- This is the basic and conceptual question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise 9.4In this question we have been given an arithmetic progression. And we have to find the 12th term from the end of the arithmetic progressions 3,8,13, … ,253
CBSE DHANPAT RAI publications
Class:- 10th
Understanding CBSE Mathematics
Question 13(ii)
Given that,
A.P = 3,8,13, … ,253
last term is 253
where, a = 3 and d = (8 – 3) = 5
an = 3 + (n – 1)5 = 253
3 + 5n – 5 = 253
5n = 253 + 2 = 255
n = 255/5
n = 51
Hence, the A.P has 51 terms.
therefore, the 12th term from the end is same as (51 – 12 + 1)th of the A.P which is the 40th term.
a40 = 3 + (40 – 1)5
= 3 + 39(5)
= 3 + 195 = 198
Hence, the 12th term from the end of the A.P is 198.