Here, First term = a = 8

Common difference = d = 10 - 8 = 2

No. of terms = 60

∴ last term will be 60^th term.

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

l = a₆₀ = a + (60 - 1) × d

∴ l = 8 + (60 - 1) × 2

⇒ l = 8 + 118 = 126

Now, sum of last 10 terms = sum of first 60 terms - sum of first 50 terms

i.e. sum of last 10 terms = S₆₀ - S₅₀

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(8) + (50 - 1)(2)]

= 25 × [16 + 98]

= 25 × 114

= 2850

Now, Sum of first 60 terms is given by:

S₆₀ = [2(8) + (60 - 1)(2)]

= 30 × [16 + 118]

= 30 × 248

= 4020

Now, S₆₀ - S₅₀ = 4020 - 2850

= 1170

∴ last term = 126, sum of last 10 terms = 1170