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 there be an A.P.
first term a and common difference d.
Also, If an denotes its nth term and Sn is the sum of first n terms,
Now we have to find n and an, if a = 2, d = 8 and Sn = 90
CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 56(vi)
Given A.P. has a=2, d=8 and Sn=90.
Now by using the formula of the sum of n terms of an A.P.
Sn = n[2a +(n – 1)d] / 2
So,
=> 90 = n[2(2) + (n – 1)8] / 2
=> 90 = n[2 + 4n – 4)]
=> 4n2 – 2n – 90 = 0
=> 4n2 – 20n + 18n – 90 = 0
=> 4n(n – 5) + 18n(n – 5) = 0
=> (n – 5) (4n + 18) = 0
=> n = 5 or n = –9/2
Ignoring n = –9/2 as n cannot be a fraction as well as negative. So, we get n = 5.
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
= 2 + (5 – 1)8
= 2 + 32
= 34
Hence, the value of n is 5 and an is 34