**Notes**
**NCERT Solutions**
**Assignments**
Given below are the

**Class 10 Maths** Important Questions(Short questions) for Arithmetic Progression

a) Concepts questions

b) Calculation problems

c) Multiple choice questions

d) Long answer questions

e) Fill in the blank's

**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
a_{n}=6n + 11

a_{k} = 6k + 11

a_{k+1} = 6(k+1) + 11

Now

a_{k+1} - a_{k}=6(k+1) + 11 -[6k + 11]=6

**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 31

^{st} 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 54

^{th} term?

**Question 6**Two A. P.’s has the same common difference. The difference between their 100

^{th} terms is 111 222 333. What is the difference between their Millionth terms?

**Question 7**The 10

^{th} and 18

^{th} terms of an A. P. are 41 and 73 respectively. Find 26

^{th} term.

**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 n

^{th} term of the A. P. 9, 7, 5… is same as the n

^{th} term of the A. P. 15, 12, 9…. find n.

**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.

**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.

**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^{2}d- md)]= [na + (n^{2}d- nd)]

[ma-na] + [m^{2}d- n^{2}d] + [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}
Other Answers
4) n =44

5) 65^{th} term is 132 more than its 54^{th} term

6) The difference between millionth terms is same as the difference between 100^{th} term i.e, 11122233

7) 105

9. 7

10. a_{2} = -8, a_{n} = 5n – 18

11. -1/2, 5/2

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