Question

answer this question

Answer 1

Heya User,

( 2 ) --> 5th term = ( a + 4d ) = 15  ---> ( i )

Now, [ 3rd term + 8th term ] = 34
=> > [ ( a + 2d ) + ( a + 7d ) ] = 34
=> > [ 2a + 9d ] = 34  --> ( ii )

Solving ( i ) and ( ii ), a = -1 ; d = 4 ;

Now, 20th term = [ a + 19d ] = [ -1 + 76 ] = 75 √√


( 11 ) Let the terms be [ a - 3d ] , [ a - d ] , [ a + d ] , [ a + 3d ]

ATQ -> [ a - 3d ] + [ a - d ] + [ a + d ] + [ a + 3d ] = 32

=> 4a = 32 => a = 8 √√

Also, ( [ a - 3d ][ a + 3d ] ) : ( [ a - d ][ a + d ] ) = 7 : 15

=> 15 ( 8 - 3d )( 8 + 3d ) = 7 ( 8 - d )( 8 + d )
=> 15 ( 64 - 9d² ) = 7 ( 64 - d² )
=> 960 - 135 d² = 448 - 7 d²
=> 512 = 128 d²
=> d = 4
=> d = ± 2

Hence, numbers are --> 2 , 6 , 10 , 14 √√√