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 that the sum of first seven terms of an A.P. is 182.

Also it is given that its 4th and 17th terms are in ratio 1: 5,

Now we have to find the A.P.

CBSE DHANPAT RAI publication

CBSE Mathematics Class 10th

Question 31

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

S

_{n}Â = n[2a + (n âˆ’ 1)d] / 2.So, S

_{7}Â = 7(2a + (7 âˆ’ 1)d) / 2=> 7(2a + 6d) = 364

=> 14a + 42d = 364

=> a + 3d = 26 .â€¦ (1)

By using the formula of nth term of an A.P.

a

_{n}Â = a + (n â€“ 1)dSo,

We are given, a

_{4Â }: a_{17}Â = 1:5=> a

_{17}Â = 5a_{4}=> a+16d = 5[a + 3d]

=> a + 16d = 5a + 15d

=> 4a = d Â â€¦. (2)

Using eq(2) in eq(1), we get,

=> a + 3(4a) = 26

=> 13a = 26

=> a = 2

On putting a = 2 in eq(2), we get,

=> d = 8

As the first term of the given A.P. is 2 andÂthe common difference is 8, So, the A.P. is 2, 10, 18, 26, â€¦..