Question

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Answer 1

Let a1, a2 be the first terms and d1,d2 the common differences of the two arithmetic progression respectively.

According to the question,

$\sf{\frac{Sum \: of \: n \: terms \: of \: first \: A.P}{Sum \: of \: n \: terms \: of \: second \: A.P} = \frac{5n + 4}{9n + 6} }$

$: \implies \frac{ \frac{n}{2} (2a_{1} + ( n - 1)d_{1})}{ \frac{n}{2} (2a_{2} + (n - 1)d_{2})} = \frac{5n + 4}{9n + 6}$

$: \implies \: \frac{2a_{1}(n - 1)d_{1}}{2a_{2}(n - 2)d_{2}} = \frac{5n + 4}{9n + 6}$

$: \implies \: \frac{a_{1} + \frac{n - 1}{2}d_{1} }{a_{2} + \frac{n - 1}{2}d_{2} } = \frac{5n + 4}{9n + 6}$

If n - 1 / 2 = 17

➠ n - 1 = 34 , n = 35

So,

$\frac{a_{1} + 17d_{1}}{a_{2} + 17d_{2}} = \frac{5 \times 35 + 4}{9 \times 35 + 6}$

$\rightarrow \: \frac{179}{321}$

Hence,

$\sf{ \frac{Eighteen \: term \: of \: first \: series}{Eighteen \: term \: of \: second \: series} = \bold{\frac{179}{321} }}$

Answer 2

Answer:

18th term of 1st series / 18th term of 2nd series = 179/321

Step-by-step explanation:

Let a1,A2 be the first terms and d1,d2 the common difference of the two arithmetic progression respectively.

A.T.Q

sum of n term of 1st A.p/ sum of n term of 2nd A.p = 5n + 4 / 9n + 6

n/2(2a1 + (n - 1)d1) / n/2(2a2 + (n - 1)d2) = 5n + 4 / 9n + 6

2a1(n - 1)d1 / 2a2(n - 2)d2 = 5n + 4 / 9n + 6

a1 + (n - 1 / 2)d1 / a2 + (n - 1 / 2)d2 = 5n + 4 / 9n + 6

if n - 1 / 2 = 17

n - 1 = 17 × 2

n - 1 = 34

n = 35

so,

a1 +17d1 / a2 + 17d2 = 5 × 35 + 4 / 9 × 35 + 6

= 179/321

18th term of 1st series / 18th term of 2nd series = 179/321

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