Question:
If $\frac{3+5+7+9+\ldots \text { up to } 35 \text { terms }}{5+8+11+\ldots . \text { up ton terms }}=7$ find the value of n.
Solution:
To find: the value of n.
We can write it as
$\frac{\frac{35}{2}(6+34(5-3))}{\frac{\mathrm{n}}{2}(10+3(\mathrm{n}-1))}=7$
$3 n^{2}+7 \times n-370=0$
Therefore $n=37 / 3,10$
Rejecting $37 / 3$ we get $n=10$
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