THE sum of n term of the series 25,22,19,16,....is 116.find the number of term and the last term.
Chapter name Arithmatic Progression
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Answered by
20
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
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
Answered by
2
the term is -3& the last term is 1.
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