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

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)

Prime numbers upto 17⇒ 2, 3, 5, 7, 11, 13, 17

Required sum = 2 + 3 + 5 + 7 + 11 + 13 + 17 = 58

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: (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

Answer: (a)

Series of first 20 odd natural numbers is an arithmetic progression with 1 as the first term and the common difference 2.

Sum of n terms in arithmetic progression is given by.

$S_n =1/2n[2a + (n – 1)d]$

Where a : First term; d : common difference

$S_{20}= 1/2×20[(2×1) + (20 -1) × 2]$

= 10 [2 + 38]=10 × 40 = 400

Note :Sum of first n consecutive odd numbers = $n^2$

Answer: (c)

22 + 24 + 26 + ... + 50

= 2 (11 + 12 + 13 + .... + 25)

= 2 [(1 + 2 + 3 + ... + 25) – (1 + 2 + 3 ... + 10)]

= 2$({25×26}/2 - {10×11}/2)$

= 2 (325 – 55) = 2 × 270 = 540

Method 2 :

Sum of first n even numbers = n (n + 1)

Required sum = Sum of 25 even numbers from 1 to 50 – sum of 10 even numbers from 1 to 20

= 25×26 – 10 × 11 = 650 – 110 = 540