Question

The sum of two numbers of an ap is 32.the ratio of the product of first and fourth to the product of second and third is 55: 63 .find the numbers

Answer 1

Hi Mate !!


Let the four consecutive terms be

( a - 3d ) , ( a - d ) , ( a + d ) , ( a + 3d )

Where a is the first term and d is the common difference of AP.


• Sum of the numbers is 32

a - 3d + a - d + a + d + a + 3d = 32

4a = 32

a = 32/4

a = 8


• The ratio of the product of first and fourth to the product of second and third is 55: 63


$\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{55}{63}$

[ Using identity :- ( x - y ) ( x + y ) = x² - y² ]


$\frac{ {a}^{2} - {(3d)}^{2} }{{a}^{2} - {d}^{2} } = \frac{55}{63}$


$\frac{( {8)}^{2} -9 {d}^{2} }{( {8)}^{2} - {d}^{2} } = \frac{55}{63}$


$\frac{64 - 9 {d}^{2} }{64 - {d}^{2} } = \frac{55}{63}$


4032 - 567d² = 3520 - 55d²

4032 - 3520 = - 55d² + 567d²

512 = 512d²

512/512 = d²

1 = d²

√1 = d

± 1 = d


__________________________

• If a = 8 and d = 1

then the AP will be :-

( a - 3d ) , ( a - d ) , ( a + d ) , ( a + 3d )

( 8 - 3 ) , ( 8 - 1 ) , ( 8 + 1 ) , ( 8 + 3 )

5 , 7 , 9 , 11 ...


• If a = 8 and d = ( - 1 )

then the AP will be :-

( a - 3d ) , ( a - d ) , ( a + d ) , ( a + 3d )

( 8 + 3 ) , ( 8 + 1 ) , ( 8 - 1 ) , ( 8 - 3 )

11 , 9 , 7 , 5 ....