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An A.P. has even number of terms, the first term of which is unity.S1 & S2 stand for the sum of its first n terms & last n terms such that S1/S2 = q. What is the common difference of the A.P.?

An A.P. has even number of terms, the first term of which is unity.S1 & S2 stand for the sum of its first n terms & last n terms such that S1/S2 = q. What is the common difference of the A.P.?

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3 Answers

Chetan Mandayam Nayakar
312 Points
9 years ago

let d be common difference. S1=(n/2)(2+(n-1)d) S2=(n/2)((1+nd)+(n-1)d), ((2n-1)d+1)/((n-1)d+2)=(1/q)

(2n-1)d+1=((n-1)/q)d+(2/q), d=((2/q)-1)/((2n-1)-((n-1)/q))

chitee gang
4 Points
9 years ago

thannkyou chetan, but it is a question from M.L. khanna, problem set 92 of progressions (Q.25 c part) and in the solutions, the ans. is given as 2. along with this a hint is also given which reads : S1/S2 = Sn/S2n - Sn = q.put n = 1,2. actually i m not getting this..would u please help...i really want to make into IITb & i m preparing myself for jee. so your explaination will be of great help to me.

Gaurav
26 Points
3 years ago
Let the number of terms be 2n and common difference be n .sn=n2[2a+(n−1)d]sn=S1=n2[2+(n−1)d]s2n=2n2[2+(2n−1)d]Sum of last n terms=S2=s2n−snS2=2n2[2+(2n−1)d]−n2[2+(n−1)d]S1S2=λS2S1=1λ2n2[2+(2n−1)d]−n2[2+(n−1)d]n2[2+(n−1)d]=1λ2n2[2+(2n−1)d]n2[2+(n−1)d]−1=1λ2n2[2+(2n−1)d]n2[2+(n−1)d]=1+1λ2 [2+(2n−1)d][2+(n−1)d]=λ+1λ2+(2n−1)d2+(n−1)d=λ+12λ=k (say)2+(2n−1)d=2k+k(n−1)d(2n−1)d−k(n−1)d=2k−2(2n−1−kn+k)d=2k−2((2−k)n+k−1)d=2k−2d=2(k−1)(2−k)n+k−1Common difference should be independent of n .So , 2−k=0k=2andd=2(2−1)(2−2)n+2−1=2 ANS...Common difference is 2 .

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