Given below are the Class 10 Maths Important Questions for Arithmetic Progression
a. Concepts questions
b. Calculation problems
c. Long answer questions
d. proof questions Question 1 The general term of a sequence is given by a_{n} = -4n + 15. Is the sequence an A. P.? If so, find its 15^{th} term and the common difference. Solution

$a_n = -4n + 15$
$a_k = -4k + 15$
$a_{k+1} = -4(k+1) + 15$
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 n^{th} term of an A. P. is 6n + 11. Find the common difference. Solution

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

a_{8} = 31
a +(8-1)d = 31
a + 7d = 31 -- (i)

a_{15} =16 + a_{11}
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? Solution

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
a_{n} = 130 + a_{31}
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? Solution

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? Solution

$a_n=a + (n-1)d$
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
So ,answer is 111222333 only

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

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. Solution

(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. Solution

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 n^{th }term of an A. P. whose 6^{th} term is 12 and the 8^{th} term is 22. Solution

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

$a + 3d + a + 7d=24$ 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 n^{th} term l show that the sum of the n^{th} term from the beginning and the m^{th} term from the end is (a + l). Question 13 If the a^{th }term of an A. P. be 1/b and b^{th} term be 1/a then show that its (ab)^{th} term is 1. Question 14 If the p^{th} term of an A. P. is q and the q^{th} term is p. Prove that its nth term is (p + q – n) Question 15 If m times the m^{th} 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. Solution

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) $1+n+n^2$ Solution

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

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