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

Become a Mathematics wizard and get above

Let there be

nterms of the AP. 9, 17, 25 …For this A.P.,

First term,

a= 9Common difference, d=a_{2}−a_{1}= 17−9 = 8As, the sum of n terms, is;

S=_{n}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)4

n^{2}+5n−636 = 04

n^{2}+53n−48n−636 = 0n(4n+ 53)−12 (4n+ 53) = 0(4

n+53)(n−12) = 0Either 4

n+53 = 0 orn−12 = 0n= (-53/4) orn= 12ncannot be negative or fraction, therefore,n= 12 only.