Question

Answer the following questions :
4 x 4 = 16
34.
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. plz answer​

Answer 1

Question :-

→ The first term of two AP's are equal and the ratio of their common difference is 1:2 ,if the 7th term of first AP and 21th term os second AP are 23 and 125 respectively .Find both AP .$\:$

Answer :-

✏First AP ( first term = 5, d = 3 )

→ 5 , 8 ,11 ,.......

✏Second AP ( first term = 5 , d = 6)

→ 5, 11 ,17 .....

To Find :-

→ Both sequences.

Solution :-

$\sf$

Given that ;

• First term of both AP s are equal.

• 7th term of first AP is 23

• 21th term of second AP is 125

Now ,

✏For first AP.

➠ First term = a ,and common difference = d

•°• 7th term = 23

→ a + 6d = 23 ......eq.(1)

✏ For second AP

➠ First term = a( Given in the question ) and let , common difference is "c"

•°• 21th term = 125

→ a + 20 c = 125 ....eq.(2)

from eq.(1)

→ 23 - 6d + 20 c = 125

→ 20 c - 6d = 125 - 23

→ 20 c - 6d = 102

or,

→ 10 c - 3d = 52 ...... eq.(3)

Now , According to the question,

→ ratio of common difference is 2:1

So ,

→ d = 1 x ( x due to ratio ; some integer )

and ,

→ c = 2x

now ,put value of "d "and "c" in eq.(3)

➠10 (2x) - 3(x) = 51

➠20 x - 3x = 51

➠ 17 x = 51

➠x = 3

hence , d = 3 and c = 2×3 = 6 ,

Now ,put value of c in eq.(2)

→ a + 20 × 6 = 125

→ a = 125 - 120

→ a = 5 ,

hence required APs are

✏First AP ( first term = 5, d = 3 )

→ 5 , 8 ,11 ,.......

✏Second AP ( first term = 5 , d = 6)

→ 5, 11 ,17 .....

Answer 2

First AP:-

First term = a

Difference = d

a7 = 23

a+6d = 23

a= 23-6d. --------------. Equation 1

Second AP:-

Given, First term = a

Difference = b

a21 = 125

a+20b = 125. --------------. Equation 2

Replace the value of a from equation 1 to equation 2

a+20b = 125

23-6d+20b = 125

20b-6d = 125-23

20b-6d =102

2(10b - 3d) = 102

10b - 3d = 51 ------------ 3

--------------------------------

d/b = 1/2

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

Let, 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,

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