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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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Chapter name Arithmatic Progression

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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