Question

$(ar + a) \times \frac{50}{3} = 100 \\ (ar ^{2} + a) \times 10 = 100$
Solve these 2 equation
find out the
Value of a=? r=?​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

$\bold{ = > (ar + a) \times \frac{50}{3} = 100.....(1)}$

$= ar + a = 100 \times \frac{3}{50} = 6$

$\bold{= a(r + 1) = 6.......(2)}$

$\bold{= > (a {r}^{2} + a) \times 10 = 100...(3)}$

$= a {r}^{2} + a = 10$

$\bold{= a( {r}^{2} + 1) = 1.....(4)}$

Equating equation 3 and 4 by eliminating method:-

$a(r + 1) = 6$

$a( {r}^{2} + 1) = 10$

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__________________

-r^2+r=-4

-r=-5

r=4

Now,put r's value in equation 2

$= > a(r + 1) = 6$

$= a(4 + 1) = 6$

$= > 5a = 6$

$= > a = \frac{6}{5}$

Therefore ,r=4 and a=6/5

Check :-in equation 1

$= > (ar + a) \times \frac{50}{3} = 100$

$= > a(r + 1) \times \frac{50}{3} = 100$

$= > \frac{6}{5} \times 5 \times \frac{50}{3}$

$= > \frac{30}{5} \times \frac{50}{3} = 10 \times 10 = 100$