# Class 10 Maths Important Questions for Arithmetic Progression

Given below are the Class 10 Maths Important Questions & extra questions for Arithmetic Progression.This exercise has variety of quetsions including tough and difficult question on AP also
a. Concepts questions
b. Calculation problems
d. proof questions

Question 1
The general term of a sequence is given by an = -4n + 15. Is the sequence an A. P.? If so, find its 15th term and the common difference.

$a_n = -4n + 15$
$a_k = -4k + 15$
$a_{k+1} = -4(k+1) + 15$
Now
$a_{k+1} - a_k=-4(k+1) + 15 -[-4k + 15]=-4$
Since difference between two terms constant.It is a AP
$a_{15} = -4(15) + 15=-45$

Question 2
The nth term of an A. P. is 6n + 11. Find the common difference.

an=6n + 11
ak = 6k + 11
ak+1 = 6(k+1) + 11
Now
ak+1 - ak=6(k+1) + 11 -[6k + 11]=6

Question 3
If the 8th term of an A. P. is 31 and the 15th term is 16 more than the 11th term, find the A. P.

a8 = 31
a +(8-1)d = 31
a + 7d = 31 -- (i)

a15 =16 + a11
a + 14d = 16 + a +10d
14d = 16 + 10d
4d = 16
d = 4
Now putting d = 4 in eq. (i) we get
a + 7(4) = 31
a + 28 = 31
a = 3

So AP
3,7,11,15...

Question 4
Which term of the arithmetic progression 5, 15, 25, ----- will be 130 more than its 31st term?

Let n th term be 130 more than the 31st term of the A.P.
First term of A.P. = 5
Common difference = 15 - 5 = 10
an = 130 + a31
5 + (n - 1) X 10 = 130 + 5 + (31 - 1) X 10
10 (n - 1) = 430
n = 44

Thus, 44th term of the A.P is 130 more than the 31st term.

Question 5
Which term of the A. P. 3, 15, 27, 39...... will be 132 more than its 54th term?

Similar question as above.
Answer is 65th term is 132 more than its 54th term

Question 6
Two A. P.’s has the same common difference. The difference between their 100th terms is 111222333. What is the difference between their Millionth terms?

$a_n=a + (n-1)d$
For Ist AP
$a_{100x}=a_x + (100-1)d$
For 2nd AP
$a_{100y}=a_y + (100-1)d$
So difference
$a_x-a_y=a_{100x}- a_{100y}$
$a_x-a_y=111222333$
This difference will remain in all the terms

Question 7
The 10th and 18th terms of an A. P. are 41 and 73 respectively. Find 26th term.

$41=a + 9d$
$73=a+ 17d$
Solving these
a=5,d=4
$a_{26} =a + 25d= 105$

Question 8
If (m + 1)th term of an A. P. is twice the (n + 1)th term. Prove that the (3m + 1)th term is twice the (m + n + 1)th term.

(m+1)th term= a + (m+1-1) d = a + m d
(n +1) th term = a+ (n+1-1) d = a + n d
now given condition is
a+ md = 2 ( a + n d ) ---- (1)

Now
(3m + 1) th term = a + (3m+1-1) d= a + 3m d -- (2)
(m+n+1) th term = a + (m+n +1 -1) d = a + (m+n) d ---(3)

Now
(3m+1)th term= a +3md
=a +md + 2md
Now we have a+md = 2 (a + nd ) from equation 1
=2(a+nd) + 2md
= 2(a + nd + md)
= 2(a +(m+n)d )
=2 (m+n+1) th ( from (3) )
Hence proved

Question 9
If the nth term of the A. P. 9, 7, 5.... is same as the nth term of the A. P. 15, 12, 9...... find n.

For 1st AP
a=9,d=-2
$a_{nx}= 9 - (n-1)2$
For 2nd AP
a=15,d=-3
$a_{ny}= 15 - (n-1)3$
Now
$9 - 2(n-1)=15 - 3(n-1)$
n=7

Question 10
Find the second term and nth term of an A. P. whose 6th term is 12 and the 8th term is 22.

$a+ 5d=12$
$a+ 7d=22$
Solving these
d=5,a=-13
$a_2= -13 + 5= -8$
$a_n= -13 + (n-1)5= 5n-18$

Question 11
The sum of 4th and 8th terms of an A. P. is 24 and the sum of 6th and 10th terms is 34. Find the first term and the common difference of the A. P.

$a + 3d + a + 7d=24$ or $2a + 10d=24$
$a+ 5d+ a + 9d=34$ or $2a + 14d =34$
Solving these
a=-1/2 ,d=5/2

Question 12
If an A. P. consists of n terms with first term a and nth term l show that the sum of the nth term from the beginning and the mth term from the end is (a + l).
Question 13
If the ath term of an A. P. be 1/b and bth term be 1/a then show that its (ab)th term is 1.
Question 14
If the pth term of an A. P. is q and the qth term is p. Prove that its nth term is (p + q – n)
Question 15
If m times the mth term of an A. P. is equal to n times its nth term. Show that the (m + n)th term of the A. P. is zero.

According to the question
$m(a_m) = n(a_n)$
Now for a,d as first term and common difference of the AP,nth term is defined as ,
$a_n= a + (n-1)d$
So
$m[a + (m-1)d] = n[a + (n-1)d]$
$[ma + (m^2 - m)d]= [na + (n^2 - n)d]$
$[ma + (m^2d- md)]= [na + (n^2d- nd)]$
$[ma-na] + [m^2d- n^2d] + [nd-md] = 0$
$a[m-n] + d[m^2 - n^2] + d[n-m] = 0$
Now we know that $x^2 - y^2 = (x+y) (x-y)$
$a[m-n] + d[(m+n) (m-n)] - d[m - n] = 0$
Divide the above equation with (m-n) ,We get :
$a + d(m+n) - d = 0$
$a + [ (m+n) - 1 ] d = 0$
So $a_{m+n} = 0$

Question 16
Justify whether it is true to say that the following are the nth terms of an AP.
(i) $2n -3$
(ii) $3n^2+5$
(iii) $1+n+n^2$

We can solve these question in three ways a. Find the Ist,2nd ,3rd and 4th term and check for the common difference. If same then in AP
b. Find d as $d=a_n - a_{n-1}$. Now if d does not depend on n, then it is in AP
c. We know that in an AP $a_n=a + (n-1)d$. We observe that $a_n$ is a linear polynomial in n. So if the expression is not linear,it is not an AP

i. $a_n=2n -3$
$a_{n-1} =2(n-1) -3=2n-5$
$d=a_n - a_{n-1}= 2n -3 - 2n + 5= 2$
So this is an nth term of A.P
ii. $a_n =3n^2+5$
This is not an linear polynomial,so it is not an nth term of A.P
iii. $a_n=1+n+n^2$
This is not an linear polynomial,so it is not an nth term of A.P

Question 17
Find the sum of 3 digit numbers which are not divisible by 7?

Sum of all three digit numbers which are not divisible by 7
= Sum of all three digit numbers - Sum of all three digit numbers which are divisible by 7

Now,lets find each of these separately
Sum of all three digit numbers
= 100 + 101 + 102 +...... + 999
Now this series is a AP with first term as 100, common difference =1 and n=900
$S_1 = \frac {n}{2} [a_1 + a_n] = \frac {900}{2} [100 + 999]= 494500$

Sum of all three digit numbers which are divisible by 7
= 105 + 112 + ..... + 994

Now this series is a AP with first term as 105, common difference=7, last term =994, n=?

Now from nth term formula
$a_n= a_1 + (n-1) \times d$
$994 = 105 + (n - 1) \times 7$
$7(n - 1) = 889$
$n = 128$

So, sum of all three digit numbers which are divisible by 7
$S_2 = \frac {n}{2} [a_1 + a_n] = \frac {128}{2} [105 + 994]= 70336$

Therefore,
Sum of all three digit numbers which are not divisible by 7 = $S_1 - S_2 = 494500 - 70336 = 424164$

Question 18
In an AP, if $S_n = 3n^2+ 5n$ and ak = 164, find the value of k.

Given: $S_n = 3n^2+ 5n$
$S_1 = 3(1)^2 + 5(1) = 3 + 5=8$
Therefore
$a_1 = 8$
$S_2 = 3(2)^2 + 5(2) = 12 + 10 = 22$
Therefore
$a_1+ a_2 =22$
$8 + a_2 = 22$
$a_2 = 14$
So, the AP is
8 ,22......
So, common difference=14
Now from nth term formula
$a_n= a_1 + (n-1) \times d$
$164 = 8 + (k-1)14$
$k = 27$

Question 19
If $S_n$ denotes the sum of first n terms of an AP, prove that
$S_{12}= 3(S_8- S_4)$

Let a is the first term of A.P and d is the common difference
$S_n=\frac {n}{2} {2a+(n-1)d}$
$S_{12}=\frac {12}{2} {2a+(12-1) d}=12a+66d$
$_S8=\frac {8}{2} {2a+7d}=8a+28d$
$S_4=\frac {4}{2} {2a+3d}=4a+6d$

Now we have to prove
$S_{12}= 3(S_8- S_4)$

Taking LHS

$LHS=S_{12}=12a+66d$

Taking RHS
$RHS=3 (S_8-S_4)=3 (8a+28d-4a-6d)=12a+66d= LHS$

Hence proved

Question 20
If the pth, qth & rth term of an AP is x, y and z respectively,
show that $x(q-r) + y(r-p) + z(p-q) = 0$

Let a is the first term of A.P and d is the common difference
$a_p= a+ (p-1) d=x$
$a_q = a + (q-1)d=y$
$a_r = a + (r-1)d=z$

Substituting these values of x,y and z in below equation

$x(q-r) + y(r-p) + z(p-q)$
$={a+ (p-1) d}(q-r) + {a+ (q-1) d} (r-p) + {a+ (r-1) d} (p-q)$=a {(q-r) + (r-p) + (p-q)} + d {(p-1)(q-r)+ (r-1) (r-p) + (r-1) (p-q)}
$= a.0 + d{p(q-r) + q(r-p) + r (p-q)- (q-r) - (r-p)-(p-q)}$= a.0 + d.0 = 0$Hence proved Question 21 Find the 20th term from the end of the AP 3, 8, 13......253 Question 22 The sum of the first n terms of an A.P whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is - 30 and the common difference is 8. Find n. Answer$S_n=\frac {n}{2}(2a + (n-1)d)$Given, a=8 and d=20$S_n=\frac {n}{2}(2(8) + (n-1)20)=n(10n-2)$When a=-30 and d=8$S_{2n}=\frac {2n}{2}(2(-30) + (2n-1)8)=n(-68 + 16n)$According to question$S_n= S_{2n}n(10n-2)=n(-68 + 16n)$n=11 Question 23 Determine k so that$k_2+ 4k + 8$,$2k_2 + 3k + 6$,$3k_2 + 4k + 4$are three consecutive terms of an AP Answer Given$t_1 = k^2 + 4k + 8t_2 = 2k^2 + 3k + 6t_3 = 3k^2 + 4k + 4$Now since they are in A.P$ t_2 - t_1 = t_3 - t_2 2k^2 + 3k + 6 - (k^2 + 4k + 8) = (3k^2 + 4k + 4) - (2k^2 + 3k + 6) 2k^2 + 3k + 6 - k^2 - 4k - 8 = 3k^2 + 4k + 4 - 2k^2 - 3k - 6 k = 0$## Summary This Class 10 Maths Important Questions for Arithmetic Progression with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail. Also Read Go back to Class 10 Main Page using below links ### Practice Question Question 1 What is$1 - \sqrt {3}\$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20