Question

If the ratio of the sum of the first n terms of the two ap's is (7n+1) : (4n+27). Find the ratio of their 9th term. (Class 10)

Answer 1

Step-by-step explanation:

 $\frac{S_{n} }{S_{n^{1} } }$        =      $\frac{7n+1}{4n+27}$

Let a and d be the first term and common difference of one A.P. and A and D be the first term and common difference for the second A.P.

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

( cancel $\frac{n}{2}$)

Divide numerator and denominator by 2 to get $a_{9}$

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

(cancel 2 in $\frac{2a}{2}$ and $\frac{2A}{2}$ )

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

taking  $\frac{(n-1)}{2}$ = 8 --------------------eq 2

n - 1 = 8 x 2

n - 1 = 16

n = 16 +1

n = 17 ----------------------------------- eq 3

Substituting eq 2 and eq 3 in eq 1

$\frac{a + 8d}{A + 8D}$ = $\frac{7 (17) +1}{4(17)+27}$

$\frac{a_{9} }{A_{9} }$     =    $\frac{120}{95}$ = $\frac{24}{19}$

∴ Ratio of their 9th term is  24:19

Hope this helps you!!

Answer 2

Step-by-step explanation:

Given that ratio of sum of n terms of two AP’s = (7n+1):(4n+27)

Let us assume the ratio of these two AP’s mth terms as am : a’m…..(2)

We know that nth term of AP formula, an = a + (n – 1)d

Hence equation (2) becomes,

am : a’m = a + (m – 1)d : a’ + (m – 1)d’

On multiplying by 2, we get

am : a’m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]

⇒ [2a + (2m – 2)d] : [2a’ + (2m – 2)d’]

= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’] = S2m – 1 : S’2m – 1

= [7(2m – 1) + 1] : [4(2m – 1) +27] [14m – 7 +1] : [8m – 4 + 27] = [14m – 6] : [8m + 23]

Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].