[Edwyne]

1. (a) if S1, S2, S3 be the sum of n, 2n, 3n terms respectively of an A.P, prove that S3= 3(S2-S1). (S1 is numaer of first n terms).
(b) find four terms in A.P whose sum is 20 and the sum of whose square is 120.
(c) the digits of a positive integer having three digits are in A.P and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
(d) if a,b,c are in A.P, prove that (I) (ab)^-1, 1/ac, 1/bc are also in A.P. (ii) a(b+c)/bc, b(a+c)/ca, c(a+b)/ab are also in A.P.
(e) if a^2, b^2, c^2 are in A.P, prove that (I) 1/(b+c), 1/(c+a), 1/(a+b) are also in A.P. (ii) a/(b+c), b/(c+a), c/(a+b) are also in A.P.
(f) the sum to infinity of a G.P series is R. The sum to infinity of the squares of the terms is 2R. The sum to infinity of the cubes of the terms is (64/13)R. Find (I) the value of R. (ii) the first term of the first original series.
2. Solve for x
(I) x(x+1)^2(x+2)=72
(ii) (x-5)(x-7)(x+4)(x+6)=504

[kreysiz]

1) a) sum of 1st n terms = (n/2)*[2a + (n-1)d] ; where n = number of terms ; a = first number of the sequence ; d = common difference between consecutive numbers
sum of 1st 2n numbers = (2n/2)*[2a + (2n-1)d]
sum of 1st 3n numbers = (3n/2)*[2a + (2n-1)d]
Use the above results to get your answer
b) Let the four numbers be ( a-3d) , (a-d) ; (a+d) ; (a+3d)
c) Let the three digits of the number be (a-d),a,(a+d)
d) Use the basic definition of AP i.e second term - first term = third term - second term = common difference
e) same as (d)
f) Sum of the infinte series in G.P is given by the eqaution:-
sum = a/(1-r) ; where a = first term of the series ; r = common ratio of the series = second term/first term