Arithmetic progression questions for class 10

This page contains arithmetic progression questions for class 10. In this link get short arithmetic progression class 10 notes. Few other links are
arithmetic progression class 10 important questions
arithmetic progression assignment
Practicing these questions on Arithmatic progression for class 10 maths can be helpful for your board exams.

Arithmetic Progression questions and answers

class 10 maths


One Marks Questions

Question 1. If k, 2k – 1 and 2k + 1 are three consecutive terms of an A.P., find the value of k.
Question 2. Find the common difference of the A.P. \(\frac{1}{2b},\frac{1-6b}{2b},\frac{1-12b}{2b},….\)
Question 3. How many two digit numbers are divisible by 3?
Question 4. What is the common difference of an A.P. in Which \(a_{21}-a_7=84\)?
Question 5. If the sum of the first p terms of an A.P. is \(ap^2+bp,\). Find its common difference.

Two Marks Questions

Question 1. If the \(17^{th}\) term of an A.P. exceeds its \(10^{th}\) term by 7, find the common difference.
Question 2. Find how many integers between 200 and 500 are divisible by 8.
Question 3. Which term of the progression \(\text{20,}19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},….\) is the first negative term?
Question 4. Find the middle term of the AP \(\text{213,205,197,……,}37\).
Question 5. Find whether -150 is a term of the A.P. \(\text{17,12,7,2,}…\)?
Question 6. If the sum of the first p terms of an A.P. is \(ap^2+bp,\). Find its common difference.
Question 7. If \(S_n\), the sum of first n terms of an A.P. is given by \(S_n=3n^2=4n,\)find the \(n^{\text{th}}\) term.
Question 8. In an A.P., if \(S_5+S_7\,\,=\,\,\text{167 and }S_{10}\,\,=\,\,\text{235,}\) then find the A.P., where denotes the sum of its first n terms.
Question 9.The sum of the first n terms of an A.P. is\(5n\,\,-\,\,n^2.\)Find the \(n^{th}\)term of this A.P.
Question 10. Find the common difference of an A.P. whose first term in 4, the last term is 49 and the sum of all its terms is 265.

Three marks questions

Question 1. If the \(m^{th}\)term of an A.P. is \(\frac{1}{n}\) and \(\text{n}^{\text{th}}\) term is \(\frac{1}{m}\) then show that its\(\left( mn \right) ^{th}\) term is 1.
Question 2. If the ratio of the sum of first n terms of two A.P’s is \((7n + 1) : (4n + 27)\), find the ratio of
their \(m^{th}\) terms.
Question 3. The digits of a positive number of three digits are in A.P. and their sum is 15. The number
obtained by reversing the digits is 594 less than the original number. Find the number.
Question 4. The\(p^{\text{th}}, q^{\text{th}}\,\,\text{and }r^{\text{th}}\) terms of an A.P. are a, b and c respectively. Show that \(a\left( q-\text{r} \right) +\text{b}\left( r-p \right) +c\left( p-q \right) =0\)
Question 5. Divide 56 in four parts in A.P. such that the ratio of the product of their extremes
\(\left( 1^{st}\,\,\text{and }4^{th} \right) \)to the product of means \(\left( 2^{\text{nd}}\,\,\text{and }3^{\text{rd}} \right) \) is 5:6.

Four or six marks questions

Question 1 The sum of four consecutive numbers in an A.P. is 32 and the ratio of the product of the first and the last term to the product of two middle terms is \(7: 15\). Find the numbers.
Question 2. The \(17^{\text{th}}\) term of an A.P. is 5 more than twice its \(8^{\text{th}}\) term. If the \(11^{th}\) term of the A.P. is 43, then find its \(n^{th}\) term.

Answers to one marks questions
Answer 1
It is given in the question that k, 2k – 1 and 2k + 1 are three consecutive terms of an A.P.
Therefore
\(\left( 2k-1 \right) -\left( k \right) =\left( 2k+1 \right) -\left( 2k-1 \right) \\\Rightarrow \,\,k-1=2\\\Rightarrow k=3\)
Answer 2
The common difference of the A.P. \(\frac{1}{2b},\frac{1-6b}{2b},\frac{1-12b}{2b},….\) is given by
\(\frac{1-6b}{2b}-\frac{1}{2b}=\frac{1-6b-1}{2b}=\frac{-6b}{2b}=-3\)
Answer 3
Two-digit numbers which are divisible by 3 are 12,15,18 ,…, 99.
This sequence forms an A.P. with first term \(\left( a \right) =12\),
common difference \(\left( d \right) =15-12=3\)
and last term \(t_n=99\)
We know that
\(t_n=a+\left( n-1 \right) d\\\therefore \,\,12+\left( n-1 \right) 3=99\\\Rightarrow 3n=99-9\\or, n=\frac{90}{3}=30\)
So,  there are  30 two-digit numbers which are divisible by 3.
Answer 4
Let \(a\) be the first term and \(d\) be the common difference of the A.P.
Given that,
\(a_{21}-a_7=84\\t_n=a+\left( n-1 \right) d\\\therefore \,\,a_{21}=a+20d\,\,and\,\,a_7=a+6d\\a+20d-\left( a+6d \right) =84\\\Rightarrow 14d=84\\\Rightarrow d=6\)

Answers to two marks questions



Answers to three marks questions

Answer 1 Let \(a\) be the first term and \(d\) be the common difference of the given A.P.
Now, \(r^{th}\) term of A.P. \(a_r=a+(r-1)d\)
According to question,
\(a_m =a+(m-1)d=\frac{1}{n} \qquad \left( i \right)\)
and,
\(a_n =a+\left( n-1 \right) d=\frac{1}{m} \qquad \left( ii \right)\)
Subtracting, \(\left( ii \right) \text{from } \left( i \right)\) , we get
\((m – n)d = \frac{m – n}{mn} \Rightarrow d = \frac{1}{mn}\)
putting \(d = \frac{1}{mn}\) in \(\left( i \right)\) , we get
\(a + (m – n) \frac{1}{mn} = \frac{1}{n} \Rightarrow a +\frac{1}{n} – \frac{1}{mn} = \frac{1}{n}\)
\(\Rightarrow a = \frac{1}{mn}\)
\(\therefore a_{mn} = a + (mn – 1)d = \frac{1}{mn} + (mn – 1) \frac{1}{mn}\)
\(= \frac{1 + mn – 1}{mn} = 1\)

Answer 2

Let \(a_1\), \(d_1\), and \(a_2\), \(d_2\) be the first term and common difference of the two A.P.’s respectively.
\(\frac{n}{2} \frac{ \left[ 2a_1 + (n – 1)d_1 \right] }{\left[ 2a_2 + (n – 1)d_2 \right]} = \frac{7n + 1}{4n + 27}  \) as given in the question
\(\Rightarrow \frac{a_1 + \frac{\left( n – 1 \right)}{2}d_1}{a_2 + \frac{\left( n – 1 \right)}{2}d_2} = \frac{7n + 1}{4n + 27}\)
put
\(\frac{n – 1}{2} = m – 1 \Rightarrow n – 1 = 2m – 2\)
\(\Rightarrow n = 2m – 2 + 1 = 2m – 1\)
\(\therefore \frac{a_1 + \left( m-1 \right) d_1}{a_2 + \left( m-1 \right) d_2}=\frac{7\left( 2m-1 \right) +1}{4\left( 2m-1 \right) +27} \)

solving it further we get the ratio as,
\( = \frac{14m – 6}{8m – 4 + 27}\)



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