model 6 operations of consecutive numbers (odd, even, square, etc.) Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Sum of all multiples of 3 upto 50

= 3 + 6 + ..... + 48

= 3 (1 + 2 + 3 + .... + 16)

= 3×${16(16+1)}/2$= 3 × ${272}/2$ = 408

[Since 1+2+3+…+n = ${n(n+1)}/2$]

Answer: (a)

$(x + 2)^2 – x^2 = 84$

or $x^2 + 4x + 4 – x^2$ = 84

4x = 84 – 4 = 80

x = $80/4$ = 20

x + 2 = 20 + 2 = 22

∴ The required sum = 20 + 22 = 42

Answer: (a)

1 + 2 + 3 + ... + n = ${n(n+1)}/2$

1 + 2 + 3 + .. + 25 = ${25(25 + 1)}/2$ = 25 × 13

Hence, 13 is a factor of required sum.

Answer: (c)

Let the three consecutive natural numbers be x, x + 1 and x + 2.

According to question,

$x^2 + (x + 1)^2 + (x + 2)^2$ = 2030

or $x^2 + x^2 + 2x + 1 + x^2 + 4x + 4$ = 2030

or $3x^2 + 6x + 5 = 2030 $

or $3x^2 + 6x – 2025$ = 0

or $x^2 + 2x – 675$ = 0

or $x^2 + 27x – 25x – 675$ = 0

$x (x + 27) – 25 (x + 27)$ = 0

or $(x – 25) (x + 27)$ = 0

$x = 25 and – 27$

∴ Required number = $x$ + 1 = 25 + 1 = 26

Answer: (c)

$10^2 + 11^2 + 12^2$ = 100 + 121 + 144 = 365

Required sum =10 + 11 + 12 = 33

Answer: (d)

Series of all natural numbers from 75 to 97 is in A.P. whose first term,

a = 75, last term, l = 97

If number of terms be n, then $a_n$ = a + (n–1)d

97 = 75 + (n – 1)

n = 97 – 74 = 23

$S_n = n/2(a+l)$

$S_23 = 23/2(75+97)$

= $23/2$× 172=1978