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Q.

If $\frac{3 + 5 + 7 + . . . . . . + n (terms)}{5 + 8 + 11 + . . . . . + 10 (terms)} = 7$ , then the value of $n$ is

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a

35

b

36

c

37

d

40

answer is A.

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Detailed Solution

Given,

\frac{3 + 5 + 7 + . . . . . . + n (terms)}{5 + 8 + 11 + . . . . . + 10 (terms)} = 7

.
We know that,

\Rightarrow S_{n} = \frac{n}{2} [2 a + (n - 1) d]

, where first term

= a

and

d =

common difference.

\frac{3 + 5 + 7 + . . . . . . + n terms}{5 + 8 + 11 + . . . . . . + 10 terms} = 7

\Rightarrow \frac{\frac{n}{2} [2 \times 3 + (n - 1) \times 2]}{\frac{10}{2} [2 \times 5 + (10 - 1) \times 3]} = 7

\Rightarrow \frac{n (2 n + 4)}{370} = 7

\Rightarrow 2 n^{2} + 4 n - 2590 = 0

\Rightarrow n^{2} + 2 n - 1295 = 0

\Rightarrow n^{2} + 37 n - 35 n - 1295 = 0

\Rightarrow n (n + 37) - 35 (n + 37) = 0

\Rightarrow (n - 35) (n + 37) = 0

\Rightarrow n = 35

Hence, the correct option is 1.

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