The sum of the first three terms and that of the last three terms of an arithmetic progression are 9 and 57 respectively. It is known that the sum of the whole sequence is 121, find the number of terms in this sequence.
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If we take a as the first term of the sequence and d as the common difference, the first three terms are a, a+ d, and a+ 2d so their sum is 3a+ 3d= 9.
I am also making use of the fact that an arithemetic sequence has the useful property that the average of all terms is the same same average of the first and last terms, so that their sum is .
If there are n terms in the sequence, the last three terms are a+ (n- 3)d, a+ (n- 2)d, and a+ (n-1)d so their sum is 3a+ 3nd- 6d= 57. Subtracting the previous equation, 3nd- 9d= 3d(n- 3)= 48. Now, we can argue that since n is an integer, so is n-2 and so 3d must be an integer divisor of 48. , so 3d must be one of 1, 3, 6, 12, 24, or 48.]
If 3d= 1 then d= 1/3 and n-3= 48 so n= 51. In that case, a= (9- 3(1/3))/3= 8/3 so that the last three terms are 8/3+ (48)(1/3)= 56/3, 8/3+ 49(1/3)= 57/3, and 8/3+ 50(1/3)= 58/3 which add to 57, of course. The sum of all 51 terms is (1/2)(8/3+ 58/3)(51)= (11)(51)= 561, not 121.
If 3d= 3 then d= 1 and n-3= 16 so n= 19. In that case, a= (9- 3)/2= 2. The last term is 2+ 18(1)= 20. The sum of all 19 terms is (1/2)(2+ 20)(19)= 11(19)= 209, not 121.
If 3d= 6, then d= 2 and n-3= 8 so n= 11. In that case, a= (9- 6)/3= 1 so the last term is 21. The sum of all 11 terms is (1/2)(1+ 21)(11)= 11(11)= 121!
If 3d is larger than 6, a= (9- 3d)/3 is not positive.