Question

Ratio of sum of n terms of two A.P.’s is 9n + 1 : 2n + 29. Find ratio of their n

th terms.​

Answer 1

Answer:

I DONT UNDERSTEND THIS Q.

Step-by-step explanation:

....... READ UP SIDE........

Answer 2

Given :

• Ratio of sum of n terms of two A.P.'s is 9 n + 1 : 2 n + 29

To find :

• Ratio of their$\rm{m_{th}}$ terms

Solution :

Let, first term of two AP's be a₁ and a₂ and common differences be d₁ and d₂

then,

As given that the ratio of sum of their n terms is 9 n + 1 : 2 n + 29

so,

$\implies\rm{\dfrac{\dfrac{n}{2}(2a_1+(n-1)d_1)}{\dfrac{n}{2}(2a_2+(n-1)d_2)}=\dfrac{9n+1}{2n+29}}$

$\implies\rm{\dfrac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\dfrac{9n+1}{2n+29}}$

dividing by 2 in both numerator and denominator in LHS we will get,

$\implies\rm{\dfrac{a_1+\dfrac{(n-1)}{2}d_1}{a_2+\dfrac{(n-1)}{2}d_2}=\dfrac{9n+1}{2n+29}}$

Now, since we need to find $\rm{m_{th}}$

term

therefore,

$\implies \rm{\dfrac{n-1}{2}=m-1}$

$\implies \rm{n=2m-2+1}$

$\implies \rm{n=2m-1}$

using this, replacing values

$\implies\rm{\dfrac{a_1+(m-1)d_1}{a_2+(m-1)d_2}=\dfrac{9(2m-1)+1}{2(2m-1)+29}}$

$\implies\rm{\dfrac{a_1+(m-1)d_1}{a_2+(m-1)d_2}=\dfrac{18m-8}{4m+27}}$

therefore,

Ratio of the $\rm{{m_{th}} }$ terms of the given AP will be 18 m - 8 : 4 m + 27.