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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 same straightline then number of triangles formed is less then(T)  125

Which of the following is only INCORRECT combination? 
  1. (I)(P)(Q)
  2. (II)(P)
  3. (III)(Q)
  4. (IV)(P)(R)(T)


Solution

The correct option is A (I)(P)(Q)
(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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