 # CBSE Board Paper Solution

Class 10 Math - 2017 Set 3 - Q27. Arithmetic Progression

#### Video Explanation

Let the first term of the 1st 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 1st 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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