Exercise 5.1 Practice
Step-by-Step Solutions Based on NCERT
Q1: Real Life Situations
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The auto fare after each km when the fare is ₹20 for the first km and ₹10 for each additional km.
Step 1: Write the Sequence
Fare for 1st km ($a_1$) = ₹20
Fare for 2nd km ($a_2$) = $20 + 10$ = ₹30
Fare for 3rd km ($a_3$) = $30 + 10$ = ₹40
Fare for 1st km ($a_1$) = ₹20
Fare for 2nd km ($a_2$) = $20 + 10$ = ₹30
Fare for 3rd km ($a_3$) = $30 + 10$ = ₹40
Step 2: Check Difference
$a_2 - a_1 = 30 - 20 = 10$
$a_3 - a_2 = 40 - 30 = 10$
Since the common difference is constant ($d=10$):
$a_2 - a_1 = 30 - 20 = 10$
$a_3 - a_2 = 40 - 30 = 10$
Since the common difference is constant ($d=10$):
Yes, it is an AP.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/5 of the air remaining in the cylinder at a time.
Step 1: Let initial air be V
$a_1 = V$
$a_1 = V$
Step 2: After 1st pump
$a_2 = V - \frac{1}{5}V = \frac{4}{5}V$
$a_2 = V - \frac{1}{5}V = \frac{4}{5}V$
Step 3: After 2nd pump
$a_3 = \frac{4}{5}V - \frac{1}{5}(\frac{4}{5}V) = \frac{16}{25}V$
$a_3 = \frac{4}{5}V - \frac{1}{5}(\frac{4}{5}V) = \frac{16}{25}V$
Step 4: Check Difference
$a_2 - a_1 = -\frac{1}{5}V$
$a_3 - a_2 = \frac{16}{25}V - \frac{20}{25}V = -\frac{4}{25}V$
Difference is not same.
$a_2 - a_1 = -\frac{1}{5}V$
$a_3 - a_2 = \frac{16}{25}V - \frac{20}{25}V = -\frac{4}{25}V$
Difference is not same.
No, it is not an AP.
(iii) The cost of digging a pit after every metre of digging, when it costs ₹200 for the first metre and rises by ₹40 for each subsequent metre.
Step 1: Write Sequence
Cost for 1m = 200
Cost for 2m = $200 + 40 = 240$
Cost for 3m = $240 + 40 = 280$
Cost for 1m = 200
Cost for 2m = $200 + 40 = 240$
Cost for 3m = $240 + 40 = 280$
Step 2: Check Difference
Difference is constant at ₹40.
Difference is constant at ₹40.
Yes, it is an AP.
(iv) The amount of money in the account every year, when ₹5000 is deposited at compound interest at 6% per annum.
Step 1: Formula
Amount $A = P(1 + \frac{R}{100})^n$
Amount $A = P(1 + \frac{R}{100})^n$
Step 2: Write Terms
$a_1 = 5000(1.06)^1$
$a_2 = 5000(1.06)^2$
$a_3 = 5000(1.06)^3$
$a_1 = 5000(1.06)^1$
$a_2 = 5000(1.06)^2$
$a_3 = 5000(1.06)^3$
Step 3: Conclusion
Terms are increasing by a factor of 1.06 (Geometric), not by a fixed number.
Terms are increasing by a factor of 1.06 (Geometric), not by a fixed number.
No, it is not an AP.
Q2: Writing Terms
Write first four terms of the AP, when the first term $a$ and the common difference $d$ are given as follows:
(i) $a = 5, d = 5$
Calculation:
$a_1 = 5$
$a_2 = 5+5=10$
$a_3 = 10+5=15$
$a_4 = 15+5=20$
$a_1 = 5$
$a_2 = 5+5=10$
$a_3 = 10+5=15$
$a_4 = 15+5=20$
5, 10, 15, 20
(ii) $a = -3, d = 0$
Calculation:
Since $d=0$, every term is the same.
Since $d=0$, every term is the same.
-3, -3, -3, -3
(iii) $a = 7, d = -2$
Calculation:
$a_1 = 7$
$a_2 = 7 + (-2) = 5$
$a_3 = 5 + (-2) = 3$
$a_4 = 3 + (-2) = 1$
$a_1 = 7$
$a_2 = 7 + (-2) = 5$
$a_3 = 5 + (-2) = 3$
$a_4 = 3 + (-2) = 1$
7, 5, 3, 1
(iv) $a = -2, d = 1/3$
Calculation:
$a_1 = -2$
$a_2 = -2 + 1/3 = -5/3$
$a_3 = -5/3 + 1/3 = -4/3$
$a_4 = -4/3 + 1/3 = -3/3 = -1$
$a_1 = -2$
$a_2 = -2 + 1/3 = -5/3$
$a_3 = -5/3 + 1/3 = -4/3$
$a_4 = -4/3 + 1/3 = -3/3 = -1$
-2, -5/3, -4/3, -1
(v) $a = -1.5, d = -0.5$
Calculation:
$a_1 = -1.5$
$a_2 = -1.5 - 0.5 = -2.0$
$a_3 = -2.0 - 0.5 = -2.5$
$a_4 = -2.5 - 0.5 = -3.0$
$a_1 = -1.5$
$a_2 = -1.5 - 0.5 = -2.0$
$a_3 = -2.0 - 0.5 = -2.5$
$a_4 = -2.5 - 0.5 = -3.0$
-1.5, -2.0, -2.5, -3.0
Q3: Finding a & d
For the following APs, write the first term and the common difference:
(i) 5, 2, -1, -4...
First Term (a): 5
Difference (d): $2 - 5 = -3$
a = 5, d = -3
(ii) -4, -2, 0, 2...
First Term (a): -4
Difference (d): $-2 - (-4) = -2 + 4 = 2$
a = -4, d = 2
(iii) 1/4, 3/4, 5/4, 7/4...
First Term (a): 1/4
Difference (d): $3/4 - 1/4 = 2/4 = 1/2$
a = 1/4, d = 1/2
(iv) 0.5, 1.6, 2.7, 3.8...
First Term (a): 0.5
Difference (d): $1.6 - 0.5 = 1.1$
a = 0.5, d = 1.1
Q4: Identify AP & Write Terms
Which of the following are APs? If they form an AP, find the common difference $d$ and write three more terms.
(i) 3, 9, 27, 81...
Check diff: $9-3=6$, $27-9=18$.
Not an AP (Diff is not constant)
(ii) 5, 5.5, 6, 6.5...
Check diff: $5.5-5=0.5$, $6-5.5=0.5$.
Constant diff $d=0.5$.
Next terms: $6.5+0.5=7$, $7+0.5=7.5$, $7.5+0.5=8$.
Yes, d=0.5. Terms: 7, 7.5, 8
(iii) -3.2, -5.2, -7.2, -9.2...
Check diff: $-5.2 - (-3.2) = -2.0$.
Constant diff $d=-2$.
Next terms: $-9.2-2=-11.2$, etc.
Yes, d=-2. Terms: -11.2, -13.2, -15.2
(iv) -6, -2, 2, 6...
Check diff: $-2 - (-6) = 4$, $2 - (-2) = 4$.
Constant diff $d=4$.
Yes, d=4. Terms: 10, 14, 18
(v) $2, 2+\sqrt{5}, 2+2\sqrt{5}, 2+3\sqrt{5}$...
Check diff: $(2+\sqrt{5}) - 2 = \sqrt{5}$.
$(2+2\sqrt{5}) - (2+\sqrt{5}) = \sqrt{5}$.
Yes, d=$\sqrt{5}$. Terms: $2+4\sqrt{5}, 2+5\sqrt{5}, 2+6\sqrt{5}$
(vi) 0.4, 0.44, 0.444, 0.4444...
$0.44 - 0.4 = 0.04$
$0.444 - 0.44 = 0.004$
Not an AP (Diff is not constant)
(vii) 0, -3, -6, -9...
Diff: $-3-0 = -3$, $-6-(-3) = -3$.
Yes, d=-3. Terms: -12, -15, -18
(viii) -1/5, -1/5, -1/5...
Diff: $-1/5 - (-1/5) = 0$.
Yes, d=0. Terms: -1/5, -1/5, -1/5
(ix) 1, 3, 9, 27...
$3-1=2$, $9-3=6$.
Not an AP.
(x) $c, 2c, 3c, 4c...$
$2c-c=c$, $3c-2c=c$.
Yes, d=c. Terms: 5c, 6c, 7c
(xi) $y, y^2, y^3, y^4...$
$y^2 - y = y(y-1)$
$y^3 - y^2 = y^2(y-1)$
Not an AP (Diff depends on y)
(xii) $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}$...
Simplify terms: $\sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3}$...
Diff: $2\sqrt{3}-\sqrt{3} = \sqrt{3}$.
Yes, d=$\sqrt{3}$. Terms: $\sqrt{75}, \sqrt{108}, \sqrt{147}$
(xiii) $\sqrt{2}, \sqrt{4}, \sqrt{6}, \sqrt{8}$...
Values: $\sqrt{2}, 2, \sqrt{6}, 2\sqrt{2}$.
Differences are clearly not same.
Not an AP.
(xiv) $2^2, 4^2, 6^2, 8^2$...
Values: 4, 16, 36, 64...
$16-4=12$, $36-16=20$.
Not an AP.
(xv) $5^2, 7^2, 73, 97$...
Values: 25, 49, 73, 97...
$49-25=24$, $73-49=24$, $97-73=24$.
Yes, d=24. Terms: 121, 145, 169