Question

The sum of first nth terms of two APs are in the ratio (3n+8)
is to (7n+15).Find the ratio of their 12th terms​

Answer 1

Answer:-

Given:

Sum of first n terms of two AP's are in the ratio (3n + 8) : (7n + 15).

Let the first terms and common differences of the two APs be a , d & a' , d'.

We know that,

Sum of first n terms of an AP (Sₙ) = n/2 * [ 2a + (n - 1)d ]

So,

⟹ n/2 * [ 2a + (n - 1)d ] : n/2 * [ 2a' + (n - 1)d' ] = 3n + 8 : 7n + 15

⟹ 2a + (n - 1)d : 2a' + (n - 1)d' = 3n + 8 : 7n + 15

Multiply numerator and denominator by 1/2 in LHS

⟹ a + (n - 1)/2 * d : a' + (n - 1)/2 * d' = 3n + 8 : 7n + 15

Now,

3n + 8 : 7n + 15 is the ratio of nth terms of the two APs.

We have to find the ratio of 12th terms.

We know that,

nth term of an AP (aₙ) = a + (n - 1)d

⟹ a₁₂ = a + (12 - 1)d

⟹ a₁₂ = a + 11d

So,

⟹ a + (n - 1)/2 * d = a + 11d

⟹ n - 1/2 = 11

⟹ n - 1 = 22

⟹ n = 23

Hence, we have to substitute n as 23 to find the ratio of their 12th terms.

⟹ a + 11d : a' + 11d' = 3(23) + 8 : 7(23) + 15

⟹ a + 11d : a' + 11d' = 77 : 176

⟹ a + 11d : a' + 11d' = 7 : 16

∴ The ratio of the 12th terms of the two APs is 7 : 16.

Answer 2

Answer:

$\begin{lgathered}\underline{\mathsf{For \: 2nd \: A.P. }} \\ \\ \\ \textsf{Given,} \\ \\ \mathsf{\implies No. \: of \: terms\: = \: n } \\ \\ \textsf{Let,} \\ \\ \mathsf{\implies First \: term \: = \: a_{2}} \\ \\ \mathsf{\implies Common \: difference \: = \: d_{2}}\end{lgathered}$

$\begin{lgathered}\textsf{Using Formula : } \\ \\ \boxed{\mathsf{\implies S_{n} \: = \: \dfrac{n}{2}[2a \: + \: ( n \: - \: 1)d]}}\end{lgathered}$

Using Formulas:

$\mathsf{\implies( S_{n})_{2} \: = \: \dfrac{n}{2}[2a_{2} \: + \: ( n \: - \: 1)d_{2}]}$

$\underline{\textsf{According to question : }}$

$\mathsf{\implies \dfrac{ (S_{n})_{1}}{(S_{n})_{2}} \: = \: \dfrac{3n \: + \: 8 }{7n \: + \: 15}}$

$\mathsf{\implies \dfrac{ \quad \cancel{\dfrac{n}{2}}[2a_{1} \: + \: (n \: - \: 1)d_{1}] \quad}{ \cancel{\dfrac{n}{2}}[2a_{2} \: + \: ( n \: - \: 1)d_{2}]} \: = \: \dfrac{3n \: + \: 8}{7n \: + \: 15}}$

$\begin{lgathered}\textsf{Substitute n = 23 : } \\ \\ \\ \mathsf{\implies \dfrac{2a_{1} \: + \: ( 23 \: - \: 1)d_{1}}{ 2a_{2} \: + \: ( 23 \: - \: 1 )d_{2}} \: = \: \dfrac{3 \: \times \: 23 \: + \: 8}{ 7 \: \times \: 23 \: + \: 15 }}\end{lgathered}$

Substitute n = 23 :

$\mathsf{\implies \dfrac{ 2a_{1} \: + \: 22d_{1}}{2a_{2} \: + \: 22 d_{2} } \: = \: \dfrac{69 \: + \: 8}{161 \: + \: 15}}$

$\mathsf{\implies \dfrac{ \cancel{2}(a_{1} \: + \: 11d_{1})}{ \cancel{2}(a_{2} \: + \: 11 d_{2})} \: = \: \dfrac{77}{176}}$

$\\ \\ \mathsf{ \implies \dfrac{a_{1} \: + \: (12 \: - \: 1)d_{1}}{a_{2} \: + \: (12 \: - \: 1)d_{2}} \: = \: \dfrac{7}{16}}$

$\mathsf{\implies \dfrac{(T_{12} )_{1}}{(T_{12})_{2}} \: = \: \dfrac{7}{16}}$

$\mathsf{\therefore \quad (T_{12})_{1} \: : \: (T_{12})_{2} \: = \: 7 \: : \: 16}$