Question

If the ratio of the sum of first $n$ terms of two A.P's is
$(7n+1)\colon(4n+26)$, find the ratio of their ${m}^{th}$ terms.

Answer 1

Question :

$\textsf{If the ratio of the sum of first n terms of two A.P's is }\frac{(7n+1)}{(4n+26)}\textsf{, find the ratio of their }\mathsf{{m}^{th} terms.}$

Answer:

$\huge{\boxed{\green{(14m-6):(8m+22)}}}$

Step-by-step explanation:

Sum of n terms is given by the formula :

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

$\bigstar \textsf{Let the first term of the first A.P be }a_1\\\\\bigstar \textsf{Let the first term of the second A.P be }a_2\\\\\bigstar \textsf{Let the common difference of first A.P be }d_1\\\\\bigstar \textsf{Let the common difference of second A.P be }d_2$

Ratio of sum of sum of n terms = $\mathsf{\frac{7n+1}{4n+26}}$

$\mathsf{\frac{\frac{n}{2}[2a_1+(n-1)d_1]}{\frac{n}{2}[2a_2+(n-1)d_2]}=\frac{7n+1}{4n+26}}\\\\\implies \mathsf{\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{7n+1}{4n+26}}$

$\implies \mathsf{\frac{\frac{2a_1+(n-1)d_1}{2}}{\frac{2a_2+(n-1)d_2}{2}}=\frac{7n+1}{4n+26}}\\\\\implies \mathsf{\frac{a_1+\frac{(n-1)d_1}{2}}{a_2+\frac{(n-1)d_2}{2}}=\frac{7n+1}{4n+26}}$

m th term is represented by a + ( m - 1 ) d

$\textbf{So put (n - 1)/2 as m - 1 :}$

$\textsf{Basically if (n-1)/2=m-1 then n-1 = 2(m-1) }\\\\\implies \mathsf{n-1=2m-2}\\\implies \mathsf{n=2m-1}$

$\implies \mathsf{\frac{a_1+\frac{(n-1)d_1}{2}}{a_2+\frac{(n-1)d_2}{2}}=\frac{7n+1}{4n+26}}\\\\\implies \mathsf{\frac{a_1+(m-1)d_1}{a_2+(m-1)d_2}=\frac{7(2m-1)+1}{4(2m-1)+26}}$

$\implies \mathsf{\frac{a_1+(m-1)d_1}{a_2+(m-1)d_2}=\frac{14m-7+1}{8m-4+26}}\\\\\implies \mathsf{\frac{a_1+(m-1)d_1}{a_2+(m-1)d_2}=\frac{14m-6}{8m+22}}$

The required solution will be :

$\huge{\boxed{\frac{14m-6}{8m+22}}}$

Answer 2

Answer:

(14m - 6) : (8m +22)

Step-by-step explanation:

Let a₁,a₂ be the first terms and d₁,d₂ be the common differences of the two given AP's. The sum of n terms is given by:

$=>\frac{S_{n}}{S'_{n}}=\frac{\frac{n}{2}[2a_{1}+(n-1)d_{1}]}{\frac{n}{2}[2a_{2}+(n-1)d_{2} ]}$

$=\frac{2a_{1}+(n-1)d_{1}}{2a_{2}+(n-1)d_{2}}$

Given that Sum of first n terms is (7n + 1):(4n + 26).

$=>\frac{2a_{1}+(n-1)d_{1}}{2a_{2}+(n-1)d_{2}}= \frac{7n+1}{4n+26}$

To find the ratio of the mth terms of the two given AP's, we replace n by (2m - 1), we get

$=>\frac{2a_{1}+(2m-1)d_{1}}{2a_{2}+(2m-2)_{2}}=\frac{7(2m - 1) + 1}{4(2m - 1) + 26}$

$=>\frac{a_{1}+(m-1)d_{1}}{a_{2}+(m-1)d_{2}} = \frac{14m-6}{8m+22}$

∴ Therefore, ratio of the mth terms of the two Ap's is (14m - 6) : (8m + 22).

Hope it helps!