Question

the sum of n terms of two AP are in the ratio 3n + 8/ 7n + 15. Find the ratio of 12th term​

Answer 1

Answer:

1

st

AP

Let first term be a common difference be d

sum of n terms s

n

=

2

n

(2a+(n−1)d)

n

th

term a

n

=a(n−1)d

Similarly for 2

nd

AP

Let first term =A common difference be D

S

n

=

2

n

(2A+(n−1)D) & n

th

term =A

n

=A+(n−1)D

We need ratio of 12

th

term

i.e.,

A

12

ofsecondAP

a

12

offirstAP

=

A+(12−1)D

a+(12−1)d

=

a+11D

a+11d

It is given that

Sumofntermsof2

nd

AP

Sumofntermsof1

st

AP

=

7n+15

3n+8

∴

2A+(n−1)D

2a+(n−1)d

=

7n+15

3n+8

………….(1)

2(A+(

2

n−1

)D)

2(a+(

2

n−1

)d)

=

7n+15

3n+8

………………(1)

we need to find

A+11D

a+11d

Hence

2

n−1

=11

n−1=22

n=23

Putting n=23 in (1)

A+(

2

23−1

)D

a+(

2

23−1

)d

=

7×23+15

3×23+18

∴

A+11D

a+11d

=

16

7

Hence ratio of their 12

th

term is

16

7

i.e., 7:16

Answer 2

Answer :

Ratio of 12th terms = 7 : 16

Step-by-step explanation :

First AP :

Let

first term be A

common difference be D

Sum of n terms,  

       $\bf S_n=\frac{n}{2}[2A+(n-1)D]$

Second AP :

Let

first term be a

common difference be d

Sum of n terms,

       $\bf S_n=\frac{n}{2}[2a+(n-1)d]$

___________________

Given,

The sum of n terms of two AP are in the ratio  $\bf \frac{3n+8}{7n+15}$

             $\bf \frac{Sum \ of \ n \ terms \ of \ first \ AP}{Sum \ of \ n \ terms \ of \ second \ AP} =\frac{3n+8}{7n+15} \\\\\\ \frac{\frac{n}{2}[2A+(n-1)D]}{\frac{n}{2}[2a+(n-1)d]} =\frac{3n+8}{7n+15} \\\\\\\frac{2A+(n-1)D}{2a+(n-1)d} =\frac{3n+8}{7n+15} \\\\\\ \frac{2[A+(\frac{n-1}{2})D]}{2[a+(\frac{n-1}{2})d]} =\frac{3n+8}{7n+15} \\\\\\ \frac{A+(\frac{n-1}{2})D}{a+(\frac{n-1}{2})d} =\frac{3n+8}{7n+15} ---[1]$

we have to find the ratio of their 12th terms.

 We know,

   nth term is given by,

      tₙ = first term + (n - 1) (common difference)

For first AP,

 12th term =  A + (12 - 1)D

                 =  A + 11D

For second AP,

 12th term =  a + (12 - 1)d

                  =  a + 11d

     Ratio of 12th terms,

           $\bf =\frac{12^{th} \ term \ of \ first \ AP}{12^{th} \ term \ of \ second \ AP} \\\\\\ =\frac{A+11D}{a+11d} ---[2]$

Comparing equation [2] with the LHS of equation [1],

         $\frac{n-1}{2} =11 \\\\ n-1=11 \times2 \\\\ n-1=22 \\\\ n=22+1 \\\\ \boxed{n=23}$

Put n = 23 in equation [1],

          $\bf \frac{A+(\frac{23-1}{2})D}{a+(\frac{23-1}{2})d} =\frac{3(23)+8}{7(23)+15} \\\\\\ \frac{A+(\frac{22}{2})D}{a+(\frac{22}{2})d} =\frac{69+8}{161+15} \\\\\\ \frac{A+11D}{a+11d} =\frac{77}{176} \\\\\\ \frac{A+11D}{a+11d} =\frac{11 \times 7}{11 \times 16} \\\\\\ \frac{A+11D}{a+11d} =\frac{7}{16}$

∴ Ratio of their 12th terms = 7 : 16