The sum of first $n$ terms of two $A . P$ .' are in the ratio $(7 n + 2) : (n + 4) .$ Find the ratio of their $5 t h$ terms.

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Solution

Let $a_{1}, a_{2}$ and $d 1, d 2$ be the first terms and common difference of the first and second arithmetic progression, respectively. According to the given condition, we have
$\frac{s u m o f n t e r m s o f f i r s t A . P}{s u m o f t h e n t e r m s o f s e c o n d A . P .} = \frac{7 n + 2}{n + 4}$
$\frac{\frac{n}{2} (2 a_{1} + (n - 1) d_{1})}{\frac{n}{2} (2 a_{2} + (n - 1) d_{2})} = \frac{7 n + 2}{n + 4}$

Cancelling out $\frac{n}{2}$ , we have

$\frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{7 n + 2}{n + 4}$ -------- Equation 1
Now, we need to find the ratio of the fifth terms
$\frac{5^{t h} t e r m o f f i r s t A . P}{5^{t h} t e r m o f s e c o n d A . P .} = \frac{a_{1} + (5 - 1) d_{1}}{a_{2} + (5 - 1) d_{2}}$
$\frac{5^{t h} t e r m o f f i r s t A . P}{5^{t h} t e r m o f s e c o n d A . P .} = \frac{a_{1} + 4 d_{1}}{a_{2} + 4 d_{2}}$

Putting $n = 9$ in Equation 1 we have
$\frac{2 a_{1} + (9 - 1) d_{1}}{2 a_{2} + (9 - 1) d_{2}} = \frac{7 \times 9 + 2}{9 + 4}$
$\frac{2 a_{1} + 8 d_{1}}{2 a_{2} + 8 d_{2}} = \frac{63 + 2}{13}$
taking 2 as common
$\frac{2 (a_{1} + 4 d_{1})}{2 (a_{2} + 4 d_{2})} = \frac{65}{13}$
Cancelling out 2 we get
$\frac{a_{1} + 4 d_{1}}{a_{2} + 4 d_{2}} = \frac{5}{1}$

So the ratio to the $5^{t h}$ term of both the AP's is $5 : 1$

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