Important Questions for CBSE Class 10 Maths Chapter 5 - Arithmetic Progressions 2 Mark Question -CoolGyan

CBSE Class 10 Maths Chapter-5 Arithmetic Progressions – Free PDF Download

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CBSE Class 10 Maths Chapter-5 Arithmetic Progressions Important Questions

CBSE Class 10 Maths Important Questions Chapter 5 – Arithmetic Progressions

2 Mark Questions

1. Find the missing variable from a, d, n and a_n, where a is the first term, d is the common difference and a_nis the nth term of AP.
(i) a= 7, d = 3, n = 8
(ii) a = –18, n= 10, a_n=0
(iii) d = –3, n = 18, a_n=−5
(iv) a = –18.9, d = 2.5, a_n=3.6
(v) a = 3.5, d = 0, n = 105
Ans. (i) a= 7, d = 3, n = 8
We need to find a_n here.
Using formula a_n=a+(n−1)d
Putting values of a, d and n,
a_n=7+(8−1)3=7+(7)3=7+21=28
(ii) a = –18, n= 10, a_n=0
We need to find d here.
Using formula a_n=a+(n−1)d
Putting values of a, a_n and n,
0 = –18 + (10 – 1)d
⇒ 0=−18+9d
⇒ 18=9d
⇒ d=2
(iii) d = –3, n = 18, a_n=−5
We need to find a here.
Using formula a_n=a+(n−1)d
Putting values of d, a_n and n,
–5 = a + (18 – 1)(–3)
⇒ −5=a+(17)(−3)
⇒ −5=a–51
⇒ a=46
(iv) a = –18.9, d = 2.5, a_n=3.6
We need to find n here.
Using formula a_n=a+(n−1)d
Putting values of d, a_n and a,
3.6 = –18.9 + (n – 1) (2.5)
⇒ 3.6=−18.9+2.5n−2.5
⇒ 2.5n=25
⇒ n=10
(v) a = 3.5, d = 0, n = 105
We need to find a_n here.
Using formula a_n=a+(n−1)d
Putting values of d, n and a,
a_n=3.5+(105−1)(0)
⇒ a_n=3.5 + 104 × 0
⇒ a_n=3.5 + 0
⇒ a_n=3.5

2. Choose the correct choice in the following and justify:
(i) 30^thterm of the AP: 10,7,4… is
(A) 97
(B) 77
(C) –77
(D) –87
(ii) 11^thterm of the AP: −3,−12,2…is
(A) 28
(B) 22
(C) –38
(D)
Ans. (i) 10,7,4…
First term = a =10, Common difference = d = 7 – 10= 4 – 7= –3
And n = 30 {Because, we need to find 30^thterm}
a_n=a+(n−1)d
⇒ a₃₀=10+(30−1)(−3)=10 – 87=−77
Therefore, the answer is (C).
(ii) −3,−½,2…
First term = a = –3, Common difference = d = −½ −(−3)=2−(−½)=
And n = 11 (Because, we need to find 11^thterm)
a_n=−3+(11 – 1)=−3+25=22

3. Which term of the AP: 3, 8, 13, 18 … is 78?
Ans. First term = a=3, Common difference = d = 8 – 3=13 – 8=5 and a_n=78
Using formula a_n=a+(n−1)d,to find n^th term of arithmetic progression,
a_n=3+(n−1)5,
⇒ 78=3+(n−1)5
⇒ 75=5n−5
⇒ 80=5n
⇒ n=16
It means 16^thterm of the given AP is equal to 78.

4. Find the number of terms in each of the following APs:
(i) 7,13,19….,205
(ii) 18,,13…, −47
Ans. (i) 7,13,19…, 205
First term = a =7, Common difference = d = 13 – 7= 19 – 13= 6
And a_n=205
Using formula a_n=a+(n−1)d, to find nth term of arithmetic progression,
205=7+(n−1)6=7+6n–6
⇒ 205=6n+1
⇒ 204=6n
⇒ n=34
Therefore, there are 34 terms in the given arithmetic progression.
(ii) 18,,13…, −47
First term = a =18, Common difference = d =
And a_n=−47
Using formula a_n=a+(n−1)d, to find nth term of arithmetic progression,
−47=18+(n−1)=36−n+
⇒ −94=36−5n+5
⇒ 5n=135
⇒ n=27
Therefore, there are 27 terms in the given arithmetic progression

5. Check whether –150 is a term of the AP: 11,8,5,2…
Ans. Let −150 is the n^th of AP 11,8,5,2… which means that a_n=−150
Here, First term = a = 11, Common difference = d = 8 – 11= –3
Using formula a_n=a+(n−1)d, to find n^th term of arithmetic progression,
−150=11+(n−1)(−3)
⇒ −150=11−3n+3
⇒ 3n=164
⇒ n=
But, n cannot be in fraction.
Therefore, our supposition is wrong. −150 cannot be term in AP.

6. An AP consists of 50 terms of which 3^rdterm is 12 and the last term is 106. Find the 29^thterm.
Ans. An AP consists of 50 terms and the 50^thterm is equal to 106 and a₃=12
Using formula a_n=a+(n−1)d, to find n^th term of arithmetic progression,
a₅₀=a+(50−1)d And a₃=a+(3−1)d
⇒ 106=a+49d And 12=a+2d
These are equations consisting of two variables.
Using equation 106=a+49d, we get a=106−49d
Putting value of a in the equation 12=a+2d,
12=106−49d+2d
⇒ 47d=94
⇒ d=2
Putting value of d in the equation, a=106−49d,
a=106 – 49(2)=106 – 98=8
Therefore, First term =a=8 and Common difference =d=2
To find 29^thterm, we use formula a_n=a+(n−1)d which is used to find n^th term of arithmetic progression,
a₂₉=8+(29−1)2=8+56=64
Therefore, 29th term of AP is equal to 64

7. How many multiples of 4 lie between 10 and 250?
Ans. The odd numbers between 0 and 50 are 1,3,5,7…49
It is an arithmetic progression because the difference between consecutive terms is constant.
First term = a = 1, Common difference = 3 – 1= 2, Last term = l=49
We do not know how many odd numbers are present between 0 and 50.
Therefore, we need to find n first.
Using formula a_n=a+(n−1)d, to find nth term of arithmetic progression, we get
49=1+(n−1)2
⇒ 49=1+2n−2
⇒ 50=2n
⇒ n=25
Applying formula, to find sum of n terms of AP, we get

8. Which term of the AP: 121, 117, 113, …..is its first negative term?
Ans. Given: 121, 117, 113, …….
Here
Now,
= = =
For the first negative term,

is an integer and
Hence, the first negative term is 32^nd term

9. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of sixteen terms of the AP.
Ans. Let the AP be
Then,

  ……….(i)
Also

Taking

=
=  =  = 76
Taking ,

=
=  =  = 20
S₁₆ = 20 and 76

10. A ladder has rungs 25 cm apart (see figure). The rungs decrease uniformly in length from 45 cm, at the bottom to 25 cm at the top. If the top and the bottom rungs are  m apart, what is the length of the wood required for the rungs?

Ans. Number of rungs  = 10
The length of the wood required for rungs = sum of 10 rungs
=  = 5 x 70 = 350 cm

11. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of  such that the sum of the numbers of the houses preceding the house numbered  is equal to the sum of the numbers of the houses following it. Find this value of

Ans. Here  and

=
=  =

=  =

=  = 49 x 25
According to question,

  = 49 x 25 –
  + = 49 x 25

  = 49 x 25

Since,  is a counting number, so negative value will be neglected.