CBSE Board Paper Solution

Explanatory Answer

Let the first term of the 1^st AP be a₁, and let its common difference be d₁.
Let the first term of the 2nd AP be b₁, and let its common difference be d₂.

Therefore, sum of the first n terms of 1^st AP = \\frac{n}{2})(2a₁ + (n – 1)d₁)
Sum of the first n terms of 2nd AP = \\frac{n}{2})(2b₁ + (n – 1)d₁)

The ratio of the sum up to first n terms of the two Aps is 7n + 1 : 4n + 27
i.e., \\frac {\frac {n} {2}\left ( {2{{a}_{1}+(n-1){d}_{1}}} \right )} {\frac {n} {2}\left ( {2{{b}_{1}+(n-1){d}_{2}}} \right )}) = \\frac{7n + 1}{4n + 27})
\\frac {\left ( {2{{a}_{1}+(n-1){d}_{1}}} \right )} {\left ( {2{{b}_{1}+(n-1){d}_{2}}} \right )}) = \\frac{7n + 1}{4n + 27}) ............ (1)

We have to compute a₉ : b₉
a₉ : b₉ = 2a₉ : 2b₉
2a₉ = 2(a₁ + 8d₁) = 2a₁ + 16d₁
2b₉ = 2(b₁ + 8d₂) = 2b₁ + 16d₂
Therefore, 2a₉ : 2b₉ = \\frac {2{{a}_{1}+16{d}_{1}}} {2{{b}_{1}+16{d}_{2}}}) ............ (2)

Comparing (1) and (2)
(n – 1) = 16 or n = 17
Therefore, a₉ : b₉ = \\frac{7(17) + 1}{4(17) + 27}) = \\frac{120}{95}) = \\frac{24}{19})

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