MATHEMATICAL LOGIC: NET JRF PART - A : Questions and Solutions : NET DEC-2018

Saturday, August 1, 2020

NET JRF PART - A : Questions and Solutions : NET DEC-2018

NET JRF PART - A : Questions and Solutions:
NET DEC-2018
1.) What is the value of 12 - 22 + 32 - 42 + 52 - ....... + 172 - 182 + 192
(a) -5 (b) 12 (c) 95 (d) 190
Ans. : (d) 190 .
Solution: 
S = 12 - 22 + 32 - 42 + 52 - 62 + ....... + 172 - 182 + 192 .
S = ( 12 - 22 ) + ( 32 - 42 ) + ( 52 - 62 ) + ....... + ( 172 - 182 ) + 192 .
S = ( 1 + 2 ) ( 1 - 2 ) + ( 3 + 4 ) ( 3 - 4 ) + ( 5 + 6 ) ( 5 - 6 ) + ( 7 + 8 ) ( 7 - 8 ) + ( 9 + 10 ) ( 9 - 10 ) + ( 11 + 12 ) ( 11 - 12 ) + ( 13 + 14 ) ( 13 - 14 ) + ( 15 + 16 ) ( 15 - 16 ) + ( 17 + 18 ) ( 17 - 18 ) + 192 .
S = ( 3 ) ( -1 ) + ( 7 ) ( -1 ) + ( 11 ) ( -1 ) + ( 15 ) ( -1 ) + ( 19 ) ( -1 ) + ( 23 ) ( -1 ) + ( 27 ) ( -1 ) + ( 31 ) ( -1 ) + ( 35 ) ( -1 ) + 192 .
S = ( -1 ) [ 3 + 7 + 11 + 15 + 19 + 23 + 27 + 31 + 35 ] + 192 .
S = ( -1 ) [ 171 ] + 192 .
S = -171 + 19 × 19 .
S = -171 + 361 = 190 .
S = 190 .



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