Question

$\: the \: value \: of \: n = ?if5+8+11....+10terms3+5+7+....+nterms​=7.thenthevalueofn=?\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ 2. \sf \: if \: \: \: \dfrac{ a + bx}{a - bx} = \dfrac{b + cx}{b - cx} = \dfrac{c + dx}{c - dx} \: .then \: a , b , c , d \: are \: inifa−bxa+bx​=b−cxb+cx​=c−dxc+dx​.thena,b,c,darein$


Note:-

1. Irrelevant answers will be reported.

2. Answer will detailed explanation.

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Answer 1

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.