Question

If the ratio of the sums of first n terms of two A.P.'s is (7n + 1): (4n + 27), find the ratio of their 9th
terms.
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Answer 1

Answer:

hope this will help you.......

Answer 2

Let the first terms and the common differences of the two AP be A, a ans D, d.

Their 9th terms are:

⇒A+(9-1)D = A+8D --------------------(i)

⇒a+(9-1)d = a+8d---------------------(ii)

A/Q:

⇒  $\frac{\frac{n}2 (2A+(n-1)D)}{\frac{n}2 (2a+(n-1)d)} = \frac{7n+1}{4n+27}$

⇒   $\frac{ (2A+(n-1)D)}{ (2a+(n-1)d)} = \frac{7n+1}{4n+27}$   (Since the number of terms in both AP is n)

 ⇒  $\frac{\frac{(2A+(n-1)D)}{2}}{(\frac{(2a+(n-1)d)}{2}} = \frac{7n+1}{4n+27}$      (Dividing the num and deno in LHS by 2)

⇒   $\frac{A+\frac{(n-1)D)}{2}}{a+\frac{(n-1)d)}{2}} = \frac{7n+1}{4n+27}$  -----------------------------------(iii)

Putting n=17 in --(iii)

  ⇒      $\frac{A+\frac{(17-1)D)}{2}}{a+\frac{(17-1)d)}{2}} = \frac{7.17+1}{4.17+27}$

  ⇒      $\frac{ (A+8D)}{ (a+8d)} = \frac{120}{95}$

  ⇒     $\frac{ (A+8D)}{ (a+8d)} = \frac{24}{19}$

Using previously formed equation (i) and (ii) we get that the ratio of their 9th terms  $=\frac{24}{19}$