Question

in the sum of first 10 terms of an A.P. is -60 and sum of first 15 terms is -165 then find the sum of first n terms​

Answer 1

let first term is a and common difference is d .

sum of n term = n/2 [ 2a +(n-1)d ]

according to questions ,

=> 10/2 [ 2a + (10-1)d ] = -60
=> 10a + 45d = -60 ------------------1 multiple by 2

=>15/2 [ 2a + (15-1)d ] = -165
=> 15a + 105d = -165 ------------------2

1st equation multiply by 3 and 2nd equation multiple by 2 equation, subtract 1st equation from 2nd equation we get ,

=> 30a + 210 d = -330 ------------------2
=> 30a + 135 d = -180 ------------------1
------------------------------------
=> 75 d = -150
=> d = -2

then , a =( -60 +90)/10 =3

therefore,

sum of n term = n/2 [ 2a +(n-1)d ]

sum of n term = n/2 [ 6 +( n-1) -2 ]

sum of n term = 3n -n² +n

sum of n term = = 4n - n² Ans