One of the most important and exam oriented question from Chapter name- Arithmetic Progression

Class 10th

Chapter number- 9

Exercise :- 9.4

This type of question has been asked in previous years exams.

In this question we have been given an arithmetic progression. And we have to find the 12th term from the end of the arithmetic progressions 1, 4, 7, 10, … ,88

CBSE DHANPAT RAI publication

Understanding CBSE Mathematics

Question 13(iii)

Given that,

A.P = 1, 4, 7, 10, … ,88

where, a = 1 and d = (4 – 1) = 3

last term is 88

an = 1 + (n – 1)3 = 88

1 + 3n – 3 = 8

3n = 90

n = 30

Hence, the A.P has 30 terms.

therefore, the 12th term from the end is same as (30 – 12 + 1)th of the A.P which is the 19th term.

= a

_{89 }= 1 + (19 – 1)3= 1 + 18(3) = 1 + 54 = 55

Hence the 12th term from the end of the A.P is 55.