Question

The sum of first n terms of two ap are in the ratio 5n+4:9n+6. find the ratio of their 18th term

Answer 1

Hope u like my process
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Ratio of sum of 2 AP's = (5n+4) :(9n+6)

$= > ( 5 + 4 + 5n - 5) : (6 + 9 + 9n - 9) \\ \\ = > (9 +(n - 1)5) : (15 + (n - 1)9)$
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Now,

1st AP.
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$= > 2 a_{1} + (n - 1) d_{1} \\ \\ = > 2( \frac{9}{2} ) + (n - 1)5$
So, A1 = 9/2 and d1 = 5

=> 18th term of A.P. = (a1 +(18-1)d1)


$= \frac{9}{2} + 17 \times 5 = \frac{9}{2} + 85 \\ \\ = \frac{179}{2}$


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2nd A.P.
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$= > 2 a_{2} + (n - 1) d_{2} \\ \\ = > 2( \frac{15}{2} ) + (n - 1)9$

So,

$= > \bf \: a_{2} = \frac{15}{2} \: \: \: \: and \: \: \: \: d_{2} = 9$


So, 18th A.P2 term is

=
$= a_{2} + (18 - 1) d_{2} \\ \\ = \frac{15}{2} + (17) \times 9 = \frac{15}{2} + 153 \\ \\ = \frac{321}{2}$
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So, the ratio of 18th term of 2 AP's is

$= > \frac{179}{2} : \frac{321}{2} = 179 : 321$
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