RD Chapter 19 Arithmetic Progressions Ex 19.6 Solutions
Question - 1 : - Find the A.M. between:
(i) 7 and 13 (ii) 12 and – 8 (iii) (x – y) and (x + y)
Answer - 1 : -
(i) Let A be theArithmetic mean
Then 7, A, 13 are inAP
Now, let us solve
A-7 = 13-A
2A = 13 + 7
A = 10
(ii) Let A be theArithmetic mean
Then 12, A, – 8 are inAP
Now, let us solve
A – 12 = – 8 – A
2A = 12 + 8
A = 2
(iii) Let A be theArithmetic mean
Then x – y, A, x + yare in AP
Now, let us solve
A – (x – y) = (x + y)– A
2A = x + y + x – y
A = x
Question - 2 : - Insert 4 A.M.s between 4 and 19.
Answer - 2 : -
Let A1, A2,A3, A4 be the 4 AM Between 4 and 19
Then, 4, A1,A2, A3, A4, 19 are in AP.
By using the formula,
d = (b-a) / (n+1)
= (19 – 4) / (4 + 1)
= 15/5
= 3
So,
A1 = a+ d = 4 + 3 = 7
A2 =A1 + d = 7 + 3 = 10
A3 =A2 + d = 10 + 3 = 13
A4 =A3 + d = 13 + 3 = 16
Question - 3 : - Insert 7 A.M.s between 2 and 17.
Answer - 3 : -
Let A1, A2,A3, A4, A5, A6, A7 bethe 7 AMs between 2 and 17
Then, 2, A1,A2, A3, A4, A5, A6, A7,17 are in AP
By using the formula,
an = a+ (n – 1)d
an =17, a = 2, n = 9
so,
17 = 2 + (9 – 1)d
17 = 2 + 9d – d
17 = 2 + 8d
8d = 17 – 2
8d = 15
d = 15/8
So,
A1 = a+ d = 2 + 15/8 = 31/8
A2 = A1 +d = 31/8 + 15/8 = 46/8
A3 = A2 +d = 46/8 + 15/8 = 61/8
A4 = A3 +d = 61/8 + 15/8 = 76/8
A5 = A4 +d = 76/8 + 15/8 = 91/8
A6 = A5 +d = 91/8 + 15/8 = 106/8
A7 = A6 +d = 106/8 + 15/8 = 121/8
∴ the 7 AMs between 2and 7 are 31/8, 46/8, 61/8, 76/8, 91/8, 106/8, 121/8
Question - 4 : - Insert six A.M.s between 15 and – 13.
Answer - 4 : -
Let A1, A2,A3, A4, A5, A6 be the 7 AMbetween 15 and – 13
Then, 15, A1,A2, A3, A4, A5, A6, – 13are in AP
By using the formula,
an = a+ (n – 1)d
an =-13, a = 15, n = 8
so,
-13 = 15 + (8 – 1)d
-13 = 15 + 7d
7d = -13 – 15
7d = -28
d = -4
So,
A1 = a+ d = 15 – 4 = 11
A2 =A1 + d = 11 – 4 = 7
A3 =A2 + d = 7 – 4 = 3
A4 =A3 + d = 3 – 4 = -1
A5 =A4 + d = -1 – 4 = -5
A6 =A5 + d = -5 – 4 = -9
Question - 5 : - There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3: 1. Find the value of n.
Answer - 5 : -
Let the series be 3, A1,A2, A3, …….., An, 17
Given, an/a1 =3/1
We know total terms inAP are n + 2
So, 17 is the (n +2)th term
By using the formula,
An = a+ (n – 1)d
An =17, a = 3
So, 17 = 3 + (n + 2 –1)d
17 = 3 + (n + 1)d
17 – 3 = (n + 1)d
14 = (n + 1)d
d = 14/(n+1)
Now,
An = 3+ 14/(n+1) = (17n + 3) / (n+1)
A1 = 3+ d = (3n+17)/(n+1)
Since,
an/a1 =3/1
(17n + 3)/ (3n+17) =3/1
17n + 3 = 3(3n + 17)
17n + 3 = 9n + 51
17n – 9n = 51 – 3
8n = 48
n = 48/8
= 6
∴ There are 6 terms inthe AP
Question - 6 : - Insert A.M.s between 7 and 71 in such a way that the 5th A.M.is 27. Find the number of A.M.s.
Answer - 6 : -
Let the series be 7, A1,A2, A3, …….., An, 71
We know total terms inAP are n + 2
So 71 is the (n + 2)thterm
By using the formula,
An = a+ (n – 1)d
An =71, n = 6
A6 = a+ (6 – 1)d
a + 5d = 27 (5th term)
d = 4
so,
71 = (n + 2)th term
71 = a + (n + 2 – 1)d
71 = 7 + n(4)
n = 15
∴ There are 15 terms inAP
Question - 7 : - If n A.M.s are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.
Answer - 7 : -
Let a and b be thefirst and last terms
The series be a, A1,A2, A3, …….., An, b
We know, Mean =(a+b)/2
Mean of A1 andAn = (A1 + An)/2
A1 =a+d
An = a– d
So, AM = (a+d+b-d)/2
= (a+b)/2
AM between A2 andAn-1 = (a+2d+b-2d)/2
= (a+b)/2
Similarly, (a + b)/2is constant for all such numbers
Hence, AM = (a + b)/2
Question - 8 : - If x, y, z are in A.P. and A1is the A.M. of x and y, and A2 isthe A.M. of y and z, then prove that the A.M. of A1 and A2 isy.
Answer - 8 : -
Given that,
A1 =AM of x and y
And A2 =AM of y and z
So, A1 =(x+y)/2
A2 =(y+x)/2
AM of A1 andA2 = (A1 + A2)/2
= [(x+y)/2 +(y+z)/2]/2
= [x+y+y+z]/2
= [x+2y+z]/2
Since x, y, z are inAP, y = (x+z)/2
AM = [(x + z/2) +(2y/2)]/2
= (y + y)/2
= 2y/2
= y
Hence proved.
Question - 9 : - Insert five numbers between 8 and 26 such that the resulting sequence is an A.P
Answer - 9 : -
Let A1, A2,A3, A4, A5 be the 5 numbers between 8 and26
Then, 8, A1,A2, A3, A4, A5, 26 are in AP
By using the formula,
An = a+ (n – 1)d
An =26, a = 8, n = 7
26 = 8 + (7 – 1)d
26 = 8 + 6d
6d = 26 – 8
6d = 18
d = 18/6
= 3
So,
A1 = a+ d = 8 + 3 = 11
A2 = A1 +d = 11 + 3 = 14
A3 = A2 +d = 14 + 3 = 17
A4 = A3 +d = 17 + 3 = 20
A5 = A4 +d = 20 + 3 = 23