Question

The first term of two A.P.s 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.P.s.​

Answer 1

AnswEr :

⋆ Let the First Term of Both APs be a.

⋆ Common Difference of First AP be d and, of Second AP be 2d.

$\underline{\bigstar\:\textsf{7th Term of First AP :}}$

$\longrightarrow\tt T_n = a + (n - 1)d\\\\\\\longrightarrow\tt T_7 = a + (7 - 1)d \\\\\\\longrightarrow\tt 23 = a + 6d\qquad\dfrac{\quad}{}eq.(1)$

$\rule{150}{1}$

$\underline{\bigstar\:\textsf{21st Term of Second AP :}}$

$\longrightarrow\tt T_n = a + (n - 1)d\\\\\\\longrightarrow\tt T_{21} = a + (21 - 1)2d \\\\\\\longrightarrow\tt 125 = a + 20 \times 2d\\\\\\\longrightarrow\tt 125 = a + 40d\qquad\dfrac{\quad}{}eq.(2)$

$\rule{170}{1}$

$\underline{\bigstar\:\textsf{Subtracting eq.(1) from eq.(2) :}}$

$:\implies\tt a + 40d = 125\\\\:\implies\tt a + 6d = 23\\\dfrac{\qquad\qquad\qquad \qquad \qquad}{}\\:\implies\tt (40d - 6d) = (125 - 23)\\\\\\:\implies\tt34d = 102\\\\\\:\implies\tt d = \cancel\dfrac{102}{34}\\\\\\:\implies\blue{\tt d = 3}$

$\underline{\bigstar\:\textsf{Putting the Value of d in eq.(1) :}}$

$:\implies\tt 23 = a + 6d\\\\\\:\implies\tt 23 = a +6(3)\\\\\\:\implies\tt 23 = a + 18\\\\\\:\implies\tt 23 - 18 = a\\\\\\:\implies\blue{\tt a = 5}$

$\rule{200}{2}$

$\bullet\:\underline\textsf{First Arithmetic Progresion :} \\\\\implies \tt a, (a + d), (a +2d), (a +3d)..\\\\\implies\tt 5, (5 + 3), (5 + 2(3)),(5 + 3(3))..\\\\\implies\large\tt\red{ 5, 8,11,14..}$

$\bullet\:\underline\textsf{Second Arithmetic Progresion :} \\\\\implies \tt a, (a + 2d), (a +2(2d)), (a +3(2d))..\\\\\implies\tt 5, (5 + 2(3)), (5 + 2(6)), (5 + 3(6))..\\\\\implies\large\tt\red{ 5, 11,17, 23..}$

Answer 2

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 to 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

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