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Step-by-step explanation:
We have to find the value of 'n' in the given expression:
\frac{3+5+7+....+n}{5+8+11+...+10 terms}=7
5+8+11+...+10terms
3+5+7+....+n
=7
since the series given in numerator and denominator is in arithmetic progression as the common difference is same throughout both the series.
Sum of Arithmetic Progression = \frac{n}{2}[2a+(n-1)d)]
2
n
[2a+(n−1)d)]
So, \frac{\frac{n}{2}[(2 \times 3)+(n-1)2)]}{\frac{10}{2}(2(5)+(9 \times 3))}= 7
2
10
(2(5)+(9×3))
2
n
[(2×3)+(n−1)2)]
=7
\frac{\frac{n}{2}[(6)+(n-1)2)]}{\frac{10}{2}(10+27)}= 7
2
10
(10+27)
2
n
[(6)+(n−1)2)]
=7
\frac{n(6+2n-2)}{2}= 7 \times 5 \times 37
2
n(6+2n−2)
=7×5×37
\frac{n(4+2n)}{2}= 7 \times 5 \times 37 \times 2
2
n(4+2n)
=7×5×37×2
4n+2n^{2}-2590=04n+2n
2
−2590=0
2n+n^{2}-1295=02n+n
2
−1295=0
n^{2}+2n-1295=0n
2
+2n−1295=0
n^{2}+37n-35n-1295=0n
2
+37n−35n−1295=0
n(n+37)-35(n+37)=0n(n+37)−35(n+37)=0
(n-35)(n+37)=0(n−35)(n+37)=0
Since 'n' can'not be negative.
So, n=35.
So, the value of 'n' in the given expression is 35.