 Hello and welcome to the session. Let us understand the following question today. How many terms of the AP 9, 17, 25 so on must be taken to give a sum of 636. Now let us write the solution. Given to us is AP as 9, 17, 25 and so on. Here A is equal to 9, B is equal to 17 minus 9 which is equal to 8 and sum of n terms is equal to 636. That is SN is equal to 636. We know SN is equal to n by 2 multiplied by 2A plus n minus 1D. Now substituting the values which implies SN is 636 which is equal to n by 2 multiplied by 2 multiplied by 9 plus n minus 1 multiplied by 8. Now which implies 636 is equal to n by 2 multiplied by 18 plus 8n minus 8 which implies 636 is equal to n by 2 multiplied by 8n plus 8 minus 8 is 10. Which implies 636 is equal to n by 2 multiplied by taking 2 common so we left with 4n plus 5. Now here this 2 and this 2 gets cancelled so we are left with 636 is equal to n multiplied by 4n plus 5. Which implies 636 is equal to 4n square plus 5n. Which implies 4n square plus 5n minus 636 is equal to 0. Now splitting the middle term and solving this further we get 4n square plus 53n minus 48n minus 636 is equal to 0. Which implies taking n common so we get 4n plus 53. Now taking minus 12 common from these 2 term we get 4n plus 53 is equal to 0. Which implies n minus 12 multiplied by 4n plus 53 is equal to 0. Therefore 12 is equal to 0 or 4n plus 53 is equal to 0. Which implies n is equal to 12 or n is equal to minus 53 by 4. But n is equal to minus 53 by 4 is not possible because number of terms cannot be negative in fractions. So n is equal to 12 hence require number of terms is equal to 12. I hope you understood the question. Bye and have a nice day.