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# Hots Questions - Arithmetic Progressions Class 10 Notes | EduRev

## 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. : Sol. Here,  ∴ d = a2 – a1 Now, an = a + (n – 1)d  Thus, the nth term is Again, we have   i.e., the 6th term is 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.     Now, replacing n by (2m-1) for getting ratio of mth terms of given APs, 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 ∴ 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. is 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, is the A.M. between ‘a’ and ‘b’
∴ By cross multiplication, we get : ⇒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]  ⇒ n =  0

Q5. Solve the equation :
1 + 4 + 7 + 10 + ... + x = 287
Sol. 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) ⇒ 2(287) = n[1 + (3n – 2)]
⇒ 574 = n[3n – 1]
⇒ 3n2 – n – 574 = 0
Solving the above quadratic equation, we get or 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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