Question

the first terms of an arithmetic sequence is 1 and the sum of first four terms is 100. find the first four terms

Answer 1

Write the terms as: a-3x, a-x,a+x,a+3x...Observe that the lowest term is a-3x and common difference is 2x.On adding, you get 4a=44 or a=11 and (a-3x)(a+(n-1)2x-3x) = 140 =(3x-11)(5x-2nx-11) =140=>x^2 (15-6n)-(33+55)x+22nx+121-140 =0=>x^2(15-6n) +x(22n-88) -19 =0The discriminant of this equationD= 484(n-4)^2  -4(19)(6n-15)=4(121n^2+16(121)-968(n) -(114n-(19)(15))=4(121n^2-1082n +16(121)+(19)(15))Now solve to find the constraint on n and how large it should be by yourself.Remember that the value of the common difference is 2x where n>4 which depends on number of terms in progression.