Question

The first term of two A.Ps are equal and the ratios of their common differences is 1:2 . If the 7th term of first A.P and 21th term of second A.P are 23 and 125 respectively. Find two A.Ps.​

Answer 1

Answer

Required APs are

5,8,11,14,.........

5,11,17,23,..........

Step by step explanation

Let us consider two APs whose first terms are a and a' and again common differences are d and d' respectively .

Now given that ,

ratio of d and d' is 1:2

d/d' = 1/2

➡d' = 2d

7th term of 1st AP = 23

➡ a + (7- 1)d =23

➡a + 6d = 23

➡a = 23- 6d

And ,

21st term of 2nd AP = 125

➡a'+ (21-1)d = 125

➡a' + 20d' = 125

➡a' = 125 - 20d'

Since the first term of both the APs are equal so ,

a = a'

➡23 - 6d = 125 - 20d'

➡ 23 - 6d = 125 - 20 (2d)

➡ 40 - 6d = 125 - 23

➡ 34d = 102

➡ d = 3

So the common difference of 1st AP is 3

and

d' = 2d

➡d' = 6

The common difference of 2nd AP is 6

Now the first of both the APs is

a =23 - 6×3

a = 23 - 18

a = 5

Now the APs are

1st AP =5 , 8 , 11 , 14,...............

and 2nd AP = 5 ,11, 17 , 23 ,............

Answer 2

$\huge\mathfrak\blue{Answer:-}$

FOR THE FIRST A.P

First term = a

Common difference = d

a7 = 23

a + 6d = 23

a = 23 - 6d ___________1

FOR THE SECOND A.P

First term = a ( given)

common difference = b

a21 = 125

a + 20b = 125 _________ 2

Replacing the value of a from equation 1 in  equation 2.

we get,

a + 20b = 125

23 - 6d + 20b = 125

20b - 6d = 125 - 23

20b - 6d = 102

2(10b - 3d) = 102

10b - 3d = 51 _________ 3

Given,

d/b = 1/2

let,

d= 1x { For x is some common positive integer}

b = 2x { for x is some common positive integer}

Replacing the value of d and b in equation 3

we get,

10b - 3d = 51

10(2x) - 3(x) = 51

20x - 3x = 51

17x = 51

x = 3

Therefore,

d = 3

b = 6

Replacing the value of d in equation 1

we get,

a + 6d = 23

a = 23 - 6(3)

a = 23 - 18

a = 5

THE FIRST A.P

5,8,11......

THE SECOND A.P

THE SECOND A.P5,11,17.......