In the previous section we learned how to write an AP when a and d are given. We must be able to work in the reverse order also. That is., if we are given a series, we must be able to do 3 things:
• Determine whether the series is an AP
• Write the common difference d of the AP
• Write the first term a of the AP
Let us see how the three tasks can be accomplished:
■ Task 1: To determine whether a given series is an AP
1. From the given series, take out the last two consecutive terms. This is our 'first set'
2. Out of the two, let the term on the left be the kth term ak. Then the term on the right will be the (k+1)th term ak+1
3. Find the difference: ak+1 - ak
4. Take a 'second set' of two consecutive terms just preceding the 'first set'. The second set should have one term in common with the first set
5. Perform the same operation. That is., find: ak+1 - ak
6. If the results in (3) and (5) are the same, the given series may be an AP
7. But to confirm all terms should be checked. So take a 'third set' of two consecutive terms just preceding the second set. This third set should have one term in common with the second set.
8. Perform the same operation. That is., find: ak+1 - ak
9. Repeat the steps until the first term is reached. If the results are the same in all the trials, the given series is an AP
[Note: We must always subtract the term on the left (ak) from the term on the right (ak+1) regardless of whether ak is smaller or larger than ak+1]
■ Task 2: To write the common difference d
If we confirm that it is an AP, we will have already completed this task 2. Because the differences that we find in each trial in task 1 above will all be the same, and it is the required d
■ Task 3: To write the first term a of the AP
This is an easy task. All we need to do is this: Write down the first term of the given series. It is the 'a'
Let us see an example:
Solved example 26.1
Which of the following list of numbers does form an AP? If they form an AP, write the next two terms :
(i) 4, 10, 16, 22, . . .
(ii) 1, -1, -3, -5, . . .
(iii) -2, 2, -2, 2, -2 , . . .
(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .
Solution:
(i) 1. Take the last two terms: ak+1 - ak = 22 - 16 = 6
2. Take the preceding two terms with one term common: ak+1 - ak = 16 - 10 = 6
3. Take the preceding two terms with one term common: ak+1 - ak = 10 - 4 = 6
[In this step we have reached the first term]
4. ak+1 - ak is same in the three tests. So it is an AP
5. The common difference d = 6
6. The term after 22 is: 22 + 6 = 28
7. The term after 28 is: 28 + 6 = 34
(ii) 1. Take the last two terms: ak+1 - ak = -5 - (-3) = -5 + 3 = -2
2. Take the preceding two terms with one term common: ak+1 - ak = -3 - (-1) = -3 + 1 = -2
3. Take the preceding two terms with one term common: ak+1 - ak = -1 - 1 = -2
[In this step we have reached the first term]
4. ak+1 - ak is same in the three tests. So it is an AP
5. The common difference d = -2
6. The term after -5 is: -5 + -2 = -7
7. The term after -7 is: -7 + -2 = -9
(iii) 1. Take the last two terms: ak+1 - ak = -2 - 2 = -4
2. Take the preceding two terms with one term common: ak+1 - ak = 2 - (-2) = 2 + 2 = 4
3. Differences are not same. So it is not an AP. We do not need to work up to the first term.
(iv) 1. Take the last two terms: ak+1 - ak = 3 - 3 = 0
2. Take the preceding two terms with one term common: ak+1 - ak = 3 - 3 = 0
3. Take the preceding two terms with one term common: ak+1 - ak = 3 - 2 = 1
3. Differences in (2) and (3) are not same. So it is not an AP. We do not need to work up to the first term.
Now we will see some solved examples based on the discussion so far in this chapter.
(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
Solution:
1. First we will write the possible terms of the series:
• Fare at the end of the 1st km = 15
• Fare at the end of the 2nd km = 15+8 = 23 [This is same as 15 + 1 × 8 = 23]
• Fare at the end of the 3rd km = 23+8 = 31 [This is same as 15 + 2 × 8 = 31]
• Fare at the end of the 4th km = 31+8 = 39 [This is same as 15 + 3 × 8 = 39]
So on . . .
2. So the series is: 15, 23, 31, 39, . . .
We have to determine whether it is an AP or not
3. From step (1), it is clear that, each term is obtained by adding a fixed number 8 to the preceding term. So it is an AP
(ii) The amount of air present in a cylinder when a vacuum pump removes 1⁄4 of the air remaining in the cylinder at a time.
Solution:
1. First we will write the possible terms of the series:
• Let the volume air present before operating the pump be V m3.
• Air remaining when one operation of the pump is completed = V - V × 1⁄4 = 3V⁄4.
• Air remaining when two operations of the pump is completed = 3V⁄4 - (3V⁄4 × 1⁄4)
[∵ 1⁄4 of the remaining 3V⁄4 is removed in the second operation]
= 3V⁄4 - 3V⁄16 = 12V⁄16 - 3V⁄16 = 9V⁄16.
• Air remaining when three operations of the pump is completed = 9V⁄16 - (9V⁄16 × 1⁄4)
[∵ 1⁄4 of the remaining 9V⁄16 is removed in the second operation]
= 9V⁄16 - 9V⁄64 = 36V⁄64 - 9V⁄64 = 27V⁄64.
so on . . .
2. So the series is: V, 3V⁄4, 9V⁄16, 27V⁄64, . . .
3. Now we will check whether it is an AP or not:
I Take the last two terms: ak+1 - ak
= 27V⁄64 - 9V⁄16
= 27V⁄64 - 36V⁄64 = - 9V⁄64
II. Take the preceding two terms with one term common: ak+1 - ak
= 9V⁄16 - 3V⁄4
= 9V⁄16 - 12V⁄16 = - 3V⁄16
III. Differences in I and II are not same. So it is not an AP
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre
Solution:
1. First we will write the possible terms of the series:
• Cost when bottom of the 1st metre is reached = 150
• Cost when bottom of the 2nd metre is reached = 150+50 = 200 [This is same as 150 + 1 × 50 = 200]
• Cost when bottom of the 3rd metre is reached = 200+50 = 250 [This is same as 150 + 2 × 50 = 250]
• Cost when bottom of the 4th metre is reached = 250+50 = 300 [This is same as 150 + 3 × 50 = 300]
so on . . .
2. So the series is: 150, 200, 250, 300, . . .
3. From step (1), it is clear that, each term is obtained by adding a fixed number 50 to the preceding term. So it is an AP
Solved example 26.3
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10
Solution:
• 1st term = a = 10
• 2nd term = a + d = 10 + 10 = 20
• 3rd term = a + 2d = 10 + 2 × 10 = 10 + 20 = 30
• 4th term = a + 3d = 10 + 3 × 10 = 10 + 30 = 40
So the series is: 10, 20, 30, 40, . . .
(ii) a = -2, d = 0
Solution:
• 1st term = a = -2
• 2nd term = a + d = -2 + 0 = -2
• 3rd term = a + 2d = -2 + 2 × 0 = -2 + 0 = -2
• 4th term = a + 3d = -2 + 3 × 0 = -2 + 0 = -2
So the series is: -2, -2, -2, -2, . . .
(iii) a = 4, d = -3
Solution:
• 1st term = a = 4
• 2nd term = a + d = 4 + (-3) = 4-3 = 1
• 3rd term = a + 2d = 4 + 2 × (-3) = 4 -6 = -2
• 4th term = a + 3d = 4 + 3 × (-3) = 4 -9 = -5
So the series is: 4, 1, -2, -5, . . .
(iv) a = -1, d = 1⁄2
Solution:
• 1st term = a = -1
• 2nd term = a + d = -1 + 1⁄2 = -1⁄2
• 3rd term = a + 2d = -1 + 2 × 1⁄2 = -1 + 1 = 0
• 4th term = a + 3d = -1 + 3 × 1⁄2 = -1 + 3⁄2 = 1⁄2
So the series is: -1, -1⁄2, 0, 1⁄2, . . .
Solved example 26.4
For the following APs, write the first term and the common difference:
(i) 3, 1, -1, -3, . . .
Solution:
• It is already given that, the series is an AP. So we do not need to check whether it is an AP or not. So we can take any two consecutive terms. Let us take the last two.
• We have: d = ak+1 - ak = -3 -(-1) = -3 + 1 = -2
• The first term a = 3
(ii) -5, -1, 3, 7, . . .
Solution:
• We have: d = ak+1 - ak = 7 -3 = 4
• The first term a = -5
(iii) 1⁄3, 5⁄3, 9⁄3, 13⁄3, . . .
Solution:
• We have: d = ak+1 - ak = 13⁄3 -9⁄3 = 4⁄3
• The first term a = 1⁄3
Solved example 26.5
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, . . .
Solution:
1. Take the last two terms: ak+1 - ak = 16 - 8 = 8
2. Take the preceding two terms with one term common: ak+1 - ak = 8 - 4 = 4
3. Differences are not same. So it is not an AP. We do not need to work up to the first term.
(ii) 2, 5⁄2, 3, 7⁄2, . . .
Solution:
1. Take the last two terms: ak+1 - ak = 7⁄2 - 3 = 3.5 - 3 = 0.5
2. Take the preceding two terms with one term common: ak+1 - ak = 3 - 5⁄2 = 3 -2.5 = 0.5
3. Take the preceding two terms with one term common: ak+1 - ak = 5⁄2 - 2 = 2.5 - 2 = 0.5
[In this step we have reached the first term]
4. ak+1 - ak is same in the three tests. So it is an AP
5. The common difference d = 0.5
6. The term after 7⁄2 is: 7⁄2 + 0.5 = 3.5 + 0.5 = 4
7. The term after 4 is: 4 + 0.5 = 4.5 = 9⁄2
8. The term after 9⁄2 is: 9⁄2 + 0.5 = 4.5 + 0.5 = 5
(iii) -1.2, -3.2, -5.2, -7.2, . . .
Solution:
1. Take the last two terms: ak+1 - ak = -7.2 - (-5.2) = -7.2 + 5.2 = -2
2. Take the preceding two terms with one term common: ak+1 - ak = -5.2 - (-3.2) = -5.2 + 3.2 = -2
3. Take the preceding two terms with one term common: ak+1 - ak = -3.2 - (-1.2) = -3.2 + 1.2 = -2
[In this step we have reached the first term]
4. ak+1 - ak is same in the three tests. So it is an AP
5. The common difference d = -2
6. The term after -7.2 is: -7.2 + (-2) = -7.2 - 2 = -9.2
7. The term after -9.2 is: -9.2 + (-2) = -9.2 - 2 = -11.2
8. The term after -11.2 is: -11.2 + (-2) = -11.2 - 2 = -13.2
(iv) -10, -6, -2, 2, . . .
Solution:
1. Take the last two terms: ak+1 - ak = 2 - (-2) = 2 + 2 = 4
2. Take the preceding two terms with one term common: ak+1 - ak = -2 - (-6) = -2 + 6 = 4
3. Take the preceding two terms with one term common: ak+1 - ak = -6 - (-10) = -6 + 10 = 4
[In this step we have reached the first term]
4. ak+1 - ak is same in the three tests. So it is an AP
5. The common difference d = 4
6. The term after 2 is: 2 + 4 = 6
7. The term after 6 is: 6 + 4 = 10
8. The term after 10 is: 10 + 4 = 14
(v) 3, 3+√2, 3+2√2, 3+3√2, . . .
Solution:
1. Take the last two terms: ak+1 - ak
= 3+3√2 - (3+2√2) = 3+3√2 - 3-2√2 = 3√2 - 2√2 = √2
2. Take the preceding two terms with one term common: ak+1 - ak
= 3+2√2 - (3+√2) = 3+2√2 - 3-√2 = 2√2 - √2 = √2
3. Take the preceding two terms with one term common: ak+1 - ak
= 3+√2 - (3) = 3+√2 - 3 = √2
[In this step we have reached the first term]
4. ak+1 - ak is same in the three tests. So it is an AP
5. The common difference d = √2
6. The term after 3+3√2 is: 3+3√2 + √2 = 3+4√2
7. The term after 3+4√2 is: 3+4√2 + √2 = 3+5√2
8. The term after 3+5√2 is: 3+5√2 + √2 = 3+6
So now we can determine whether a given series is an AP or not. We can also write it's first term and common difference. In the next section we will see how to find the nth term of an AP.
• Determine whether the series is an AP
• Write the common difference d of the AP
• Write the first term a of the AP
Let us see how the three tasks can be accomplished:
■ Task 1: To determine whether a given series is an AP
1. From the given series, take out the last two consecutive terms. This is our 'first set'
2. Out of the two, let the term on the left be the kth term ak. Then the term on the right will be the (k+1)th term ak+1
3. Find the difference: ak+1 - ak
4. Take a 'second set' of two consecutive terms just preceding the 'first set'. The second set should have one term in common with the first set
5. Perform the same operation. That is., find: ak+1 - ak
6. If the results in (3) and (5) are the same, the given series may be an AP
7. But to confirm all terms should be checked. So take a 'third set' of two consecutive terms just preceding the second set. This third set should have one term in common with the second set.
8. Perform the same operation. That is., find: ak+1 - ak
9. Repeat the steps until the first term is reached. If the results are the same in all the trials, the given series is an AP
[Note: We must always subtract the term on the left (ak) from the term on the right (ak+1) regardless of whether ak is smaller or larger than ak+1]
■ Task 2: To write the common difference d
If we confirm that it is an AP, we will have already completed this task 2. Because the differences that we find in each trial in task 1 above will all be the same, and it is the required d
■ Task 3: To write the first term a of the AP
This is an easy task. All we need to do is this: Write down the first term of the given series. It is the 'a'
Let us see an example:
Solved example 26.1
Which of the following list of numbers does form an AP? If they form an AP, write the next two terms :
(i) 4, 10, 16, 22, . . .
(ii) 1, -1, -3, -5, . . .
(iii) -2, 2, -2, 2, -2 , . . .
(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .
Solution:
(i) 1. Take the last two terms: ak+1 - ak = 22 - 16 = 6
2. Take the preceding two terms with one term common: ak+1 - ak = 16 - 10 = 6
3. Take the preceding two terms with one term common: ak+1 - ak = 10 - 4 = 6
[In this step we have reached the first term]
4. ak+1 - ak is same in the three tests. So it is an AP
5. The common difference d = 6
6. The term after 22 is: 22 + 6 = 28
7. The term after 28 is: 28 + 6 = 34
(ii) 1. Take the last two terms: ak+1 - ak = -5 - (-3) = -5 + 3 = -2
2. Take the preceding two terms with one term common: ak+1 - ak = -3 - (-1) = -3 + 1 = -2
3. Take the preceding two terms with one term common: ak+1 - ak = -1 - 1 = -2
[In this step we have reached the first term]
4. ak+1 - ak is same in the three tests. So it is an AP
5. The common difference d = -2
6. The term after -5 is: -5 + -2 = -7
7. The term after -7 is: -7 + -2 = -9
(iii) 1. Take the last two terms: ak+1 - ak = -2 - 2 = -4
2. Take the preceding two terms with one term common: ak+1 - ak = 2 - (-2) = 2 + 2 = 4
3. Differences are not same. So it is not an AP. We do not need to work up to the first term.
(iv) 1. Take the last two terms: ak+1 - ak = 3 - 3 = 0
2. Take the preceding two terms with one term common: ak+1 - ak = 3 - 3 = 0
3. Take the preceding two terms with one term common: ak+1 - ak = 3 - 2 = 1
3. Differences in (2) and (3) are not same. So it is not an AP. We do not need to work up to the first term.
Solved example 26.2
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
Solution:
1. First we will write the possible terms of the series:
• Fare at the end of the 1st km = 15
• Fare at the end of the 2nd km = 15+8 = 23 [This is same as 15 + 1 × 8 = 23]
• Fare at the end of the 3rd km = 23+8 = 31 [This is same as 15 + 2 × 8 = 31]
• Fare at the end of the 4th km = 31+8 = 39 [This is same as 15 + 3 × 8 = 39]
So on . . .
2. So the series is: 15, 23, 31, 39, . . .
We have to determine whether it is an AP or not
3. From step (1), it is clear that, each term is obtained by adding a fixed number 8 to the preceding term. So it is an AP
(ii) The amount of air present in a cylinder when a vacuum pump removes 1⁄4 of the air remaining in the cylinder at a time.
Solution:
1. First we will write the possible terms of the series:
• Let the volume air present before operating the pump be V m3.
• Air remaining when one operation of the pump is completed = V - V × 1⁄4 = 3V⁄4.
• Air remaining when two operations of the pump is completed = 3V⁄4 - (3V⁄4 × 1⁄4)
[∵ 1⁄4 of the remaining 3V⁄4 is removed in the second operation]
= 3V⁄4 - 3V⁄16 = 12V⁄16 - 3V⁄16 = 9V⁄16.
• Air remaining when three operations of the pump is completed = 9V⁄16 - (9V⁄16 × 1⁄4)
[∵ 1⁄4 of the remaining 9V⁄16 is removed in the second operation]
= 9V⁄16 - 9V⁄64 = 36V⁄64 - 9V⁄64 = 27V⁄64.
so on . . .
2. So the series is: V, 3V⁄4, 9V⁄16, 27V⁄64, . . .
3. Now we will check whether it is an AP or not:
I Take the last two terms: ak+1 - ak
= 27V⁄64 - 9V⁄16
= 27V⁄64 - 36V⁄64 = - 9V⁄64
II. Take the preceding two terms with one term common: ak+1 - ak
= 9V⁄16 - 3V⁄4
= 9V⁄16 - 12V⁄16 = - 3V⁄16
III. Differences in I and II are not same. So it is not an AP
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre
Solution:
1. First we will write the possible terms of the series:
• Cost when bottom of the 1st metre is reached = 150
• Cost when bottom of the 2nd metre is reached = 150+50 = 200 [This is same as 150 + 1 × 50 = 200]
• Cost when bottom of the 3rd metre is reached = 200+50 = 250 [This is same as 150 + 2 × 50 = 250]
• Cost when bottom of the 4th metre is reached = 250+50 = 300 [This is same as 150 + 3 × 50 = 300]
so on . . .
2. So the series is: 150, 200, 250, 300, . . .
3. From step (1), it is clear that, each term is obtained by adding a fixed number 50 to the preceding term. So it is an AP
Solved example 26.3
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10
Solution:
• 1st term = a = 10
• 2nd term = a + d = 10 + 10 = 20
• 3rd term = a + 2d = 10 + 2 × 10 = 10 + 20 = 30
• 4th term = a + 3d = 10 + 3 × 10 = 10 + 30 = 40
So the series is: 10, 20, 30, 40, . . .
(ii) a = -2, d = 0
Solution:
• 1st term = a = -2
• 2nd term = a + d = -2 + 0 = -2
• 3rd term = a + 2d = -2 + 2 × 0 = -2 + 0 = -2
• 4th term = a + 3d = -2 + 3 × 0 = -2 + 0 = -2
So the series is: -2, -2, -2, -2, . . .
(iii) a = 4, d = -3
Solution:
• 1st term = a = 4
• 2nd term = a + d = 4 + (-3) = 4-3 = 1
• 3rd term = a + 2d = 4 + 2 × (-3) = 4 -6 = -2
• 4th term = a + 3d = 4 + 3 × (-3) = 4 -9 = -5
So the series is: 4, 1, -2, -5, . . .
(iv) a = -1, d = 1⁄2
Solution:
• 1st term = a = -1
• 2nd term = a + d = -1 + 1⁄2 = -1⁄2
• 3rd term = a + 2d = -1 + 2 × 1⁄2 = -1 + 1 = 0
• 4th term = a + 3d = -1 + 3 × 1⁄2 = -1 + 3⁄2 = 1⁄2
So the series is: -1, -1⁄2, 0, 1⁄2, . . .
Solved example 26.4
For the following APs, write the first term and the common difference:
(i) 3, 1, -1, -3, . . .
Solution:
• It is already given that, the series is an AP. So we do not need to check whether it is an AP or not. So we can take any two consecutive terms. Let us take the last two.
• We have: d = ak+1 - ak = -3 -(-1) = -3 + 1 = -2
• The first term a = 3
(ii) -5, -1, 3, 7, . . .
Solution:
• We have: d = ak+1 - ak = 7 -3 = 4
• The first term a = -5
(iii) 1⁄3, 5⁄3, 9⁄3, 13⁄3, . . .
Solution:
• We have: d = ak+1 - ak = 13⁄3 -9⁄3 = 4⁄3
• The first term a = 1⁄3
Solved example 26.5
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, . . .
Solution:
1. Take the last two terms: ak+1 - ak = 16 - 8 = 8
2. Take the preceding two terms with one term common: ak+1 - ak = 8 - 4 = 4
3. Differences are not same. So it is not an AP. We do not need to work up to the first term.
(ii) 2, 5⁄2, 3, 7⁄2, . . .
Solution:
1. Take the last two terms: ak+1 - ak = 7⁄2 - 3 = 3.5 - 3 = 0.5
2. Take the preceding two terms with one term common: ak+1 - ak = 3 - 5⁄2 = 3 -2.5 = 0.5
3. Take the preceding two terms with one term common: ak+1 - ak = 5⁄2 - 2 = 2.5 - 2 = 0.5
[In this step we have reached the first term]
4. ak+1 - ak is same in the three tests. So it is an AP
5. The common difference d = 0.5
6. The term after 7⁄2 is: 7⁄2 + 0.5 = 3.5 + 0.5 = 4
7. The term after 4 is: 4 + 0.5 = 4.5 = 9⁄2
8. The term after 9⁄2 is: 9⁄2 + 0.5 = 4.5 + 0.5 = 5
(iii) -1.2, -3.2, -5.2, -7.2, . . .
Solution:
1. Take the last two terms: ak+1 - ak = -7.2 - (-5.2) = -7.2 + 5.2 = -2
2. Take the preceding two terms with one term common: ak+1 - ak = -5.2 - (-3.2) = -5.2 + 3.2 = -2
3. Take the preceding two terms with one term common: ak+1 - ak = -3.2 - (-1.2) = -3.2 + 1.2 = -2
[In this step we have reached the first term]
4. ak+1 - ak is same in the three tests. So it is an AP
5. The common difference d = -2
6. The term after -7.2 is: -7.2 + (-2) = -7.2 - 2 = -9.2
7. The term after -9.2 is: -9.2 + (-2) = -9.2 - 2 = -11.2
8. The term after -11.2 is: -11.2 + (-2) = -11.2 - 2 = -13.2
(iv) -10, -6, -2, 2, . . .
Solution:
1. Take the last two terms: ak+1 - ak = 2 - (-2) = 2 + 2 = 4
2. Take the preceding two terms with one term common: ak+1 - ak = -2 - (-6) = -2 + 6 = 4
3. Take the preceding two terms with one term common: ak+1 - ak = -6 - (-10) = -6 + 10 = 4
[In this step we have reached the first term]
4. ak+1 - ak is same in the three tests. So it is an AP
5. The common difference d = 4
6. The term after 2 is: 2 + 4 = 6
7. The term after 6 is: 6 + 4 = 10
8. The term after 10 is: 10 + 4 = 14
(v) 3, 3+√2, 3+2√2, 3+3√2, . . .
Solution:
1. Take the last two terms: ak+1 - ak
= 3+3√2 - (3+2√2) = 3+3√2 - 3-2√2 = 3√2 - 2√2 = √2
2. Take the preceding two terms with one term common: ak+1 - ak
= 3+2√2 - (3+√2) = 3+2√2 - 3-√2 = 2√2 - √2 = √2
3. Take the preceding two terms with one term common: ak+1 - ak
= 3+√2 - (3) = 3+√2 - 3 = √2
[In this step we have reached the first term]
4. ak+1 - ak is same in the three tests. So it is an AP
5. The common difference d = √2
6. The term after 3+3√2 is: 3+3√2 + √2 = 3+4√2
7. The term after 3+4√2 is: 3+4√2 + √2 = 3+5√2
8. The term after 3+5√2 is: 3+5√2 + √2 = 3+6
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