Question

If the ratio of the sum of first n terms of two A.p's is (7n+1):(4n+27),find the ratio of their M'th terms.​

Answer 1

Let the first term of 1st and 2nd AP be "a" and "A" respectively

Let the common difference of 1st and 2nd AP be "d" and "D" respectively

According to the question,

$\mathsf{\frac{Sum\:of\:n\:terms\:in\:1st\:AP}{Sum\:of\:n\:terms\:in\:2nd\:AP} = \frac{7n + 1}{4n + 27}}$

$\mathsf{\implies\:\frac{\cancel{\frac{n}{2}}\:[2a + (n -1)d}{\cancel{\frac{n}{2}}\:[2A + (n - 1)D} = \frac{7n + 1}{4n + 27}}$

Cancelling n/2 and taking 2 as common

$\mathsf{\implies\:\frac{\cancel{2}[a + (\frac{n -1}{2})d}{\cancel{2}[A + (\frac{n - 1}{2})D} = \frac{7n + 1}{4n + 27} \:\rightarrow\:(1)}$

We know that

$\mathsf{a_{m} = a + (m - 1)d}$

$\mathsf{\implies\: \frac{n - 1}{2} = (m -1) \: \rightarrow\:(2)}$

$\mathsf{\implies\: n - 1 = 2m - 2}$

$\mathsf{\implies\: n = 2m - 1\:\rightarrow\:(3)}$

From equation (1):

Replacing (n-1)/2 from LHS and (n) from RHS

$\mathsf{\implies\:\frac{a + (m - 1)d}{A + (m - 1)D} = \frac{7(2m - 1) + 1}{4(2m - 1) + 27}\rightarrow{Using\:(2)\:and\:(3)}}$

$\mathsf{\implies\:\frac{a + (m - 1)d}{A + (m - 1)D} = \frac{14m - 7 + 1}{8m - 4 + 27}}$

$\mathsf{\implies\:\frac{a + (m - 1)d}{A + (m - 1)D} = \frac{14m - 6}{8m + 23}}$

$\fbox{\mathsf{\implies\:\frac{a_{m}}{A_{m}} = \frac{14m - 6}{8m + 23}}}$

Answer 2

Question:-

If the ratio of the sum of first n terms of two A.p's is (7n+1):(4n+27), find the ratio of their $\displaystyle\sf {m^{th}}$ terms.

Answer:-

$\displaystyle\Large\boxed {\sf {T_{m_1}:T_{m_2}=(14m-6):(8m+23)}}$

Solution:-

If n is odd then the sum of first n terms of an AP is,

$\displaystyle\longrightarrow\sf{S_n=n\cdot T_{\left (\frac {n+1}{2}\right)}}$

[Recall the formula "Sum = No. of terms × Average"]

Ratio of sums,

$\displaystyle\longrightarrow\sf{S_{n_1}:S_{n_2}=(7n+1):(4n+27)}$

$\displaystyle\longrightarrow\sf{n\cdot T_{\left (\frac {n+1}{2}\right)_1}:n\cdot T_{\left (\frac {n+1}{2}\right)_2}=(7n+1):(4n+27)}$

$\displaystyle\longrightarrow\sf{T_{\left (\frac {n+1}{2}\right)_1}:T_{\left (\frac {n+1}{2}\right)_2}=(7n+1):(4n+27)}$

To find ratio of $\displaystyle\sf {m^{th}}$ terms, let,

• $\displaystyle\sf{m=\dfrac {n+1}{2}}$

• $\displaystyle\sf {n=2m-1}$

Then,

$\displaystyle\longrightarrow\sf{T_{m_1}:T_{m_2}=(7(2m-1)+1):(4(2m-1)+27)}$

$\displaystyle\longrightarrow\sf{\underline {\underline {T_{m_1}:T_{m_2}=(14m-6):(8m+23)}}}$