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Sn =116 sn = n/2 [2a + (n-1) d ] is the formula to sum to n terms in an AP

a=25 , d = -3 {22-25= -3} 116 =n/2 [ 2(25) +(n-1) -3] 116 = n/2 [50 -3n +3] 116*2 =50n -3n² +3n 232 = 50n -3n² +3n 3n²-53n +232 =0 by applying d formula -b +/- √b²-4ac /2a we get roots as 9.6 & 8 thus n=8 to get the last term we use d formula sn =n/2 [a + l ] where l is the last term 116 =8/2 [25+l ] 116 = 4 [25 +l] 116/4 = 25 +l 29 = 25 +l ∴ l = 4