NCERT Solutions Class 10th Maths Chapter – 5 Arithmetic Progressions Examples

Solution: Here, a = 3/2, d = 1/2 – 3/2 = -1.

Remember that we can find d using any two consecutive terms, once we know that the numbers are in AP.

Example 2. Which of the following list of numbers 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) We havea₂ – a₁ = 10-4 = 6
a₃ – a₂ = 16 – 10 = 6
a₄ – a₃ = 22 – 16 = 6i.e., a_{k + 1} a_k is the same every time.
So, the given list of numbers forms an AP with the common difference d = 6.
The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.

(ii) a₂ – a₁ = – 1 – 1 = -2
a₃ – a₂ = 3 – (-1) = – 3 + 1 = -2
a₄ – a₃ = – 5 – (-3) = – 5 + 3 = -2
i.e., a_{k + 1} a_k is the same every time.
So, the given list of numbers forms an AP with the common difference d = -2.
The next two terms are:
– 5 + (-2) = -7 and – 7 + (-2) = -9

(iii) a₂ – a₁ = – 2 – (-2) = 2 + 2 = 4
a₃ – a₂ = – 2 – 2 = -4As a_₂ – a₁ ≠ a₃ – a₂, the given list of numbers does not form an AP.(iv) a₂ – a₁ = 1 – 1 = 0
a₃ – a₂ = 1 – 1 = 0
a₄ – a₃ = 2 – 1 = 1
Here, a₂ – a₁ = a₃ – a₂ ≠ a₄ – a₃
So, the given list of numbers does not form an AP.

Example 3. Find the 10th term of the AP: 2, 7, 12, . . .

Solution: Here, a = 2,
d = 7 – 2 = 5 and n = 10.
We have a_n = a + (n – 1) d
So, a₁₀ = 2 + (10 – 1) x 5 = 2 + 45 = 47
Therefore, the 10th term of the given AP is 47.

Example 4. Which term of the AP: 21, 18, 15,…is – 81? Also, is any term 0? Give reason for your answer.

Solution: Here, a = 21, d= 18 – 21 = −3 and a_n = – 81, and we have to find n.
As a_n = a + (n-1) d,
we have-81 = 21+ (n − 1)(-3)
-81 = 24 – 3n
-105 = – 3nSo, n = 35
Therefore, the 35th term of the given AP is – 81.
Next, we want to know if there is any n for which a,, = 0. If such an n is there, then
21+ (n-1) (-3)= 0,
i.e., 3(n-1)= 21
i.e., n = 8
So, the eighth term is 0.

Example 5. Determine the AP whose 3rd term is 5 and the 7th term is 9.

Solution: We have
a = a + (3 – 1) d = a + 2d = 5
and a = a + (7 – 1) d = a + 6d = 9
Solving the pair of linear equations (1) and (2), we get
a = 3, d = 1
Hence, the required AP is 3, 4, 5, 6, 7, . . .

Example 6. Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .

Solution: We have
a₂ – a₁ = 11 – 5 = 6, a₃ – a₂ = 17 – 11 = 6, a₄ – a₃ = 23 – 17 = 6
As a_{k + 1} – a_k is the same for k = 1, 2, 3, etc.,

the given list of numbers is an AP.
Now, a = 5 and d = 6.
Let 301 be a term, say, the nth term of this AP.
We know that
a_n = a + (n – 1) d
So, 301 = 5 + (n – 1) × 6
i.e., 301 6n – 1
301 = 6n – 1
So, n = 302/6 = 151/2
But n should be a positive integer (Why?). So, 301 is not a term of the given list of numbers.

Example 7. How many two-digit numbers are divisible by 3?

Solution: The list of two-digit numbers divisible by 3 is:
12, 15, 18, …….,99
Is this an AP? Yes it is. Here, a = 12, d= 3, a = 99.
As an = a + (n – 1)
we have 99 = 12 + (n – 1) × 3
i.e., 87 = (n – 1) × 3
i.e., n – 1 = 87/3 = 29
i.e., n = 29 + 1 = 30
n = 29 + = 130
So, there are 30 two-digit numbers divisible by 3.

Example 8. Find the 11th term from the last term (towards the first term) of the
AP: 10, 7, 4, . . ., -62.

Solution: Here, a = 10, d = 7 – 10 = – 3, 1 = -62,
where l = a + (n – 1) d
To find the 11th term from the last term, we will find the total number of terms in the AP.
So, -62 = 10 + (n – 1)(-3)
i.e., -72 = (n – 1)(-3)
i.e., n – 1 = 24
or n = 25
So, there are 25 terms in the given AP.
The 11th term from the last term will be the 15th term. (Note that it will not be the 14th term. Why?)
So, a₁₅ = 10 + (15 – 1)(-3) = 10 – 42 = -32
i.e., the 11th term from the last term is – 32.

Alternative Solution : 

If we write the given AP in the reverse order, then a = -62 and d = 3 (Why?)
So, the question now becomes finding the 11th term with these a and d.
So,
a₁₁ = -62 + (11 – 1) × 3 = – 62 + 30 = -32
So, the 11th term, which is now the required term, is -32.

Example 9. A sum of 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact.

Solution: We know that the formula to calculate simple interest is given by
Simple Interest = P × R × T/100So, the interest at the end of the 1st year = ₹ 1000 × 8 × 1/100 = ₹ 80
The interest at the end of the 2nd year = ₹ 1000 × 8 × 2/100 = ₹ 80
The interest at the end of the 3rd year = ₹ 1000 × 8 × 3/100 = ₹ 80Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on.
So, the interest (in) at the end of the 1st, 2nd, 3rd, . . . years, respectively are
80, 160, 240,…It is an AP as the difference between the consecutive terms in the list is 80, i.e.,
d = 80. Also, a = 80.
So, to find the interest at the end of 30 years, we shall find a_30.
Now, a₃₀ = a + (30 – 1) d = 80 + 29 × 80 = 2400
So, the interest at the end of 30 years will be ₹2400.

Example 10. In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?

Solution: The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are : 23, 21, 19,…, 5
It forms an AP (Why?). Let the number of rows in the flower bed be n.
a = 23, d = 21 – 23 = −2, a = 5
5 = 23 + (n – 1) (-2)
– 18 = (n – 1)(-2)
n = 10
So, there are 10 rows in the flower bed.

Example 11. Find the sum of the first 22 terms of the AP: 8, 3, -2, …

Solution: Here, a = 8, d = 3 – 8 = -5, n = 22.
We know that
S = 2 [2a + (n − 1) d]
Therefore,
S =22/2 [16 + 21 (-5)] = 11(16 – 105) = 11(-89) = -979
So, the sum of the first 22 terms of the AP is – 979.

Example 12. If the sum of the first 14 terms of an AP is 1050 and its first term is 10 find the 20th term.

Solution: Here, S14 = 1050, n = 14, a = 10.
As Sn = n/2 [2a + (n – 1) d],
so, 1050 = 14/2[20 + 13d] = 140 + 91d
i.e., 910 = 91d
or, d = 10
Therefore, a20 = 10 + (20 – 1) × 10 = 200 i.e. 20th term is 200.

Example 13: How many terms of the AP: 24, 21, 18, . . . must be taken so that their sum is 78?

Solution: Here, a = 24, d = 21 – 24 = -3, S_n = 78. We need to find n.
We know that Sn = n/2 [2a + (n – 1) d]
So, 78 = n/2[48 + (n – 1)(-3)] = n/2[51 – 3n]
or 3n² 51n + 156 = 0
or n² 17n + 52 = 0
or (n – 4)(n – 13) = 0
or n = 4 or 13
Both values of n are admissible. So, the number of terms is either 4 or 13.

Example 14. Find the sum of:

(i) the first 1000 positive integers
(ii) the first n positive integers

Solution:
(i) Let S = 1 + 2 + 3 +…+ 1000
Using the formula S_n = n/2(a + l) for the sum of the first n terms of an AP, we
have
S1000 = 1000/2(1+1000) 500 1001 = 500500
So, the sum of the first 1000 positive integers is 500500.

(ii) Let Sn = 1 + 2 + 3 +…+ n
Here a = 1 and the last term l is n.

Therefore, S_n = n(l + n)/2 or S_n = n (n + 1)/2

So, the sum of first n positive integers is given by S_n = n(n + 1)/2

Example 15. Find the sum of first 24 terms of the list of numbers whose nth term is given by
an = 3 + 2n

Solution:
a_n = 3 + 2n,
a₁ = 3 + 2 = 5
a₂ = 3 + 2 × 2 = 7
a₃ = 3 + 2 × 3 = 9
List of numbers becomes 5, 7, 9, 11,…
7 – 5 = 9 – 7 = 11 – 9 = 2 and so on.
So, it forms an AP with common difference d = 2.
To find S24 we have n = 24, a = 5, d = 2.
S₂₄ = 24/2[2 x 5 + (24 – 1) × 2] = 12 [10 + 46] = 672
So, sum of first 24 terms of the list of numbers is 672.

Example 16. A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find:

(i) the production in the 1st year
(iii) the total production in first 7 years
(ii) the production in the 10th year

Solution: (i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP.
Let us denote the number of TV sets manufactured in the nth year by a
Then, a₃ = 600 and a₇ = 700
or, a + 2d = 600
and a + 6d = 700
Solving these equations, we get d = 25 and a = 550.
Therefore, production of TV sets in the first year is 550.

(ii) Now,
a₁₀ = a + 9d = 550 + 9 × 25 = 775
So, production of TV sets in the 10th year is 775.

(iii) Also,
S₇ = 7/2 [2 × 550 + (7 – 1) × 25]
= 7/2 [1100 + 150] = 4375
Thus, the total production of TV sets in first 7 years is 4375.