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

Answer: (a)

Let the numbers be 3x, 3x + 3 and 3x + 6

3x + 3x + 3 + 3x + 6 = 72

9x + 9 = 72

9x = 72 – 9 = 63

x =$63/9$ = 7

∴ Largest number = 3x + 6 = 3 × 7 + 6 = 27

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

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: (d)

S = 1 + 3 + 5 + ..... to 50 terms

Here, a = 1; d = 3 – 1 = 2; n = 50

S = $n/2[2a + (n - 1)d]$

= $50/2[2×1+ (50 - 1) ×2]$

= 25 (2 + 98) = 25 × 100 = 2500

Answer: (d)

According to the question,

First number = a = 103; Last number = l = 998

If the number of such numbers be n, then,

998 = 103 + (n – 1) × 5

(n – 1) × 5= 998 – 103 = 895

n – 1 = $895/5$ = 179

n = 180

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

=$180/2$(103 +998)

= 90 × 1101 = 99090

Answer: (b)

x + x + 1 + x + 2 = 27

3x + 3 = 27

3x = 24

x = 8

Three consecutive no's whose sum is 27 are 8, 9,10.

Hence, next 3 consecutive no's having 36 as sum are 11, 12 and 13