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Sequences and Series question

2004 · Shift 0 · Q94
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Sequences and Series question

2004 · Shift 0 · Q94

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Let Tr{{T_r}}Tr​ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, men,  Tm=1n  and Tn=1m, m e n,\,\,{T_m} = {1 \over n}\,\,and\,{T_n} = {1 \over m},\,men,Tm​=n1​andTn​=m1​, then a - d equals
  1. A
    1m+1n{1 \over m} + {1 \over n}m1​+n1​
  2. B
    1
  3. C
    1m n{1 \over {m\,n}}mn1​
  4. D
    0
View written solutionFree

Correct answer: D

  1. For an A.P., the rrrth term is Tr=a+(r−1)d.T_r=a+(r-1)d.Tr​=a+(r−1)d.

So, Tm=a+(m−1)d=1n...(1)T_m=a+(m-1)d=\frac{1}{n} \quad ...(1)Tm​=a+(m−1)d=n1​...(1) Tn=a+(n−1)d=1m...(2)T_n=a+(n-1)d=\frac{1}{m} \quad ...(2)Tn​=a+(n−1)d=m1​...(2)

  1. Subtract (1) from (2): [a+(n−1)d]−[a+(m−1)d]=1m−1n[a+(n-1)d]-[a+(m-1)d]=\frac{1}{m}-\frac{1}{n}[a+(n−1)d]−[a+(m−1)d]=m1​−n1​ (n−m)d=n−mmn.(n-m)d=\frac{n-m}{mn}.(n−m)d=mnn−m​.

Since m≠nm\ne nm=n, divide by (n−m)(n-m)(n−m): d=1mn.d=\frac{1}{mn}. d=mn1​.

  1. Substitute this into (1): a+(m−1)⋅1mn=1na+(m-1)\cdot \frac{1}{mn}=\frac{1}{n}a+(m−1)⋅mn1​=n1​ a=1n−m−1mna=\frac{1}{n}-\frac{m-1}{mn}a=n1​−mnm−1​ a=m−(m−1)mn=1mn.a=\frac{m-(m-1)}{mn}=\frac{1}{mn}. a=mnm−(m−1)​=mn1​.

Thus, a=d=1mn.a=d=\frac{1}{mn}. a=d=mn1​.

Therefore, a−d=0.a-d=0.a−d=0.

  1. Checking options:
  • A: 1m+1n\frac{1}{m}+\frac{1}{n}m1​+n1​ — incorrect
  • B: 111 — incorrect
  • C: 1mn\frac{1}{mn}mn1​ — incorrect
  • D: 000 — correct

Hence the correct answer is D.

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