EXERCISE 5.1
Arithmetic Progressions • 4 Questions
Question 1
Hint available
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 ` 15 for the first km and ` 8 for each additional km. (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. (iii) The cost of digging a well after every metre of digging, when it costs ` 150 for the first metre and rises by ` 50 for each subsequent metre. (iv) The amount of money in the account every year, when ` 10000 is deposited at compound interest at 8 % per annum.
Key Idea
An arithmetic progression (AP) is a sequence of numbers in which the difference between any two successive terms is constant. If the successive terms are $a, a+d, a+2d, \dots$, the constant $d$ is called the common difference. To decide whether a given situation forms an AP, we examine whether the successive terms differ by the same amount.
Step-by-Step Solution
1. Recall the definition: A sequence $\{T_n\}$ is an AP if $T_{n+1}-T_n = d$ (constant) for all $n$.
2. Examine each situation:
- (i) Taxi fare:
*First km*: $\text{Fare}_1 = 15$ rupees.
*Each additional km*: an extra $8$ rupees is added.
Hence the fare after $n$ km is $\text{Fare}_n = 15 + (n-1)\times 8$.
The difference between successive terms is $\text{Fare}_{n+1}-\text{Fare}_n = 8$, a constant. Therefore the list of fares forms an AP with first term $a=15$ and common difference $d=8$.
- (ii) Air in a cylinder:
Let $A_n$ be the amount of air remaining after the $n^{\text{th}}$ pump. The pump removes $\frac{1}{4}$ of the air present, so $A_{n+1}=\frac{3}{4}A_n$.
The ratio $\frac{A_{n+1}}{A_n}=\frac{3}{4}$ is constant, but the *difference* $A_{n+1}-A_n$ is not constant (it changes with $A_n$). Hence the sequence is geometric, not arithmetic.
- (iii) Cost of digging a well:
Cost for the first metre = $150$ rupees.
Cost for each subsequent metre rises by $50$ rupees.
Thus the cost for the $n^{\text{th}}$ metre is $C_n = 150 + (n-1)\times 50$.
The successive difference $C_{n+1}-C_n = 50$ is constant. Hence the list of costs per metre is an AP with $a=150$, $d=50$.
- (iv) Money in a bank account with compound interest:
Initial amount $M_0 = 10000$ rupees.
Each year the amount grows by $8\%$, i.e., $M_{n+1}=1.08\,M_n$.
This gives a constant *ratio* $1.08$, not a constant *difference*. Therefore the yearly amounts form a geometric progression, not an AP.
3. Conclusion: Situations (i) and (iii) produce arithmetic progressions because the successive terms differ by a fixed amount (₹8 and ₹50 respectively). Situations (ii) and (iv) produce geometric progressions because the successive terms are obtained by multiplying by a constant factor.
4. Answer in short: (i) and (iii) are AP; (ii) and (iv) are not.
2. Examine each situation:
- (i) Taxi fare:
*First km*: $\text{Fare}_1 = 15$ rupees.
*Each additional km*: an extra $8$ rupees is added.
Hence the fare after $n$ km is $\text{Fare}_n = 15 + (n-1)\times 8$.
The difference between successive terms is $\text{Fare}_{n+1}-\text{Fare}_n = 8$, a constant. Therefore the list of fares forms an AP with first term $a=15$ and common difference $d=8$.
- (ii) Air in a cylinder:
Let $A_n$ be the amount of air remaining after the $n^{\text{th}}$ pump. The pump removes $\frac{1}{4}$ of the air present, so $A_{n+1}=\frac{3}{4}A_n$.
The ratio $\frac{A_{n+1}}{A_n}=\frac{3}{4}$ is constant, but the *difference* $A_{n+1}-A_n$ is not constant (it changes with $A_n$). Hence the sequence is geometric, not arithmetic.
- (iii) Cost of digging a well:
Cost for the first metre = $150$ rupees.
Cost for each subsequent metre rises by $50$ rupees.
Thus the cost for the $n^{\text{th}}$ metre is $C_n = 150 + (n-1)\times 50$.
The successive difference $C_{n+1}-C_n = 50$ is constant. Hence the list of costs per metre is an AP with $a=150$, $d=50$.
- (iv) Money in a bank account with compound interest:
Initial amount $M_0 = 10000$ rupees.
Each year the amount grows by $8\%$, i.e., $M_{n+1}=1.08\,M_n$.
This gives a constant *ratio* $1.08$, not a constant *difference*. Therefore the yearly amounts form a geometric progression, not an AP.
3. Conclusion: Situations (i) and (iii) produce arithmetic progressions because the successive terms differ by a fixed amount (₹8 and ₹50 respectively). Situations (ii) and (iv) produce geometric progressions because the successive terms are obtained by multiplying by a constant factor.
4. Answer in short: (i) and (iii) are AP; (ii) and (iv) are not.
Question 2
Hint available
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 (ii) a = –2, d = 0 (iii) a = 4, d = – 3 (iv) a = – 1, d = 1 2 (v) a = – 1.25, d = – 0.25
Key Idea
In an arithmetic progression (AP) the nth term is given by \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference. The first four terms correspond to \(n = 1,2,3,4\).
Step-by-Step Solution
1. Write the general formula for the nth term of an AP: \[a_n = a + (n-1)d.\]
2. Substitute \(n = 1,2,3,4\) to obtain the first four terms:
- \(a_1 = a + 0\cdot d = a\)
- \(a_2 = a + 1\cdot d = a + d\)
- \(a_3 = a + 2\cdot d = a + 2d\)
- \(a_4 = a + 3\cdot d = a + 3d\)
3. For each given pair \((a,d)\), compute the four terms using the above expressions.
(i) \(a = 10, d = 10\):
\(a_1 = 10\)
\(a_2 = 10 + 10 = 20\)
\(a_3 = 10 + 2\times10 = 30\)
\(a_4 = 10 + 3\times10 = 40\)
(ii) \(a = -2, d = 0\):
\(a_1 = -2\)
\(a_2 = -2 + 0 = -2\)
\(a_3 = -2 + 0 = -2\)
\(a_4 = -2 + 0 = -2\)
(iii) \(a = 4, d = -3\):
\(a_1 = 4\)
\(a_2 = 4 + (-3) = 1\)
\(a_3 = 4 + 2(-3) = -2\)
\(a_4 = 4 + 3(-3) = -5\)
(iv) \(a = -1, d = \frac{1}{2}\):
\(a_1 = -1\)
\(a_2 = -1 + \frac{1}{2} = -\frac{1}{2}\)
\(a_3 = -1 + 2\times\frac{1}{2} = 0\)
\(a_4 = -1 + 3\times\frac{1}{2} = \frac{1}{2}\)
(v) \(a = -1.25, d = -0.25\):
\(a_1 = -1.25\)
\(a_2 = -1.25 + (-0.25) = -1.50\)
\(a_3 = -1.25 + 2(-0.25) = -1.75\)
\(a_4 = -1.25 + 3(-0.25) = -2.00\)
4. List the results for each part.
2. Substitute \(n = 1,2,3,4\) to obtain the first four terms:
- \(a_1 = a + 0\cdot d = a\)
- \(a_2 = a + 1\cdot d = a + d\)
- \(a_3 = a + 2\cdot d = a + 2d\)
- \(a_4 = a + 3\cdot d = a + 3d\)
3. For each given pair \((a,d)\), compute the four terms using the above expressions.
(i) \(a = 10, d = 10\):
\(a_1 = 10\)
\(a_2 = 10 + 10 = 20\)
\(a_3 = 10 + 2\times10 = 30\)
\(a_4 = 10 + 3\times10 = 40\)
(ii) \(a = -2, d = 0\):
\(a_1 = -2\)
\(a_2 = -2 + 0 = -2\)
\(a_3 = -2 + 0 = -2\)
\(a_4 = -2 + 0 = -2\)
(iii) \(a = 4, d = -3\):
\(a_1 = 4\)
\(a_2 = 4 + (-3) = 1\)
\(a_3 = 4 + 2(-3) = -2\)
\(a_4 = 4 + 3(-3) = -5\)
(iv) \(a = -1, d = \frac{1}{2}\):
\(a_1 = -1\)
\(a_2 = -1 + \frac{1}{2} = -\frac{1}{2}\)
\(a_3 = -1 + 2\times\frac{1}{2} = 0\)
\(a_4 = -1 + 3\times\frac{1}{2} = \frac{1}{2}\)
(v) \(a = -1.25, d = -0.25\):
\(a_1 = -1.25\)
\(a_2 = -1.25 + (-0.25) = -1.50\)
\(a_3 = -1.25 + 2(-0.25) = -1.75\)
\(a_4 = -1.25 + 3(-0.25) = -2.00\)
4. List the results for each part.
Question 3
Hint available
For the following APs, write the first term and the common difference: (i) 3, 1, – 1, – 3, . . . (ii) – 5, – 1, 3, 7, . . . (iii) 1 5 9 13 , , , , 3 3 3 3 . . . (iv) 0.6, 1.7, 2.8, 3.9, . . .
Key Idea
In an arithmetic progression (AP) the first term $a$ is the first listed number. The common difference $d$ is obtained by subtracting any term from the next term, i.e., $d = t_{n+1} - t_n$. Once $a$ and $d$ are known, the AP is completely determined.
Step-by-Step Solution
1. Identify the first term
The first term $a$ is simply the first number of the given sequence.
2. Find the common difference
Compute $d = \text{second term} - \text{first term}$. (The same value will be obtained for any consecutive pair if the sequence is truly an AP.)
(i) 3, 1, –1, –3, …
- First term: $a = 3$
- Common difference: $d = 1 - 3 = -2$
(ii) –5, –1, 3, 7, …
- First term: $a = -5$
- Common difference: $d = (-1) - (-5) = 4$
(iii) 1, 5, 9, 13, …
- First term: $a = 1$
- Common difference: $d = 5 - 1 = 4$
(iv) 0.6, 1.7, 2.8, 3.9, …
- First term: $a = 0.6$
- Common difference: $d = 1.7 - 0.6 = 1.1$
Thus for each AP we have identified $a$ and $d$ as required.
The first term $a$ is simply the first number of the given sequence.
2. Find the common difference
Compute $d = \text{second term} - \text{first term}$. (The same value will be obtained for any consecutive pair if the sequence is truly an AP.)
(i) 3, 1, –1, –3, …
- First term: $a = 3$
- Common difference: $d = 1 - 3 = -2$
(ii) –5, –1, 3, 7, …
- First term: $a = -5$
- Common difference: $d = (-1) - (-5) = 4$
(iii) 1, 5, 9, 13, …
- First term: $a = 1$
- Common difference: $d = 5 - 1 = 4$
(iv) 0.6, 1.7, 2.8, 3.9, …
- First term: $a = 0.6$
- Common difference: $d = 1.7 - 0.6 = 1.1$
Thus for each AP we have identified $a$ and $d$ as required.
Question 4
Hint available
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, . . . (ii) 5 7 2, , 3, , 2 2 . . . (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . (iv) – 10, – 6, – 2, 2, . . . (v) 3, 3 2 , 3 2 2 , 3 3 2 , . . . (vi) 0.2, 0.22, 0.222, 0.2222, . . . (vii) 0, – 4, – 8, –12, . . . (viii) – 1 2 , – 1 2 , – 1 2 , – 1 2 , . . . 56 (ix) 1, 3, 9, 27, . . . (x) a, 2a, 3a, 4a, . . . (xi) a, a2, a3, a4, . . . (xii) 2, 8, 18 , 32, . . . (xiii) 3, 6, 9 , 12 , . . . (xiv) 12, 32, 52, 72, . . . (xv) 12, 52, 72, 73, . . .
Key Idea
An arithmetic progression (AP) is a sequence of numbers in which the difference between any two successive terms is constant. This constant difference is called the common difference \(d\). For a sequence \(a_1, a_2, a_3, \dots\) to be an AP, we must have \(a_{k+1} - a_k = d\) for all \(k\). If the sequence is an AP, the next three terms are obtained by repeatedly adding \(d\) to the last given term.
Step-by-Step Solution
1. Check each given sequence
- Compute the successive differences.
- If all the differences are equal, the sequence is an AP; otherwise it is not.
- For an AP, record the common difference \(d\) and generate three more terms by adding \(d\) to the last term repeatedly.
2. Individual items
- (i) 2, 4, 8, 16, …
\[\Delta_1 = 4-2 = 2,\; \Delta_2 = 8-4 = 4,\; \Delta_3 = 16-8 = 8\]
The differences are not equal ⇒ Not an AP.
- (ii) 5, \frac{7}{2}, 3, \frac{2}{2}, … (interpreted as 5, 3.5, 3, 1)
\[\Delta_1 = 3.5-5 = -1.5,\; \Delta_2 = 3-3.5 = -0.5,\; \Delta_3 = 1-3 = -2\]
Differences are not equal ⇒ Not an AP.
- (iii) –1.2, –3.2, –5.2, –7.2, …
\[\Delta = (-3.2)-(-1.2) = -2,\; (-5.2)-(-3.2) = -2,\; (-7.2)-(-5.2) = -2\]
All differences are \(-2\). Hence it is an AP with \(d = -2\).
Next three terms: \(-9.2, -11.2, -13.2\).
- (iv) –10, –6, –2, 2, …
\[\Delta = (-6)-(-10) = 4,\; (-2)-(-6) = 4,\; 2-(-2) = 4\]
Common difference \(d = 4\). Next three terms: \(6, 10, 14\).
- (v) 3, 3\frac{2}{+}, 3\frac{2}{2}+ , 3\frac{3}{2}+ , … (the notation is ambiguous; the successive differences are not constant) ⇒ Not an AP.
- (vi) 0.2, 0.22, 0.222, 0.2222, …
\[\Delta_1 = 0.22-0.2 = 0.02,\; \Delta_2 = 0.222-0.22 = 0.002,\; \Delta_3 = 0.2222-0.222 = 0.0002\]
Differences are not equal ⇒ Not an AP.
- (vii) 0, –4, –8, –12, …
\[\Delta = -4-0 = -4,\; -8-(-4) = -4,\; -12-(-8) = -4\]
AP with \(d = -4\). Next three terms: \(-16, -20, -24\).
- (viii) –\frac{1}{2}, –\frac{1}{2}, –\frac{1}{2}, –\frac{1}{2}, …
All terms are equal, so the common difference \(d = 0\). It is an AP.
Next three terms: \(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}\).
- (ix) 1, 3, 9, 27, …
Ratios are constant (3) but differences are 2, 6, 18 … not equal ⇒ Not an AP.
- (x) a, 2a, 3a, 4a, …
\[\Delta = 2a-a = a,\; 3a-2a = a,\; 4a-3a = a\]
AP with common difference \(d = a\). Next three terms: \(5a, 6a, 7a\).
- (xi) a, a^2, a^3, a^4, …
Successive differences are \(a^2-a, a^3-a^2, a^4-a^3\); they are equal only for special values of \(a\) (e.g., \(a=1\) or \(a=0\)). In general, Not an AP.
- (xii) 2, 8, 18, 32, …
\[\Delta_1 = 6,\; \Delta_2 = 10,\; \Delta_3 = 14\]
Differences are not equal ⇒ Not an AP.
- (xiii) 3, 6, 9, 12, …
\[\Delta = 3\]
AP with \(d = 3\). Next three terms: \(15, 18, 21\).
- (xiv) 12, 32, 52, 72, …
\[\Delta = 20\]
AP with \(d = 20\). Next three terms: \(92, 112, 132\).
- (xv) 12, 52, 72, 73, …
Differences: 40, 20, 1 – not constant ⇒ Not an AP.
3. Summary
- APs: (iii), (iv), (vii), (viii), (x), (xiii), (xiv)
- For each AP, the common difference \(d\) and three additional terms are listed above.
- All other listed sequences are not arithmetic progressions.
- Compute the successive differences.
- If all the differences are equal, the sequence is an AP; otherwise it is not.
- For an AP, record the common difference \(d\) and generate three more terms by adding \(d\) to the last term repeatedly.
2. Individual items
- (i) 2, 4, 8, 16, …
\[\Delta_1 = 4-2 = 2,\; \Delta_2 = 8-4 = 4,\; \Delta_3 = 16-8 = 8\]
The differences are not equal ⇒ Not an AP.
- (ii) 5, \frac{7}{2}, 3, \frac{2}{2}, … (interpreted as 5, 3.5, 3, 1)
\[\Delta_1 = 3.5-5 = -1.5,\; \Delta_2 = 3-3.5 = -0.5,\; \Delta_3 = 1-3 = -2\]
Differences are not equal ⇒ Not an AP.
- (iii) –1.2, –3.2, –5.2, –7.2, …
\[\Delta = (-3.2)-(-1.2) = -2,\; (-5.2)-(-3.2) = -2,\; (-7.2)-(-5.2) = -2\]
All differences are \(-2\). Hence it is an AP with \(d = -2\).
Next three terms: \(-9.2, -11.2, -13.2\).
- (iv) –10, –6, –2, 2, …
\[\Delta = (-6)-(-10) = 4,\; (-2)-(-6) = 4,\; 2-(-2) = 4\]
Common difference \(d = 4\). Next three terms: \(6, 10, 14\).
- (v) 3, 3\frac{2}{+}, 3\frac{2}{2}+ , 3\frac{3}{2}+ , … (the notation is ambiguous; the successive differences are not constant) ⇒ Not an AP.
- (vi) 0.2, 0.22, 0.222, 0.2222, …
\[\Delta_1 = 0.22-0.2 = 0.02,\; \Delta_2 = 0.222-0.22 = 0.002,\; \Delta_3 = 0.2222-0.222 = 0.0002\]
Differences are not equal ⇒ Not an AP.
- (vii) 0, –4, –8, –12, …
\[\Delta = -4-0 = -4,\; -8-(-4) = -4,\; -12-(-8) = -4\]
AP with \(d = -4\). Next three terms: \(-16, -20, -24\).
- (viii) –\frac{1}{2}, –\frac{1}{2}, –\frac{1}{2}, –\frac{1}{2}, …
All terms are equal, so the common difference \(d = 0\). It is an AP.
Next three terms: \(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}\).
- (ix) 1, 3, 9, 27, …
Ratios are constant (3) but differences are 2, 6, 18 … not equal ⇒ Not an AP.
- (x) a, 2a, 3a, 4a, …
\[\Delta = 2a-a = a,\; 3a-2a = a,\; 4a-3a = a\]
AP with common difference \(d = a\). Next three terms: \(5a, 6a, 7a\).
- (xi) a, a^2, a^3, a^4, …
Successive differences are \(a^2-a, a^3-a^2, a^4-a^3\); they are equal only for special values of \(a\) (e.g., \(a=1\) or \(a=0\)). In general, Not an AP.
- (xii) 2, 8, 18, 32, …
\[\Delta_1 = 6,\; \Delta_2 = 10,\; \Delta_3 = 14\]
Differences are not equal ⇒ Not an AP.
- (xiii) 3, 6, 9, 12, …
\[\Delta = 3\]
AP with \(d = 3\). Next three terms: \(15, 18, 21\).
- (xiv) 12, 32, 52, 72, …
\[\Delta = 20\]
AP with \(d = 20\). Next three terms: \(92, 112, 132\).
- (xv) 12, 52, 72, 73, …
Differences: 40, 20, 1 – not constant ⇒ Not an AP.
3. Summary
- APs: (iii), (iv), (vii), (viii), (x), (xiii), (xiv)
- For each AP, the common difference \(d\) and three additional terms are listed above.
- All other listed sequences are not arithmetic progressions.