Answer :

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 =* a_{50} = 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 = S_{50} - S_{35}

Now, Sum of first n terms of an AP is

S_{n} = [2a + (n - 1)d]

∴ Sum of first 50 terms is given by:

S_{50} = [2(5) + (50 - 1)(7)]

= 25 × [10 + 343]

= 25 × 353

= 8825

Now, Sum of first 35 terms is given by:

S_{35} = [2(5) + (35 - 1)(7)]

= (35/2) × [10 + 238]

= (35/2) × 248

= 35 × 124

= 4340

Now, S_{50} - S_{35} = 8825 - 4340

= 4485

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

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