Question

3+5+7+.....+n/5+8+11+....+10 terms=7. find value of n

Answer 1

Answer:

n=35

Step-by-step explanation:

$\frac{sn}{s10} = 7$

$\frac{ \frac{n}{2}(6 + (n - 1)2) }{5(10 + 9(3))} = 7$

$\frac{n(n + 2)}{185} = 7$

${n}^{2} + 2n = 1295$

$(n - 35)(n + 37) = 0$

n=35

Hope it helps you

Answer 2

Answer:

There are two APs given in the above question. One AP is in the numerator and the other is in the denominator.

Given Equation: $\frac{3+5+7+------+n}{5+8+11+-----+10 terms} = 7$

$\frac{S_n}{S_{10}} = 7\\\\\frac{\frac{n}{2}[2a + (n-1)d] }{\frac{10}{2} [2(5) + (10-1)3]} = 7\\\\\frac{\frac{n}{2}[2(3) + (n-1)2] }{5[10+9 \times 3]} = 7\\\\\frac{\frac{n}{2}[6 + 2n - 2] }{5[10+27]} = 7\\\\\frac{3n + n^2-n}{5\times 37} = 7\\\\\frac{n^2 + 2n}{185} = 7\\\\n^2 + 2n = 7 \times 185\\\\n^2 + 2n = 1295\\\\n^2 + 2n - 1295 = 0\\\\n^2 + 37n - 35 n -1295 = 0\\\\n (n + 37) - 35(n + 37) = 0\\\\(n - 35) (n + 37) = 0\\\\n = 35, -37$

Neglecting negative root, we get

n = 35

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