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

Class 10th

Chapter number- 9

Exercise :- 9.6

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 number of terms

Also it is given that if 1 is added to each term of this A.P., then we have to find the sum of all terms of the A.P. thus obtained.

CBSE DHANPAT RAI publication

CBSE Mathematics Class 10th

Question 37

If 1 is added to each term of the A.P. then the new A.P. is –11, –8, –5, . . . , 22.

First term, a = –11 and common difference, d = – 8 – (–11) = 3

And, we know that nth term = a

_{n}= a + (n – 1)d=> 22 = –11 + (n – 1)3

=> 3n = 36

=> n = 12

By using the formula of the sum of first n terms of an A.P.

S

_{n}= n[a + a_{n}] / 2.=> S

_{12}= 12[–11 + 22]/2= 6[11]

= 66

Hence, the sum after adding 1 to each of the terms in the A.P is 66.