The given question is a hard 3-mark question from the chapter Arithmetic Progressions. Applying concepts related to sum of the first 'n' terms of an AP and using that to find the ratio of the nth term of two sequences in Arithmetic Progression. It is an important question and appeared in Section C of the 2016 CBSE class 10 board paper.

Question 15: 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

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Let the first term of the first AP be a_{1} and let its common difference be d_{1}

Let the first term of the second AP be b_{1} and let its common difference be d_{2}

Sum of 1st n term of first AP = \\frac{n}{2}) (2a_{1} + (n-1) d_{1})

Sum of 1st n term of second AP = \\frac{n}{2}) (2b_{1} + (n-1) d_{2})

**The ratio of the sums of first n terms of the 2 APs is (7n + 1) : (4n + 27)**

So, \\frac{\frac{n}{2}(2a_1+(n-1) d_1))}{\frac{n}{2}(2b_1+(n-1) d_2)})= \\frac{7n+1}{4n+27})

Or \\frac{2a_1+(n-1) d_1}{2b_1+(n-1) d_2}) = \\frac{7n+1}{4n+27}) ----- (1)

The m^{th} term of the first AP a_{m} = a_{1} + (m-1) d_{1}

The m^{th} term of the second AP b_{m} = b_{1} + (m-1) d_{2}

Ratio of the m^{th} terms of the 2 APs, \\frac{a_m}{b_m}) = \\frac{a_1+(m-1) d_1}{b_1+(m-1) d_2})

**Multiply the numerator and denominator of the fraction by 2**

\\frac{a_m}{b_m}) = \\frac{2a_1+2(m-1) d_1}{2b_1+2(m-1) d_2}) ---- (2)

**Compare LHS of (1) and (2)**

2(m - 1) = n - 1

2m - 2 = n - 1

Or n = 2m - 1

**Substitute the value of n as (2m - 1) in equation (1)**

\\frac{2a_1+(2m-1-1) d_1}{2b_1+(2m-1-1) d_2}) = \\frac{7(2m-1)+1}{4(2m-1)+27})

\\frac{a_m}{b_m}) = \\frac{2a_1+2(m-1) d_1}{2b_1+2(m-1) d_2})= \\frac{14m - 7 + 1}{8m - 4 + 27}) = \\frac{14m - 6}{8m + 23})

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