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.
Sn = n[2a + (n − 1)d] / 2.
So, S7 = 7(2a + (7 − 1)d) / 2
=> 7(2a + 6d) = 364
=> 14a + 42d = 364
=> a + 3d = 26 .… (1)
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
We are given, a4 : a17 = 1:5
=> a17 = 5a4
=> a+16d = 5[a + 3d]
=> a + 16d = 5a + 15d
=> 4a = d …. (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 and
the common difference is 8, So, the A.P. is 2, 10, 18, 26, …..