Question

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

Answer 1

Correct question :

Ratio of sum of n terms of two AP's is 9n + 1 : 2n + 29. Find the ratio between their $\sf{\bold{m_{th}}}$ terms.

Given :

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

To find :

• Ratio of their $\sf{m_{th}}$ 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,

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

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

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

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

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

therefore,

$\mapsto \sf{\dfrac{n-1}{2}=m-1}$

$\mapsto \sf{n=2m-2+1}$

$\mapsto \red{\sf{n=2m-1}}$

using this, replacing values

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

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

therefore,

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

Answer 2

Answer

Given

Ratio of sum of n terms of two

A.P's is 9 n+1:2n+29

To find:

Ratio of their mth terms

Solution :

Let, first term of two AP's be a,

and a and common differencees

be d, and dz

then,

As given that the ratio of sum of

their n terms is 9 n+1:2n+29

SO,

(2a1+(n- 1)dı)

9n+1

5(2a2 +(n -1)da)

2n+29

2a1 +(n-1]d