Question

the sumof n terms of two
Arithmetic series are in
ratio 2n+3:6n+3 then find the ratio of their 13th term​

Answer 1

AnswEr:-

• The ratio is 53 : 153

GivEn:-

• The sum of nth term of two arithmetic series are in ratio of 2n + 3 : 6n + 3

To Find:-

• The ratio of their 13th term

SolutiOn:-

Let the sum of one A.P. be Sn and and another one be S'n.

According to the given question,

$\implies \sf \: \dfrac{S_{n}}{S'_{n}} = \dfrac{2n + 3}{6n + 3}$

$\implies \sf \: \dfrac{ \dfrac{n}{2}[ 2a_{1} + (n - 1)d_{1}]}{ \dfrac{n}{2} [2a_{2} + (n - 1)d_{2}]} = \dfrac{2n + 3}{6n + 3}$

$\implies \sf \: \dfrac{a_{1} + ( \frac{n - 1}{2})d_{1} }{a_{2} + ( \frac{n - 1}{2})d_{2} } = \dfrac{2n + 3}{6n + 3}$

Now, we have to find the 13th term of each A.P.

Therefore,

$\sf \: (\dfrac{n - 1}{2} ) = 12$

$\implies \sf \: n - 1 = 12 \times 2$

$\implies \sf \: n = 24 + 1$

$\boxed { \red{\sf{ \implies \: n = 25}}}$

Now, putting the value of n we get,

$\implies \sf \: \dfrac{a_{1} + 12d_{1} }{a_{2} + 12d_{2} } = \dfrac{(2 \times 25)+ 3}{(6 \times 25) + 3}$

$\implies \sf \: \dfrac{a_{1} + 12d_{1} }{a_{2} + 12d_{2} } = \dfrac{50+ 3}{150 + 3}$

$\boxed{ \red{ \sf{ \implies \: \frac{t_{n}}{t'_{n}} = \dfrac{53}{153} }}}$

Therefore, the ratio is 53 : 153

Answer 2

Answer:

AnswEr:-

The ratio is 53 : 153

GivEn:-

The sum of nth term of two arithmetic series are in ratio of 2n + 3 : 6n + 3

To Find:-

The ratio of their 13th term

SolutiOn:-

Let the sum of one A.P. be Sn and and another one be S'n.

According to the given question,

\implies \sf \: \dfrac{S_{n}}{S'_{n}} = \dfrac{2n + 3}{6n + 3}⟹

S

n

′

S

n

=

6n+3

2n+3

\implies \sf \: \dfrac{ \dfrac{n}{2}[ 2a_{1} + (n - 1)d_{1}]}{ \dfrac{n}{2} [2a_{2} + (n - 1)d_{2}]} = \dfrac{2n + 3}{6n + 3}⟹

2

n

[2a

2

+(n−1)d

2

]

2

n

[2a

1

+(n−1)d

1

]

=

6n+3

2n+3

\implies \sf \: \dfrac{a_{1} + ( \frac{n - 1}{2})d_{1} }{a_{2} + ( \frac{n - 1}{2})d_{2} } = \dfrac{2n + 3}{6n + 3}⟹

a

2

+(

2

n−1

)d

2

a

1

+(

2

n−1

)d

1

=

6n+3

2n+3

Now, we have to find the 13th term of each A.P.

Therefore,

\sf \: (\dfrac{n - 1}{2} ) = 12(

2

n−1

)=12

\implies \sf \: n - 1 = 12 \times 2⟹n−1=12×2

\implies \sf \: n = 24 + 1⟹n=24+1

\boxed { \red{\sf{ \implies \: n = 25}}}

⟹n=25

Now, putting the value of n we get,

\implies \sf \: \dfrac{a_{1} + 12d_{1} }{a_{2} + 12d_{2} } = \dfrac{(2 \times 25)+ 3}{(6 \times 25) + 3}⟹

a

2

+12d

2

a

1

+12d

1

=

(6×25)+3

(2×25)+3

\implies \sf \: \dfrac{a_{1} + 12d_{1} }{a_{2} + 12d_{2} } = \dfrac{50+ 3}{150 + 3}⟹

a

2

+12d

2

a

1

+12d

1

=

150+3

50+3

\boxed{ \red{ \sf{ \implies \: \frac{t_{n}}{t'_{n}} = \dfrac{53}{153} }}}

⟹

t

n

′

t

n

=

153

53

Therefore, the ratio is 53 : 153