A sequence is obtained by deleting all perfect squares from the set of

1.	A sequence is obtained by deleting all perfect squares from the set of all natural numbers starting from 1, then
Answer» Correct option is (C) The reminder is 1 when 931^th term is divided by 2 Given sequence is 2,3,5,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26...... 2003^rd term = ? 1892^th term = 1937 1980^th term = 2026 2003^rd term = 2026 + (2003 - 1980) = 2026 + 23 = 2049 870^th term = 901 930^th term = 962 \(\therefore\) 931^st term = 963 (a, b) \(\frac{T_{2003}}{2048}=\frac{2049}{2048}\) the reminder left 1 option (a), and (b) are false (c, d) \(\frac{T_{931}}2=481+\frac12\) Thus, reminder left 1 option (c) is correct

A sequence is obtained by deleting all perfect squares from the set of all natural numbers starting from 1, then

Answer»

Correct option is (C) The reminder is 1 when 931^th term is divided by 2

Given sequence is

2,3,5,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26......

2003^rd term = ?

1892^th term = 1937

1980^th term = 2026

2003^rd term = 2026 + (2003 - 1980)

= 2026 + 23 = 2049

870^th term = 901

930^th term = 962

\(\therefore\) 931^st term = 963

(a, b) \(\frac{T_{2003}}{2048}=\frac{2049}{2048}\) the reminder left 1

option (a), and (b) are false

(c, d) \(\frac{T_{931}}2=481+\frac12\)

Thus, reminder left 1

option (c) is correct

Discussion

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