Question

the first terms of two aps are equal and the ratio of their common difference is 1:2. if the 7th term of the first ap and 21st term of 2nd ap are 23 and 125 respectively. find two aps​

Answer 1

Answer:

FIRST A.P = 5,8,11......

SECOND A.P = 5,11,17.....

Step-by-step explanation:

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...........

GOOD DAY BRO ;)

Answer 2

The required first AP is 5,8,11,... and the required second AP is 5,11,17,...

Given :

The first terms of the two APs are equal. The ratio of their common differences is 1:2. The 7th term of the first AP is 23. The 21st term of the second AP is 125.

To Find :

the two APs.

Solution :

We can find the solution to this problem in the following way.

We know that the arithmetic progression which has 'a' as its first term and 'd' as the common difference; has its nth term as follows.

$t_{n} =a+(n-1)d$

We can indicate the terms of the first and the second APs using the suffix 1 and 2 respectively.

Now we can write the following equation for the nth terms in the first and the second APs.

$t_{n1} =a_1+(n_1-1)d_1\\t_{n2} =a_2+(n_2-1)d_2$

Thus we can write the following equation for the first terms in the first and the second APs.

$a_{1} =a_{2} = a(say)$

We can rewrite the equations for the nth terms in the first and the second APs as follows.

$t_{n1} =a+(n_1-1)d_1\\t_{7} =a+(7-1)d_1\\23=a+6d_1\\and\\t_{n2} =a+(n_2-1)d_2\\t_{21} =a+(21-1)d_2\\125 =a+20d_2$

We can now eliminate 'a' from the above two equations and get the following.

$a+20d_2-a-6d_1=125-23\\20d_2-6d_1=102\\10d_2-3d_1=51$

We also have the following ratio.

$\frac{d_1}{d_2} =\frac{1}{2} \\\frac{d_1}{1} =\frac{d_2}{2}=d(say)\\ d_1=d, d_2=2d$

We can now write the following equation.

$10\times 2d-3\times d=51\\20d-3d=51\\17d=51\\d=3$.

we can find the common differences and the first term of the APs in the following way

$d_1=d=3, \\d_2=2d=6\\23=a+6d_1\\23=a+6\times3\\23=a+18\\a=5$

So, we can have the following equation for the nth terms in the first and the second APs.

$t_{n1} =5+(n_1-1)3\\and\\t_{n2} =5+(n_2-1)6$

Thus we have the first AP as 5,8,11,... and the second AP as 5,11,17,... which is the required answer.

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