Question

If the ratio of the sum of the first n terms of two AP is (7n+1) :(4n+27), then find the ratio of their 9th terms​

Answer 1

Answer:

24 : 19

Step-by-step explanation:

Given,

Ratio of  the sum of the first 'n' terms of two A.P is :-

7n + 1 : 4n + 27

To Find :-

Ratio of their 9th term's.

Solution :-

sum of n terms in an A.P is :-

$S_n = \frac{n}{2}[2a +(n - 1)d]$

Let,

First A.P be A.P_1

first term of A.P_1 is a_1

common difference of A.P_1 is d_1

$\implies S_n(A.P_1)= \frac{n}{2}[2a_1+(n - 1)d_1]$

Second A.P be A.P_2

first term of A.P_2 is a_2

common difference of A.P_2 is d_2

$\implies S_n(A.P_2) = \frac{n}{2}[2a_2+(n - 1)d_2]$

Given,

$\frac{S_n(A.P)_1}{S_n(A.P)_2} = \frac{7n + 1}{4n + 27}$

Equating both :-

$\sf\frac{S_n(A.P)_1}{S_n(A.P)_2} = \frac{7n + 1}{4n + 27} = \frac{\frac{n}{2}[2a_1+(n-1)d_1]}{\frac{n}{2}[2a_2+(n-1)d_2]}$

$\sf\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2} = \frac{7n + 1}{4n + 27}$

$\sf\dfrac{2(a_1+\dfrac{(n - 1)}{2}d_1)}{2(a_2+\dfrac{n-1}{2}d_2)} =  \frac{7n + 1}{4n + 27}$

$\sf\dfrac{a_1+\dfrac{(n - 1)}{2}d_1}{a_2+\dfrac{n-1}{2}d_2}=  \frac{7n + 1}{4n + 27}$

[let it be equation (1)]

Hence, This is applicable to any value for A.P

We, need to find the ratio of their 9th terms​ so,

a_9 = a+(9 - 1)d

= a + 8d

Since, we need to equate both "(n-1)\2 and 8"

$\implies\frac{n - 1}{2} = 8$

n - 1 = 8(2)

n - 1 = 16

n = 16 + 1

n = 17.

Substituting values in eq(1):-

$\sf\dfrac{a_1+8d_1}{a_2+8d_2}=  \dfrac{7(17) + 1}{4(17) + 27}$

$\sf\dfrac{t_9(A.P)_1}{t_9(A.P)_2} = \dfrac{119 + 1}{68 + 27}$

= $\sf\dfrac{120}{95}$

= $\sf\dfrac{24}{19}$

$\sf\therefore\dfrac{t_9(A.P)_1}{t_9(A.P)_2} = \dfrac{24}{19}$

Answer 2

$\huge{\bf{\green{\mathfrak{\dag{\underline{\underline{Question:-}}}}}}}$

$\bullet{\leadsto}$ If the ratio of the sum of the first n terms of two AP is (7n + 1) : (4n + 27), then find the ratio of their 9th terms.

$\huge {\bf{\orange{\mathfrak{\dag{\underline{\underline{Answer:-}}}}}}}$

$\bullet{\leadsto}$ This is a very good question, just don’t get complicated or confused.

$\bullet{\leadsto} \: \textsf{Ratio of sum of 1st n terms of 2 A.P is (7n + 1) : (4n + 27).}$

$\bullet{\leadsto} \: \textsf{Ratio of 9th term?}$

$\bullet{\leadsto} \: \sf Let \: the\: 1st \:terms \:of \:2\: A.P’s\: be\: a_{1} \: and \: a’_{1}.$

$\bullet{\leadsto} \: \sf Let \: the\: common\: differences\: \:2\: A.P’s\: be\: d \: and \: d’.$

$\bullet{\leadsto} \: \sf Let \: the\: sums \:of \:2\: A.P’s\: be\: S_{1} \: and \: S’_{1}.$

$\bullet{\leadsto} \: \sf Let \: the\: 2 \: A.P’s\: be\: T_{1} \: and \: T’_{1}.$

$\bullet{\leadsto} \: \textsf{According to the question,}$

$\bullet{\leadsto} \: \boxed{\pink{\sf \dfrac{T_{1}}{T’_{1}} \: = \: \dfrac{7n \: + \: 1}{4n \: + \: 27} \: = \: \dfrac{\dfrac{n}{2} (2a \: + \: (n \: - \: 1) \: d}{\dfrac{n}{2} (2a’ \: + \: (n \: - \: 1) \: d’}}}$

$\bullet{\leadsto} \: \sf \dfrac{T_{1}}{T’_{1}} \: = \: \dfrac{7n \: + \: 1}{4n \: + \: 27} \: = \: \dfrac{\cancel{\dfrac{n}{2}} (2a \: + \: (n \: - \: 1) \: d}{\cancel{\dfrac{n}{2}} (2a’ \: + \: (n \: - \: 1) \: d’}$

$\bullet{\leadsto} \: \sf \dfrac{T_{1}}{T’_{1}} \: = \: \dfrac{7n \: + \: 1}{4n \: + \: 27} \: = \: \dfrac{2a \: + \: (n \: - \: 1) \: d}{2a’ \: + \: (n \: - \: 1) \: d’}$

$\bullet{\leadsto} \: \textsf{Taking LCM and removing 2 out,}$

$\bullet{\leadsto} \: \sf \dfrac{T_{1}}{T’_{1}} \: = \: \dfrac{7n \: + \: 1}{4n \: + \: 27} \: = \: \dfrac{2 (a \: + \: \dfrac{(n \: - \: 1)}{2} \: d)}{2 (a’ \: + \: \dfrac{(n \: - \: 1)}{2} \: d’)}$

$\bullet{\leadsto} \: \sf \dfrac{T_{1}}{T’_{1}} \: = \: \dfrac{7n \: + \: 1}{4n \: + \: 27} \: = \: \dfrac{\cancel{2} (a \: + \: \dfrac{(n \: - \: 1)}{2} \: d)}{\cancel{2} (a’ \: + \: \dfrac{(n \: - \: 1)}{2} \: d’)}$

$\bullet{\leadsto} \: \sf \dfrac{T_{1}}{T’_{1}} \: = \: \dfrac{7n \: + \: 1}{4n \: + \: 27} \: = \: \dfrac{a \: + \: \dfrac{(n \: - \: 1)}{2} \: d}{a’ \: + \: \dfrac{(n \: - \: 1)}{2} \: d’}$

$\bullet{\leadsto} \: \textsf{Now,}$

$\bullet{\leadsto} \: \sf{We \: need \: to \: find \: out \: a_{9} \: term.}$

$\bullet{\leadsto} \: \sf Let \: the\: both \: 9th \: terms \: of \: A.P \: be\: a_{9} \: and \: a’_{9}.$

$\bullet{\leadsto} \: \textsf{We know that,}$

$\bullet{\leadsto} \: \sf a_{9} \: = \: a \: + \: 8d$

$\bullet{\leadsto} \: \sf Taking, \: \dfrac{a \: + \: \dfrac{(n \: - \: 1)}{2} \: d}{a’ \: + \: \dfrac{(n \: - \: 1)}{2} \: d’}$

$\bullet{\leadsto} \: \sf Comparing \: with \: a_{9},$

$\bullet{\leadsto} \: \sf \dfrac{a \: + \: \dfrac{(n \: - \: 1)}{2} \: d}{a’ \: + \: \dfrac{(n \: - \: 1)}{2} \: d’} \: = \: \dfrac{a \: + \: 8d}{a’ \: + \: 8d’}$

$\bullet{\leadsto} \: \implies{\sf \dfrac{n \: - \: 1}{2} \: = \: 8}$

$\bullet{\leadsto} \: \sf n \: - \: 1 \: = \: 8 \: * \: 2$

$\bullet{\leadsto} \: \sf n \: - \: 1 \: = \: 16$

$\bullet{\leadsto} \: \sf n \: = \: 16 \: + \: 1$

$\bullet{\leadsto} \: \underline{\boxed{\blue{\sf n \: = \: 17.}}}$

$\bullet{\leadsto}$ As we have found the value of n, no need to take the sum of A.P’s.

$\bullet{\leadsto}$ We can take only the ratio and substitute it with n and find out.

$\bullet{\leadsto} \: \sf \dfrac{a_{9}}{a’_{9}} \: = \: \dfrac{7n \: + \: 1}{4n \: + \: 27}$

$\bullet{\leadsto} \: \sf \dfrac{a_{9}}{a’_{9}} \: = \: \dfrac{7 \: * \: 17 \: + \: 1}{4 \: * \: 17 \: + \: 27}$

$\bullet{\leadsto} \: \sf \dfrac{a_{9}}{a’_{9}} \: = \: \dfrac{119 \: + \: 1}{68 \: + \: 27}$

$\bullet{\leadsto} \: \sf \dfrac{a_{9}}{a’_{9}} \: = \: \dfrac{120}{95}$

$\bullet{\leadsto} \: \textsf{Simplifying,}$

$\bullet{\leadsto} \: \sf \dfrac{a_{9}}{a’_{9}} \: = \: \dfrac{\cancel{120}}{\cancel{95}}$

$\bullet{\leadsto} \: \underline{\boxed{\purple{\sf \dfrac{a_{9}}{a’_{9}} \: = \: \dfrac{24}{19}}}}$

$\huge{\bf{\red{\mathfrak{\dag{\underline{\underline{Conclusion:-}}}}}}}$

$\bullet{\leadsto} \: \boxed{\therefore{\textsf{Ratio of their 9th terms is 24:19.}}}$