 Now those students has work of the following problem. It says the sixth term of an arithmetic progression that is AP is minus 10 and the tenth term is minus 26. Determine the fifteenth term of the AP. So let's now move on to the solution. To find the fifteenth term we should know the journal formula for finding the nth term of the AP. So the nth term of the AP that is arithmetic progression is given by a n which is equal to a plus n minus 1 into d where a is the first term of the AP and d is the common difference. Whenever you do such problems you need to write what is a what is d and that is very important. Now we are given that the sixth term the arithmetic progression that is AP is minus 10 that is a6 is equal to a plus 6 minus 1 into d. Now a6 is minus 10 so this implies a plus 6 minus 1 that is 5 into d is minus 10. Let's call this as equation 1. Now we are given that the nth term of AP is minus 26. So again using this formula we have minus 26 is equal to a plus 10 minus 1 d that is plus 9 d is equal to minus 26. Let's name this as 2 and here you can also write that nth term is 8n which is given by a plus 10 minus 1 into d and 8n is minus 26. Now we will solve these two equations for a and d because to find the 15th term we need to know the first term and the common difference. So we subtract equation 2 from equation 1 like this a plus 5 d is equal to minus 10 and a plus 9 d is equal to minus 26. Now subtracting whenever we subtract sign changes so plus a gets cancelled with minus a plus 5 d minus 9 d is minus 4 d and 26 minus 10 is 16. So this implies d is equal to 16 upon minus 4 and this implies d is equal to minus 4. Now we will substitute the value of d in any one of the two equations to get the value of a. So we have from equation 1 we have a plus 5 d which is minus 4 is equal to minus 10. So this implies a minus 20 is equal to minus 10 and this implies a is equal to minus 10 plus 20 and this implies a is equal to 10. So the first term is 10. Now since we have obtained the first term and the common difference of the arithmetic progression we can easily find out the 15th term and the 15th term a15 is given by a plus 15 minus 1 into d. Now substitute the values of a and d here. So a15 is equal to 10 plus 14 into minus 4. So this is equal to minus 46. So the 15th term of the arithmetic progression is minus 46. So this completes the question and the session. Bye for now. Take care. Have a very nice day.