Here, First term = a = 5

Common difference = d = 12 - 5 = 7

No. of terms = 50

∴ last term will be 50^th term.

Using the formula for finding n^th term of an A.P.,

l = a₅₀ = a + (50 - 1) × d

∴ l = 5 + (50 - 1) × 7

⇒ l = 5 + 343 = 348

Now, sum of last 15 terms = sum of first 50 terms - sum of first 35 terms

i.e. sum of last 15 terms = S₅₀ - S₃₅

Now, Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ Sum of first 50 terms is given by:

S₅₀ = [2(5) + (50 - 1)(7)]

= 25 × [10 + 343]

= 25 × 353

= 8825

Now, Sum of first 35 terms is given by:

S₃₅ = [2(5) + (35 - 1)(7)]

= (35/2) × [10 + 238]

= (35/2) × 248

= 35 × 124

= 4340

Now, S₅₀ - S₃₅ = 8825 - 4340

= 4485

∴ last term = 348, sum of last 15 terms = 4485