Question

The sums of n terms of 2A.P.'s are in the ratio 7n+1:4n+27. Show that ratio of their 11th term is 4:3.​

Answer 1

Answer:

Step-by-step explanation:

Given,

Ratio of  the sum of the first 'n' terms of two A.P is :-

7n + 1 : 4n + 27

To Find :-

Ratio of their 11th term's.

Solution :-

 

sum of n terms in an A.P is :-

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

Let,

First A.P be A.P_1

first term of A.P_1 is a_1

common difference of A.P_1 is d_1

$\implies S_n(A.P_1)=\dfrac{n}{2}[2a_1+(n-1)d_1]$

Second A.P be A.P_2

first term of A.P_2 is a_2

common difference of A.P_2 is d_2

$\implies S_n(A.P)_2=\dfrac{n}{2}[2a_2+(n-1)d_2]$

According to question :-

$\dfrac{S_n(A.P_1)}{S_n(A.P_2}=\dfrac{7n+1}{4n+27}$

$\implies \dfrac{\dfrac{n}{2}[2a_1+(n-1)d_1]}{\dfrac{n}{2}[2a_2+(n-1)d_2]}=\dfrac{7n+1}{4n+27}$

$\dfrac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\dfrac{7n+1}{4n+27}$

$\dfrac{2\bigg(a_1+\dfrac{n-1}{2}d_1\bigg)}{2\bigg(a_2+\dfrac{n-1}{2}d_2\bigg)}=\dfrac{7n+1}{4n+27}$

$\dfrac{a_1+\dfrac{n-1}{2}d_1}{a_2+\dfrac{n-1}{2}d_2}=\dfrac{7n+1}{4n+27}$

[let it be equation (1)]

Hence, This is applicable to any value for A.P

We, need to find the ratio of their 11th terms​ so,

a_11 = a+(11 - 1)d

= a + 10d

Since, we need to equate both "(n-1)\2 and 10"

$\implies \dfrac{n-1}{2}=10$

n - 1 = 2(10)

n - 1 = 20

n = 20 + 1

n = 21

Substituting values in eq(1):-

$\dfrac{a_1+10d_1}{a_2+10d_2}=\dfrac{7(21)+1}{4(21)+27}$

$\dfrac{t_{11}(A.P_1)}{t_{11}(A.P_2)}=\dfrac{147+1}{84+27}$

$=\dfrac{148}{111}$

$=\dfrac{4}{3}$

= 4 : 3

$\therefore \dfrac{t_{11}(A.P_1)}{t_{11}(A.P_2}=\dfrac{4}{3}$