In the previous section we saw the method of square completion. We also saw a solved example. In this section we will see a few more solved examples.
Solved example 29.6
9 added to the two consecutive multiples of 6 gives 729. What are the numbers?
Solution:
1. Let x be the first of the two consecutive multiples of 6. Then the second will be (x+6)
2. Product of the two numbers is x(x+6)
3. 9 added to the product is [1+x(x+6)]
4. So we can write: [9+x(x+6)] = 729 ⇒ [9+x2+6x] = 729 ⇒ x2+6x = 720
5. The coefficient of x is 6.
• Half of the coefficient is 6⁄2 = 3
• Square of this is 32 = 9
6. Add this square to both sides. We get:
x2+6x+9 = 720+9 ⇒ x2+6x+1 = 729
7. But x2+6x+9 is (x+3)2. So we can write:
8. (x+3)2 = 729 ⇒ (x+3) = √729 = 27 ⇒ x = 24
• So the first multiple of 6 is 24, and the next multiple is 30
Check: 9 + (24 × 30) = 9 + 720 = 729
Solved example 29.7
16 added to the sum of the first few terms of the arithmetic sequence 9, 11, 13, . . . gave 256. How many terms were added?
Solution:
1. Let first n terms be added.
We have: Sum of first n terms of an arithmetic sequence = n⁄2[2a+(n-1)d]. See details here.
2. In this problem, a = 9 and d = 2. So we can write:
Sum of first n terms = n⁄2[2×9+(n-1)×2] = n⁄2[18+2n-2] = n⁄2[16+2n]
3. When 16 is added to this sum, we get 256. So we can write:
n⁄2[16+2n] + 16 = 256 ⇒ n⁄2[16+2n] = 256-16 = 240
⇒ n[16+2n] = 480 ⇒ 16n+2n2 = 480 ⇒ 2n2 + 16n = 480
4. The coefficient of the second degree term must be 1. So we divide both sides by 2. We get:
n2 + 8n = 240
5. The coefficient of n is 8.
• Half of the coefficient is 8⁄2 = 4
• Square of this is 42 = 16
6. Add this square to both sides. We get:
n2+8x+16 = 240+16 ⇒ n2+8n+16 = 256
7. But n2+8n+16 is (n+4)2. So we can write:
8. (n+4)2 = 256 ⇒ (n+4) = √256 = 16 ⇒ n = 12
• So the number of terms added is 12
Check: Sum of first 12 terms = 12⁄2[2×9+(12-1)×2] = 6×[18+11×2] = 6×[18+22] = 6×[40] = 240
• 16 added to 240 is 256
Solved example 29.8
How many terms of the arithmetic sequence 5, 7, 9, . . . must be added to get 140?
Solution:
1. Let the number of terms to be added be n.
We have: Sum of first n terms of an arithmetic sequence = n⁄2[2a+(n-1)d]. See details here.
2. In this problem, a = 5 and d = 2. So we can write:
Sum of first n terms = n⁄2[2×5+(n-1)×2] = n⁄2[10+2n-2] = n⁄2[8+2n]
3. Given that sum = 140. So we can write:
n⁄2[8+2n] = 140
⇒ n[8+2n] = 280 ⇒ 8n+2n2 = 280 ⇒ 2n2 + 8n = 280
4. The coefficient of the second degree term must be 1. So we divide both sides by 2. We get:
n2 + 4n = 140
5. The coefficient of n is 4.
• Half of the coefficient is 4⁄2 = 2
• Square of this is 22 = 4
6. Add this square to both sides. We get:
n2+4n+4 = 140+4 ⇒ n2+4n+4 = 144
7. But n2+4n+4 is (n+2)2. So we can write:
8. (n+2)2 = 144 ⇒ (n+2) = √144 = 12 ⇒ n = 10
• So the number of terms to be added is 10
Check: Sum of first 10 terms = 10⁄2[2×5+(10-1)×2] = 5×[10+9×2] = 5×[10+18] = 5×[28] = 140
Solved example 29.9
A mathematician travelled 300 kms to attend a conference. During his talk he said: “Had my average speed been increased by 10 km per hour, I could have reached here 1 hour earlier”. What was the average speed?
Solution:
1. Let the average speed be 'x' km/h
2. Distance travelled = 300 km.
3. So time taken 't' = Distance⁄speed = 300⁄x hours
4. This time would be reduced by 1 h. So the new time = (300⁄x-1) hours
5. For that, the average speed must increase by 10 km/h. So the new speed = (x+10)
6. The distance remains the same. We have: Distance = time × speed = (300⁄x-1) × (x+10)
⇒ Distance = 300 = (300⁄x-1) × (x+10) ⇒ 300 = [(300-x)⁄x]×(x+10)
⇒ 300x = (300-x)(x+10) ⇒ 300x = 300x - x2 +3000 -10x
⇒ 0 = -x2+3000 -10x ⇒ x2+10x = 3000
7. The coefficient of x is 10.
• Half of the coefficient is 10⁄2 = 5
• Square of this is 52 = 25
8. Add this square to both sides. We get:
x2+10x+25 = 3000+25 ⇒ x2+10x+25 = 3025
9. But x2+10x+25 is (x+5)2. So we can write:
10. (x+5)2 = 3025 ⇒ (x+5) = √3025 = 55 ⇒ x = 50
• So the average speed is 50 km/h
Check: Put x = 50 in (6). We get: 502 + (10 × 50) = 2500 + 500 = 3000
Solved example 29.6
9 added to the two consecutive multiples of 6 gives 729. What are the numbers?
Solution:
1. Let x be the first of the two consecutive multiples of 6. Then the second will be (x+6)
2. Product of the two numbers is x(x+6)
3. 9 added to the product is [1+x(x+6)]
4. So we can write: [9+x(x+6)] = 729 ⇒ [9+x2+6x] = 729 ⇒ x2+6x = 720
5. The coefficient of x is 6.
• Half of the coefficient is 6⁄2 = 3
• Square of this is 32 = 9
6. Add this square to both sides. We get:
x2+6x+9 = 720+9 ⇒ x2+6x+1 = 729
7. But x2+6x+9 is (x+3)2. So we can write:
8. (x+3)2 = 729 ⇒ (x+3) = √729 = 27 ⇒ x = 24
• So the first multiple of 6 is 24, and the next multiple is 30
Check: 9 + (24 × 30) = 9 + 720 = 729
Solved example 29.7
16 added to the sum of the first few terms of the arithmetic sequence 9, 11, 13, . . . gave 256. How many terms were added?
Solution:
1. Let first n terms be added.
We have: Sum of first n terms of an arithmetic sequence = n⁄2[2a+(n-1)d]. See details here.
2. In this problem, a = 9 and d = 2. So we can write:
Sum of first n terms = n⁄2[2×9+(n-1)×2] = n⁄2[18+2n-2] = n⁄2[16+2n]
3. When 16 is added to this sum, we get 256. So we can write:
n⁄2[16+2n] + 16 = 256 ⇒ n⁄2[16+2n] = 256-16 = 240
⇒ n[16+2n] = 480 ⇒ 16n+2n2 = 480 ⇒ 2n2 + 16n = 480
4. The coefficient of the second degree term must be 1. So we divide both sides by 2. We get:
n2 + 8n = 240
5. The coefficient of n is 8.
• Half of the coefficient is 8⁄2 = 4
• Square of this is 42 = 16
6. Add this square to both sides. We get:
n2+8x+16 = 240+16 ⇒ n2+8n+16 = 256
7. But n2+8n+16 is (n+4)2. So we can write:
8. (n+4)2 = 256 ⇒ (n+4) = √256 = 16 ⇒ n = 12
• So the number of terms added is 12
Check: Sum of first 12 terms = 12⁄2[2×9+(12-1)×2] = 6×[18+11×2] = 6×[18+22] = 6×[40] = 240
• 16 added to 240 is 256
Solved example 29.8
How many terms of the arithmetic sequence 5, 7, 9, . . . must be added to get 140?
Solution:
1. Let the number of terms to be added be n.
We have: Sum of first n terms of an arithmetic sequence = n⁄2[2a+(n-1)d]. See details here.
2. In this problem, a = 5 and d = 2. So we can write:
Sum of first n terms = n⁄2[2×5+(n-1)×2] = n⁄2[10+2n-2] = n⁄2[8+2n]
3. Given that sum = 140. So we can write:
n⁄2[8+2n] = 140
⇒ n[8+2n] = 280 ⇒ 8n+2n2 = 280 ⇒ 2n2 + 8n = 280
4. The coefficient of the second degree term must be 1. So we divide both sides by 2. We get:
n2 + 4n = 140
5. The coefficient of n is 4.
• Half of the coefficient is 4⁄2 = 2
• Square of this is 22 = 4
6. Add this square to both sides. We get:
n2+4n+4 = 140+4 ⇒ n2+4n+4 = 144
7. But n2+4n+4 is (n+2)2. So we can write:
8. (n+2)2 = 144 ⇒ (n+2) = √144 = 12 ⇒ n = 10
• So the number of terms to be added is 10
Check: Sum of first 10 terms = 10⁄2[2×5+(10-1)×2] = 5×[10+9×2] = 5×[10+18] = 5×[28] = 140
Solved example 29.9
A mathematician travelled 300 kms to attend a conference. During his talk he said: “Had my average speed been increased by 10 km per hour, I could have reached here 1 hour earlier”. What was the average speed?
Solution:
1. Let the average speed be 'x' km/h
2. Distance travelled = 300 km.
3. So time taken 't' = Distance⁄speed = 300⁄x hours
4. This time would be reduced by 1 h. So the new time = (300⁄x-1) hours
5. For that, the average speed must increase by 10 km/h. So the new speed = (x+10)
6. The distance remains the same. We have: Distance = time × speed = (300⁄x-1) × (x+10)
⇒ Distance = 300 = (300⁄x-1) × (x+10) ⇒ 300 = [(300-x)⁄x]×(x+10)
⇒ 300x = (300-x)(x+10) ⇒ 300x = 300x - x2 +3000 -10x
⇒ 0 = -x2+3000 -10x ⇒ x2+10x = 3000
7. The coefficient of x is 10.
• Half of the coefficient is 10⁄2 = 5
• Square of this is 52 = 25
8. Add this square to both sides. We get:
x2+10x+25 = 3000+25 ⇒ x2+10x+25 = 3025
9. But x2+10x+25 is (x+5)2. So we can write:
10. (x+5)2 = 3025 ⇒ (x+5) = √3025 = 55 ⇒ x = 50
• So the average speed is 50 km/h
Check: Put x = 50 in (6). We get: 502 + (10 × 50) = 2500 + 500 = 3000
In the next section, we will see another type of square completion.
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