the sum of first 9 terms of an AP is 171 and that of the first 24 is 996.find the AP Share with your friends Share 1 Varun.Rawat answered this Let a be the first term and d be the common difference.Now, sum of first n terms of an AP is given by,Sn = n22a+n-1dNow, S9 = 171⇒922a+8d = 171⇒2a+8d = 38 .....1S24 = 996⇒2422a+23d = 996⇒2a+23d = 83 .....2Subtracting 1 from 2, we get 15d = 45⇒d = 3Now, from 1, we get2a+24 = 38⇒2a = 14⇒a = 7So, the required AP is :7, 10, 13, 16, .......... 2 View Full Answer Tanush Jain answered this S9-171 and S24-996 S9= 9/2(2a+(9-1)d) = 9/2(2a+8d)... = 9a +36d . S24= 12(2a+23d)...=24a+276d. 24a + 276d and 9a + 36d....now solve both the equations and get a and d!! 0 Arshik Anilkumar answered this we have given that S9 = 171 and S24 = 996 we know that Sn = n/2 [2a+ (n-1) d] by applying this formula on sum of 9 and 24 S9 = 9/2 [2a + (9-1)*d] 171 = 9/2 [ 2a + 8d ] 2a + 8d = 38 .....(1) now S24 = 24/2 [2a + (24-1) d] 996 = 12 [2a + 23d] 2a + 23d = 83 ........(2) now subtracting eq.(1) from eq.(2) 2a + 23d = 83 2a + 8d = 38 - - - 15d = 45 d = 3 substituting the value of d in eq(1) 2a + 8*3 = 38 2a = 38-24 a = 14/2 = 7 as we have got a and d we can form the AP 3,10,13,16 ...... 2 Anjana Satheesh answered this S9 = 171 9/2 [2a + (9-1)d] = 171 18a + 72d = 342 a + 4d = 19 ......... (i) S24 = 996 24/2 [2a + (24-1)d] = 996 24a + 276d = 996 2a + 23d = 83 ......... (ii) Multiply (i) by 2 2a + 8d = 38 ........... (iii) Solve (ii) and (iii) 2a + 23d = 83 2a + 8d = 38 d = 3 Substitute in (i) a + 4 (3) = 19 a = 7 Hence , the A.P. is 7, 10, 13,..... 0
