Question

# List- IList-II(I)Number of integral solutions of(P)  132x+y+z=1, x≥−4, y≥−4, z≥−4is less than(Q)  99(II)Greatest term in the expression of 43√2(1+1√2)12 is(R)  120(III)If a1,a2,a3...a100 are in H.P. thenvalue of ∑99i=1aiai+1a1a100 is -(S)  100(IV)If 8 points out of 11 are in same straightline then number of triangles formed is less then(T)  125 Which of the following is only INCORRECT combination? (I)→(P)(Q)(II)→(P)(III)→(Q)(IV)→(P)(R)(T)

Solution

## The correct option is A (I)→(P)(Q)(I)→x+y+z=1x+4=x1≥0y+4=y1≥0z+4=z1≥0⎫⎪⎬⎪⎭x1+y1+z1=13 So number of solutions =13+3−1C3−1=15C2=105 (II)(1+1√2)12→|Tr+1|≥|Tr|12Cr(1√2)r≥ 12Cr−1(1√2)r−112−r+1r≥√2r≤13√2+1r≤5.385 So r=5 T6 is numerically greatest term. T6=12C5(1√2)5∴43√2×12C5×(1√2)5=132 (III) 1a1,1a2,1a3,...1a100 in A.P. 1a2−1a1=1a3−1a2=...=1a100−1a99a1−a2a1a2=d1a1a100∑99i=1aiai+1=1a1a100(a1a2+a2a3+...+a99a100)=1a1a100(a1−a2d+a2−a3d+...+a99−a100d)=a1−a100a1a100×d Now 1a100=1a1+99d⇒d=a1−a100a1a100×199⇒a1−a100a1a100×d=99 (IV) Number of triangles =11C3−8C3=109

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