 Hi and welcome to the session I am Shashin and I am going to help you to solve the following question. Question is the sum of 4th and 8th terms of an AP is 24 and the sum of 6th and 10th terms is 44. Find the first three terms of the AP. First of all let us understand the key idea to solve the given question. The nth term of an AP with first term a and common difference d is given by a n is equal to a plus n minus 1 multiplied by d. Let us now start with the solution. Let the first three terms of the AP be a a plus g a plus 2 d. Now clearly we can see to find the three terms of the AP we need to find a and d. Now we know according to the question 4th term of AP plus 8th term of AP must be equal to 24. We know 4th term of AP is denoted by a4 and 8th term of AP is a8 is equal to 24. Let us name this equation as 1. Now we know by key idea n that is a nth term of the AP is given by a plus n minus 1 multiplied by d where a is the first term of the AP and d is the common difference. So 4th term is equal to a plus 4 minus 1 multiplied by d. We had substituted 4 for n. Now we get 4th term is equal to a plus 3 d. Similarly we can find out 8th term. Now 8th term is equal to a plus 8 minus 1 multiplied by d. So we get 8th term is equal to a plus 7 d. Now substituting the value of 4th term and 8th term in equation 1 we get a plus 3 d plus 8 plus 7 d is equal to 24. This implies 2 a plus 10 d is equal to 24. Now this further implies 2 multiplied by a plus 5 d is equal to 24. This implies a plus 5 d is equal to 24 divided by 2. This implies a plus 5 d is equal to 12. Here we have taken 2 as a common factor. Now dividing the both sides by 2 we get a plus 5 d is equal to 12. Also the second condition given in the question is the sum of the 6th term and the 10th term of the AP is equal to 44. Now therefore we can write 6th term as a6 and 10th term as 8n. So we get a6 plus 8n is equal to 44. Now let us find out the values for a6 and 8n below a6 is equal to a plus 6 minus 1 multiplied by d which implies a6 is equal to a plus 5 d. Now we know nth term of the AP is given by a plus n minus 1 multiplied by d. Now here we to find the 6th term we have substituted for n6. So we get 6th term of the AP is equal to a plus 5 d. Similarly we can find out the 10th term of the AP. So we get 10th term of the AP is equal to a plus 9 d. Now let us name this equation as 2. Now we will substitute the corresponding values of a6 and 8n in the equation 2. Now substituting for a6 and 8n in equation 2 we get a plus 5 d plus a plus 9 d is equal to 44. This implies 2a plus 14 d is equal to 44. This implies 2 multiplied by a plus 7 d is equal to 44. Here we have taken 2 as a common factor on left hand side. Now dividing both sides by 2 we get the equation a plus 7 d is equal to 22. Now we have already proved above that a plus 5 d is equal to 12. So we can rewrite the equation a plus 5 d is equal to 12. Let us name this equation as 3 and this equation as 4. Now subtracting equation 4 from equation 3 we get a plus 7 d minus a plus 5 d is equal to 22 minus 12. Now this implies a plus 7 d minus a minus 5 d is equal to 10, a and a will get cancelled. Now we get 2d is equal to 10. This implies d is equal to 10 divided by 2 or d is equal to 5. Therefore we get d is equal to 5. Now substituting d is equal to 5 in equation 4 we get a plus 5 divided by 5 is equal to 12. This implies a plus 25 is equal to 12. This implies a is equal to 12 minus 25 or we can write a is equal to minus 13. Therefore we get a is equal to minus 13. As for a is equal to minus 13 and d is equal to 5, first term of a p that is a is equal to minus 13, second term is equal to a plus d is equal to minus 13 plus 5 is equal to minus 8, third term of the a p is equal to a plus 2d that is equal to minus 13 plus 2 multiplied by 5 which is equal to minus 3. So the required three terms of the a p are minus 13, minus 8, minus 3. So the first three terms of a p are minus 13, minus 8, minus 3. This is our required answer. Take care and good bye.