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List- IList-II(I)Number of integral solutions of(P) 132x+y+z=1, x4, y4, z4is less than(Q) 99(II)Greatest term in the expression of 432(1+12)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 smae straightline then number of triangles formed is less then(T) 125

Which of the following is only CORRECT combination?

A
(I)(P)(Q)
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B
(II)(Q)
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C
(III)(R)
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D
(IV)(P)(R)(T)
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Solution

The correct option is D (IV)(P)(R)(T)
(I)x+y+z=1x+4=x10y+4=y10z+4=z10x1+y1+z1=13
So number of solutions =13+31C31=15C2=105

(II)(1+12)12|Tr+1||Tr|12Cr(12)r 12Cr1(12)r112r+1r2r132+1r5.385
So r=5
T6 is numerically greatest term.
T6=12C5(12)5432×12C5×(12)5=132

(III) 1a1,1a2,1a3,...1a100 in A.P.
1a21a1=1a31a2=...=1a1001a99a1a2a1a2=d1a1a10099i=1aiai+1=1a1a100(a1a2+a2a3+...+a99a100)=1a1a100(a1a2d+a2a3d+...+a99a100d)=a1a100a1a100×d
Now 1a100=1a1+99dd=a1a100a1a100×199a1a100a1a100×d=99

(IV) Number of triangles =11C38C3=109

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