how many terms of the ap 25,22,19.....are needed to give the sum 116 ? also find the last term
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Solution :
Given A.P : 25,22,19, ....,
first term = a = 25
Common difference = a2 - a1
=> d = 22 - 25 = -3
Let sum of n terms = Sn = 116
=> n/2[ 2a + ( n -1 )d ] = 116
=> n/2[ 2×22 + ( n-1)(-3) ] = 116
=> n[ 44 - 3n + 3 ] = 232
Here , some error in the question ,
we don't get 116 as sum , plz check again
=> n( 47 -3n ) = 232
=> 47n - 3n² = 232
=> 3n² - 47n + 232 = 0
=>
Given A.P : 25,22,19, ....,
first term = a = 25
Common difference = a2 - a1
=> d = 22 - 25 = -3
Let sum of n terms = Sn = 116
=> n/2[ 2a + ( n -1 )d ] = 116
=> n/2[ 2×22 + ( n-1)(-3) ] = 116
=> n[ 44 - 3n + 3 ] = 232
Here , some error in the question ,
we don't get 116 as sum , plz check again
=> n( 47 -3n ) = 232
=> 47n - 3n² = 232
=> 3n² - 47n + 232 = 0
=>
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