In this question - How many terms of an AP: 9,17,25 must be taken - Maths - Arithmetic Progressions

Question

In this question -> How many terms of an AP: 9,17,25 must be
taken to give a sum of 636 I know how to solve it but the problem
arises when I reach this step -> 4n2
​+5n-636=0 ; I know how to factorize but its
diffcult to find the factors of large number like this So kindly
help and explain to how to find factors easily for these kind of
sums and also using quadratic wont do any good , both methods take
time

Brijendra Pal · Accepted Answer

Dear Student, Agree with your point that quadratic formula and finding D is not good here as it will take enough time
 First of all follow middle term split in such questions
And, keep in mind that when you will factorize 636 , you will reach a point that will indicate that from that point middle term split is applicable
 here 4n2+5n-636=0 Now product of ac = 4×636Now follow the steps as below and keep in mind that you have to takesame steps when similar question comes, = 4×4×159=4×4×3×53Now 53 cannot be factorized further so now we have to adjust 4×4×3×53 in two terms such that their sum of differenceis 5 but sum will be more than 5 53 itself is greater than 5 so we have 4×4×3×53= 48×53and 53-48 = 5 so 48 and 53 are the two split = 4n2+53n-48n-636=0 =n4n+53-124n+53=4n+53n-12so n = 12 taking integral value only