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

$\Large{\boxed{\underline{\overline{\mathfrak{\star \: QuEsTiOn :- \: \star}}}}}$

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.

$\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}$

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

$\Large{\underline{\underline{\bf{GiVen:-}}}}$

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

$\Large{\underline{\underline{\bf{Find:-}}}}$

What's the Ratio of their $\sf{m_{th}}$ terms ?

$\Large{\underline{\underline{\bf{SoLuTion:-}}}}$

Here we go !

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

then,

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

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

now we got after dividing 2

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

Hence we now find $\sf{m_{th}}$ term

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

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

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

using the , replacing values

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

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

So,

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

Additional Information :-

Sum of n terms in AP ↠ n/2[2a + (n – 1)d]

Sum of natural numbers ↠ n(n+1)/2

Sum of square of ‘n’ natural numbers ↠ [n(n+1)(2n+1)]/6

Sum of Cube of ‘n’ natural numbers. ↠ [n(n+1)/2]²