Q. 344.1( 9 Votes )

An AP 5, 12, 19, ... has 50 terms. Find its last term. Hence, find the sum of its last 15 terms.

Answer :

Here, First term = a = 5

Common difference = d = 12 - 5 = 7


No. of terms = 50


last term will be 50th term.


Using the formula for finding nth 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


Sn = [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


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