What is the solution of question 4 of class 10 arithmetic progressions. Best and easy way to solve the question its very important for class 10. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Share
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Let there be n terms of the AP. 9, 17, 25 …
For this A.P.,
First term, a = 9
Common difference, d = a2−a1 = 17−9 = 8
As, the sum of n terms, is;
Sn = n/2 [2a+(n -1)d]
636 = n/2 [2×a+(8-1)×8]
636 = n/2 [18+(n-1)×8]
636 = n [9 +4n −4]
636 = n (4n +5)
4n2 +5n −636 = 0
4n2 +53n −48n −636 = 0
n (4n + 53)−12 (4n + 53) = 0
(4n +53)(n −12) = 0
Either 4n+53 = 0 or n−12 = 0
n = (-53/4) or n = 12
n cannot be negative or fraction, therefore, n = 12 only.