Question

If 3+5+7+... to n terms/
5+8+11+... to 10 terms
=7 then find the value of n​

Answer 1

$\huge{\mid{\red{\underline{\boxed{\tt{AnSweR}}}}}}$

_________________

Given:

$\dfrac{3+5+7.....n}{5+8+11......a_n}=7</p><p></p><h3>For Series in numerator</h3><p>a=3</p><p>d=5-3=2</p><p>[tex]a_n = n$

For Series in numerator

a=5

d=8-5=3

$a_n = ?$

Here In the denominator we have 10 terms

_________________

Solution:

Here first we will find the last term of the denominator series

$\implies<strong> </strong>a_n= a+(n-1)d \\ \implies 5+(10-1)3 \\ \implies 5+27 \\ \leadsto \boxed{a_n=32}$

Now,

Sum of series

$S_n=\dfrac{n}{2} [a+a_n] \\ \implies \dfrac{10}{2} \times [5+32] \\ \implies 5 \times 37 \\ \leadsto\boxed{185}$

$\dfrac{3+5+8......n}{185}=7 \\ \implies \dfrac{n}{2} \times [2a +(n-1)d ] = 185 \times 7 \\ \implies \dfrac{n}{2} [2(3) + (n-1)2] = 1295 \\ \implies \dfrac{n}{2} \times [6 + 2n -2 ] = 1295 \\ \implies \\ \implies \dfrac{n}{\cancel{2}} [\cancel{4} + \cancel{2}n]= 1295 \\ \implies 2n + n^2 = 1295 \\ \implies n^2 -35n+37n+1295 \\ \implies n(n-35) + 37(n-35) \\ \implies (n-35)(n+37) \\ n=-37,35 \\ \\ \huge\leadsto \boxed{\boxed{n=35}}$

Answer 2

Answer:

Given:

\dfrac{3+5+7.....n}{5+8+11......a_n}=7 For Series in numerator a=3 d=5-3=2 [tex]a_n = n

5+8+11......a

n

3+5+7.....n

=7ForSeriesinnumeratora=3d=5−3=2[tex]a

n

=n

For Series in numerator

a=5

d=8-5=3

a_n = ?a

n

=?

Here In the denominator we have 10 terms

_________________

Solution:

Here first we will find the last term of the denominator series

\begin{lgathered}\implies a_n= a+(n-1)d \\ \implies 5+(10-1)3 \\ \implies 5+27 \\ \leadsto \boxed{a_n=32}\end{lgathered}

⟹a

n

=a+(n−1)d

⟹5+(10−1)3

⟹5+27

⇝

a

n

=32

Now,

Sum of series

\begin{lgathered}S_n=\dfrac{n}{2} [a+a_n] \\ \implies \dfrac{10}{2} \times [5+32] \\ \implies 5 \times 37 \\ \leadsto\boxed{185}\end{lgathered}

S

n

=

2

n

[a+a

n

]

⟹

2

10

×[5+32]

⟹5×37

⇝

185

\begin{lgathered}\dfrac{3+5+8......n}{185}=7 \\ \implies \dfrac{n}{2} \times [2a +(n-1)d ] = 185 \times 7 \\ \implies \dfrac{n}{2} [2(3) + (n-1)2] = 1295 \\ \implies \dfrac{n}{2} \times [6 + 2n -2 ] = 1295 \\ \implies \\ \implies \dfrac{n}{\cancel{2}} [\cancel{4} + \cancel{2}n]= 1295 \\ \implies 2n + n^2 = 1295 \\ \implies n^2 -35n+37n+1295 \\ \implies n(n-35) + 37(n-35) \\ \implies (n-35)(n+37) \\ n=-37,35 \\ \\ \huge\leadsto \boxed{\boxed{n=35}}\end{lgathered}

185

3+5+8......n

=7

⟹

2

n

×[2a+(n−1)d]=185×7

⟹

2

n

[2(3)+(n−1)2]=1295

⟹

2

n

×[6+2n−2]=1295

⟹

⟹

2

n

[

4

+

2

n]=1295

⟹2n+n

2

=1295

⟹n

2

−35n+37n+1295

⟹n(n−35)+37(n−35)

⟹(n−35)(n+37)

n=−37,35

⇝

n=35