Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev

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Class 10 : Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev

The document Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
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Hots questions
Q1. Find the ‘6th’ term of the A.P. : Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Sol. Here,
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev

∴ d = a2 – a1
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Now, an = a + (n – 1)d
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Thus, the nth term is Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Again, we have
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
i.e., the 6th term is Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Q2. 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 mth terms.
Sol. Let the first terms of given AP's be a1 and a2, common differences be d1 and d2 and   let Sn and  the sum of n terms.
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Now, replacing n by (2m-1) for getting ratio of mth terms of given APs,
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Thus, the required ratio of mth term of given AP's is (14m – 6) : (8m + 23)

Q3. If the numbers a, b, c, d and e form an A.P., then find the value of a – 4b + 6c – 4d + e
Sol. We have the first term of A.P. as ‘a’.
Let D be the common difference of the given A.P.,
Then :
b = a + D,  c = a + 2D, d= a + 3D and e = a + 4D
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
∴ a – 4b + 6c – 4d + e
=   a – 4(a + D ) + 6 (a + 2D) – 4 (a + 3D) + (a + 4D)
= a – 4a + 6a – 4a + a – 4D + 12D – 12D + 4D
= 8a – 8a + 16D – 16D = 0
Thus, a – 4b + 6c – 4d + e = 0

Q4. Hots Questions - Arithmetic Progressions Class 10 Notes | EduRevis the arithmetic mean between ‘a’ and ‘b’, then, find the value of ‘n’.
Sol.  Note : A.M., between ‘a’ and ‘b ’= 1/2
(a + b)
We know that :
A.M. between ‘a’ and ‘b’ = a+b/2
It is given that,
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev  is the A.M. between ‘a’ and ‘b’
∴  Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
By cross multiplication, we get :
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
⇒2an + 1 + 2bn + 1 = an + 1 + abn + anb + bn + 1
⇒ 2an + 1 – an + 1 + 2bn + 1– bn + 1= abn + anb
⇒ an + 1 + bn + 1 = abn +anb
⇒ an+1 – anb = abn – bn+1
⇒ an[a–b] = bn[a –b]
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev

⇒ n =  0

Q5. Solve the equation :
1 + 4 + 7 + 10 + ... + x = 287 
Sol.
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
Since,
∴ a = 1,d = 3 and an = x
∴ an = a + (n – 1)d
⇒ x = 1 + (n – 1) 3   or   x = 3n – 2
Also, Sn = n/2 (a+l)
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
⇒ 2(287) = n[1 + (3n – 2)]
⇒ 574 = n[3n – 1]
⇒ 3n2 – n – 574 = 0
Solving the above quadratic equation, we get
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
or
Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev
But, negative n is not desirable.
∴ n = 14
x = 3n – 2
Now, x = 3(14) – 2 = 42 – 2 = 40
Thus, x = 40

Q6. Find three numbers in A.P. whose sum is 21 and their product is 231.
Sol. Let the three numbers in A.P. are:
a – d,    a,   a + d
∴ (a – d) + a + (a + d) = 21
⇒ a – d + a + a + d = 21
or 3a = 21 ⇒   a = 7
Also, (a – d) × a × (a + d) = 231
∴ (7 – d) × 7 × (7 + d) = 231
⇒ (7 – d) (7 + d) × 7 = 231
⇒ 72 – d2 = 231 /7 = 33
⇒ 49 – d2 = 33
or d2 = 49 – 33 = 16
⇒ d = ± 4
Now, when d = 4, then three numbers in AP are : (7 – 4), 7,  (7 + 4) i.e. 3, 7, 11.
When d = –4, then three numbers in AP are : [7 – (–4)],  7,  [7 + (–4)]
or  11, 7, 3

Q7. The ninth term of an A.P. is equal to seven times the second term and twelfth term exceeds five times the third term by 2. Find the first term and common difference.
Sol. Let 'a' be the first term and 'd' be the common difference
Since, an = a + (n–1)d  
∴ a9 = a + 8d  and a12 = a + 11d
Also a2 = a + d   and a3 = a + 2d
since a9 = 7a2   and a12 = 5a3 + 2
or a9 = 7 (a + d)  ⇒ a9  = 7a + 7d     ... (1)
a12 = 5 (a + 2d) + 2
⇒ a12 = 5a + 10d + 2           ... (2)
Now, a + 8d = 7a + 7d            [From (1)]
or – 6a +   d = 0  ... (3)
Also a + 11d = 5a + 10d + 2    [From (2)]
or –4a + d = 2         ... (4)
Subtracting (4) from (3), we have
–2a = –2   ⇒  a = 1
Now, from (3), –6 + d = 0  ⇒  d = 6
Thus, a = 1   and  d = 6

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