IIT JEE , AIEEE Forum - Mathematics , Algebra

kaushalcool

_{^{The number of terms of A.P.is even:the sum of odd terms is 24,of the even termsis 30.and the last term exceeds the first by 21\2,find thenumbers of terms and the series?}}

sankydreams

let no. of trms = 2n
first trm = a
common diff = d

first odd term = a
last odd term =a + (2n - 2)d
common difference between odd terms = 2d
no of odd terms = 2n/2 = n
sum of odd terms = n/2[a + a + (n-2)d]
=> 24= n/2(2a + (n - 2)d)..............(1)

last term = a + (2n -1)d
according to ques
a+(2n-1)d -a =21/2
=>(2n-1)d = 21/2..........(2)

first even term = a+d
last even term = a + (2n-1)d = a + 21/2
common diff bw even trms = 2d
no of even terms = n
sum of even terms= n/2(a+a+(2n-1)d)
30=n/2(2a + 21/2).........(3)

solve(1)(2)(3) to get answer
rate if this helps u...
cheers

magiclko

if no of terms is n, first term is t and common difference is d, then we have...

thus first odd term = t

first even term = t + d

commomn difference between two odd/even terms = 2d

therefore,

24 = [ n {2t + (n-1)2d} ] / 2

=> 24 = n {t + (n-1)d} ...............(1)

also,

30 = [ n {2 (t+d) + (n-1)2d} ] / 2

=> 30 = n {(t+d) + (n-1)d} .................(2)

subtracting 1 from 2, we have

nd = 6

also,

T_n = t + 21/2

=> t + (2n-1) d = t +21/2

=> (2n-1) d = 21/2

=> 2nd - d = 21/2

=> d = 12 -21/2

= 3/2

thrfore n = 4 ...

and using this in (1), we have t = 3/2

thrfore the series is 3/2 , 3 , 9/2 , 6 , 15/2, 9, 21/2, 12

lazycol

let a b 1st term,d common diff., n b no :of terms

consider d odd terms of ap,it again is an ap wid

1st term = a; common diff. = 2d; no of terms = n/2

sum = 24 = n/4 [ 2a + (n/2 - 1)2d ] = n/4[ 2a+(n-2)d]....................(1)

consider d even terms of ap,it again is an ap wid

1st term = a+d; common diff. = 2d; no of terms = n/2

sum = 30 = n/4 [ 2(a+d) + (n/2 - 1)2d ] = n/4[ 2a+nd]....................(2)

last term of series = a+(n-1)d

diff. = 21/2 = a+(n-1)d - a = (n-1)d; nd = 21/2 + d..........................(3)

(2) - (1) gives 6 = (n/4)* (2d) = nd/2

12 = 21/2 + d { by (3) } ie d = 3/2

(3) implies n = (21/2 + d)/d = 8

(2) gives 2a = (30 *4 /n) - nd = 3 ie a = 3/2

hence no : of trems = 8

series is 3/2, 3, 9/2, 6, 15/2, 9, 21/2, 12

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