)
S15 = (15/2)(108)
S15 = 15 * 54
S15 = 810

Therefore, the sum of the last 15 terms of the AP is 810.

Summary:
The last term of the given arithmetic progression is 348. The sum of the last 15 terms of the AP is 810.