Question

please guys need your help ​

Answer 1

Answer:

148/111

Step-by-step explanation:

ratio of the sum of the terms=7n+1/4n+27

n/2(2A+(n-1)D)=7n+1

-------------------    ------

n/2(2a+(n-1)d)    4n+27

2A+(n-1)D=7n+1

--------------  -------

2a+(n-1)d   4n+27

A+(n-1)D/2=7n+1

--------------   ------

a+(n-1)d/2   4n+27

now  we require the LHS to be

A+10D

--------

a+10d

so ,(n-1)D/2=10D

or, n-1=20

or,n=21

so if we substite the value n=21,we will get the ratio of 11th term

ratio of the 11th terms=7*21+1)/(4*21+27)

=148/111

aliter;

to get the ratio of the mth term from the ratio of the given nth terms

use the formula

n=2m-1

like in this case n=2*11-1=21

for futher queries,(ICSE or cbse questions) do inbox me

Answer 2

SOLUTION:-

There are 2 A.P. with different first term & common difference for the first A.P.

For the first A.P.

⚫Let first term be a

⚫common difference be d

⚫sum of nth term

$= > {}^{S} n = \frac{n}{2} [2a + (n - 1)d]$

&

nth term=an =a+(n-1)d

For the second A.P.

⚫Let first term=A

⚫common difference= D

So,

⚫Sn=n/2[2A+(n-1)D]

⚫nth term= An= A+(n-1)D

We need to find ratio of 11th term:

$= > \frac{ {}^{a} 11 \: of \: first \: A.P.}{ {}^{a}11 \: of \: second \: A.P } \\ \\ = > \frac{a + (11 - 1)d}{A + (11 - 1)D} \\ \\ = > \frac{a + 10d}{A+ 10D}$

Given:

$\frac{Sum \: of \: n \: terms \: of \: 1st \: A.P}{Sum \: of \: n \: terms \: of \: 2nd \: A.P.} = \frac{7n + 1}{4n + 27}$

$= > \frac{ \frac{n}{2} [2a + (n - 1)d]}{ \frac{n}{2}[2A + (n - 1)D]} = \frac{7n + 1}{4n + 27} \\ \\ = > \frac{[2a + (n - 1)d]}{[2A + (n - 1)D]} = \frac{7n + 1}{4n + 27 } \\ \\ = > \frac{2[a + ( \frac{n - 1}{2})d] }{2[A + ( \frac{n - 1}{2} )D]} = \frac{7n + 1}{4n + 27} \\ \\ = > \frac{[a + ( \frac{n - 1}{2} )d]}{[a + ( \frac{n - 1}{2})D] } = \frac{7n + 1}{4n + 27} ...............(1)$

We need to find:

$\frac{a + 10d}{A + 10D}$

Hence,

$\frac{n - 1}{2} = 10 \\ \\ = > n - 1 = 20 \\ \\ = > n = 20 + 1 \\ \\ = > n = 21$

Putting the value of n in equation (1), we get;

$\frac{a + ( \frac{21 - 1}{2} )d}{A + ( \frac{21 - 1}{2})D} = \frac{7 \times 21 + 1}{4 \times 21 + 27} \\ \\ = > \frac{a + ( \frac{20}{2})d }{A + (\frac{20}{2} )D} = \frac{147 + 1}{84 + 27} \\ \\ = > \frac{a + 10d}{A+ 10D} = \frac{148}{111}$

Hence,

The ratio of 11th term is 148:111.

Hope it helps ☺️