 Hi and welcome to the session. Let's work out the following question. The question says the 8th term of an arithmetic progression is 0. Prove that its 38th term is triple its 18th term. So let us see the solution to this question. Let the first term be a common difference. Now we have a8 that is 8th term is 0 or we can say a plus 8 minus 1 into d is equal to 0 because an that is nth term is given by a plus n minus 1 into d. This implies a plus 7d is equal to 0 and this we call 1. Now a18 that is 18th term is a plus 18 minus 1 into d that is equal to a plus 17d. Now from 1 we get a plus 7d is equal to 0. This implies a is equal to minus 7d. Therefore a18 that is the 18th term is equal to minus 7d plus 17d that is equal to 10d and a38 that is the 38th term is equal to a plus 38 minus 1 is 37d. Again we put a to be equal to minus 70 and we get minus 7d plus 37d is equal to 30d. So a38 that is a38th term is equal to 3 times 10d and 10d is the 18th term. So we can say that a38 is equal to 3 into a18. This is what we were supposed to prove in this question. Hence proved. I hope that you understood the solution and enjoyed the session. Have a good day.