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Match the following ( array l Column - I & Column - II array l (A) If n =210, then n ^2 is divisible by the greatest prime number which is greater than array & (p) 16 array l (B) Between 4 and 2916 is inserted odd number (2 n+1) G.M's, then the ( n +1) th G.M is divisible by the greatest odd integer which is less than array & (q) 10 array l (C) In a certain progression four consecutive terms are 40, 30, 24, 20, then

Options

  1. A( A ) ( r ) ;( B ) ( p ) ;( C ) ( q ) ;( D ) ( s )
  2. B( A ) ( p,q,r,s,t ) ;( B ) ( r,s ) ;( C ) ( p,q,t ) ;( D ) ( r,s )
  3. C( A ) ( q ) ;( B ) ( r ) ;( C ) ( s ) ;( D ) ( p )
  4. D( A ) ( s ) ;( B ) ( q ) ;( C ) ( p ) ;( D ) ( r )

Correct answer

B. ( A ) ( p,q,r,s,t ) ;( B ) ( r,s ) ;( C ) ( p,q,t ) ;( D ) ( r,s )

Step-by-step solution

(A) aligned & n=210 & n ( n +1)=420 & n =20 & n ^2=(10)(7)(41) aligned (B) 4, G₁, G₂, G_ n+1 , G_ 3 n , G_ 3-1 , 2916 G _ n ,-1 will be the middle mean of (2 n +1) odd means aligned & G_ n+1 ^2=4 2916=4 9 324=4 9 4 81 & G_ n+1 =2 3 2 9=108 aligned Greatest odd number by which G _ a +1 is divisible is 27 (C) Terms are 40, 30, 24, 20 aligned & 1 30 - 1 40 = 1 120 & 1 24 - 1 30 = 1 24 30 = 1 120 & 1 20 - 1 24 = 4 20 24 = 1 120 aligned 1 30 1 24 1 20 are in AB. witi common difference d = 1 120 Next term is 1 20 + 1 120

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