Arithmetic Progression: 8 Important Questions with Solutions
Practice these 8 selected questions covering the $n^{th}$ term, sum of terms ($S_n$), and application-based word problems. These are curated based on frequent Board Exam patterns.
Question 1
Find the 10th term of the Arithmetic Progression (A.P.): 2, 7, 12, ...
Solution:
Here, first term \(a = 2\)
Common difference \(d = 7 - 2 = 5\)
We need to find the 10th term (\(t_{10}\)).
Common difference \(d = 7 - 2 = 5\)
We need to find the 10th term (\(t_{10}\)).
Formula for the \(n^{th}\) term:
$$t_n = a + (n - 1)d$$
Substituting the values:
$$t_{10} = 2 + (10 - 1)5$$
$$t_{10} = 2 + (9)5$$
$$t_{10} = 2 + 45$$
$$t_{10} = 47$$
Answer: The 10th term of the A.P. is 47.
Question 2
Which term of the A.P. 21, 18, 15, ... is -81?
Solution:
Given A.P.: 21, 18, 15, ...
\(a = 21\)
\(d = 18 - 21 = -3\)
Let the \(n^{th}\) term be -81, so \(t_n = -81\).
\(a = 21\)
\(d = 18 - 21 = -3\)
Let the \(n^{th}\) term be -81, so \(t_n = -81\).
$$t_n = a + (n - 1)d$$
$$-81 = 21 + (n - 1)(-3)$$
$$-81 - 21 = (n - 1)(-3)$$
$$-102 = (n - 1)(-3)$$
Dividing both sides by -3:
$$\frac{-102}{-3} = n - 1$$
$$34 = n - 1$$
$$n = 34 + 1$$
$$n = 35$$
Answer: The 35th term of the A.P. is -81.
Question 3
Find the sum of all 3-digit natural numbers which are divisible by 5.
Solution:
The 3-digit numbers divisible by 5 are: 100, 105, 110, ..., 995.
This is an A.P. where \(a = 100\), \(d = 5\), and the last term \(t_n = 995\).
This is an A.P. where \(a = 100\), \(d = 5\), and the last term \(t_n = 995\).
Step 1: Find the number of terms (n)
$$t_n = a + (n - 1)d$$ $$995 = 100 + (n - 1)5$$ $$995 - 100 = (n - 1)5$$ $$895 = (n - 1)5$$ $$n - 1 = \frac{895}{5}$$ $$n - 1 = 179 \Rightarrow n = 180$$
$$t_n = a + (n - 1)d$$ $$995 = 100 + (n - 1)5$$ $$995 - 100 = (n - 1)5$$ $$895 = (n - 1)5$$ $$n - 1 = \frac{895}{5}$$ $$n - 1 = 179 \Rightarrow n = 180$$
Step 2: Find Sum (\(S_n\))
Using formula: \(S_n = \frac{n}{2} [a + t_n]\) $$S_{180} = \frac{180}{2} [100 + 995]$$ $$S_{180} = 90 \times 1095$$ $$S_{180} = 98,550$$
Using formula: \(S_n = \frac{n}{2} [a + t_n]\) $$S_{180} = \frac{180}{2} [100 + 995]$$ $$S_{180} = 90 \times 1095$$ $$S_{180} = 98,550$$
Answer: The sum is 98,550.
Question 4
Find the value of \(k\) such that \(2k\), \(k + 10\), and \(3k + 2\) are three consecutive terms of an A.P.
Solution:
If \(a, b, c\) are in A.P., then the common difference is constant.
i.e., \(b - a = c - b\) or \(2b = a + c\).
i.e., \(b - a = c - b\) or \(2b = a + c\).
Here, \(a = 2k\), \(b = k + 10\), \(c = 3k + 2\).
$$2(k + 10) = (2k) + (3k + 2)$$ $$2k + 20 = 5k + 2$$ $$20 - 2 = 5k - 2k$$ $$18 = 3k$$ $$k = \frac{18}{3} = 6$$
$$2(k + 10) = (2k) + (3k + 2)$$ $$2k + 20 = 5k + 2$$ $$20 - 2 = 5k - 2k$$ $$18 = 3k$$ $$k = \frac{18}{3} = 6$$
Answer: The value of k is 6.
Question 5
For an A.P., if the first term is 6 and the common difference is -3, find \(S_{12}\).
Solution:
Given: \(a = 6\), \(d = -3\), \(n = 12\).
Formula for Sum:
$$S_n = \frac{n}{2} [2a + (n - 1)d]$$
Substituting values:
$$S_{12} = \frac{12}{2} [2(6) + (12 - 1)(-3)]$$
$$S_{12} = 6 [12 + (11)(-3)]$$
$$S_{12} = 6 [12 - 33]$$
$$S_{12} = 6 [-21]$$
$$S_{12} = -126$$
Answer: The sum of the first 12 terms is -126.
Question 6
The sum of the 4th and 8th terms of an A.P. is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.
Solution:
We know \(t_n = a + (n-1)d\).
Condition 1: \(t_4 + t_8 = 24\) $$(a + 3d) + (a + 7d) = 24$$ $$2a + 10d = 24 \Rightarrow a + 5d = 12 \quad \dots(1)$$
Condition 2: \(t_6 + t_{10} = 44\) $$(a + 5d) + (a + 9d) = 44$$ $$2a + 14d = 44 \Rightarrow a + 7d = 22 \quad \dots(2)$$
Condition 1: \(t_4 + t_8 = 24\) $$(a + 3d) + (a + 7d) = 24$$ $$2a + 10d = 24 \Rightarrow a + 5d = 12 \quad \dots(1)$$
Condition 2: \(t_6 + t_{10} = 44\) $$(a + 5d) + (a + 9d) = 44$$ $$2a + 14d = 44 \Rightarrow a + 7d = 22 \quad \dots(2)$$
Subtract (1) from (2):
$$(a + 7d) - (a + 5d) = 22 - 12$$
$$2d = 10 \Rightarrow d = 5$$
Substitute \(d = 5\) in equation (1): $$a + 5(5) = 12$$ $$a + 25 = 12$$ $$a = 12 - 25 = -13$$
Substitute \(d = 5\) in equation (1): $$a + 5(5) = 12$$ $$a + 25 = 12$$ $$a = 12 - 25 = -13$$
Answer: The A.P. is -13, -8, -3, ...
Question 7
If the sum of first 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of first \(n\) terms.
Solution:
Given: \(S_7 = 49\) and \(S_{17} = 289\).
Formula: \(S_n = \frac{n}{2}[2a + (n-1)d]\)
Formula: \(S_n = \frac{n}{2}[2a + (n-1)d]\)
For \(n=7\):
$$49 = \frac{7}{2}[2a + 6d] \Rightarrow 49 = 7[a + 3d] \Rightarrow 7 = a + 3d \dots(1)$$
For \(n=17\): $$289 = \frac{17}{2}[2a + 16d] \Rightarrow 289 = 17[a + 8d] \Rightarrow 17 = a + 8d \dots(2)$$
For \(n=17\): $$289 = \frac{17}{2}[2a + 16d] \Rightarrow 289 = 17[a + 8d] \Rightarrow 17 = a + 8d \dots(2)$$
Subtract (1) from (2):
$$5d = 10 \Rightarrow d = 2$$
From (1): \(a + 3(2) = 7 \Rightarrow a = 1\).
Sum of \(n\) terms:
$$S_n = \frac{n}{2}[2(1) + (n-1)2]$$
$$S_n = \frac{n}{2}[2 + 2n - 2] = \frac{n}{2}[2n] = n^2$$
Answer: The sum of the first n terms is \(n^2\).
Question 8 (Word Problem)
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 the production in the 1st year.
Solution:
Since production increases uniformly, it forms an A.P.
Let production in 1st year be \(a\) and annual increase be \(d\).
Given: \(t_3 = 600\) and \(t_7 = 700\).
Let production in 1st year be \(a\) and annual increase be \(d\).
Given: \(t_3 = 600\) and \(t_7 = 700\).
Using \(t_n = a + (n-1)d\):
$$t_3 = a + 2d = 600 \quad \dots(1)$$
$$t_7 = a + 6d = 700 \quad \dots(2)$$
Subtract (1) from (2):
$$4d = 100 \Rightarrow d = 25$$
Substitute \(d\) in (1):
$$a + 2(25) = 600$$
$$a + 50 = 600$$
$$a = 550$$
Answer: The production in the 1st year was 550 TV sets.