Series and Progressions

harsh.champ wrote:Assume the given sum of the series, 7.5 + 15.5 + 10.5 + 13.0 + 13.5 + 10.5....+ x – y + z is
20634, (x, y, z > 0) then what is the value of the expression (2x + 3y + 5z)?

(A)1084
(B)1284
(C)1464
(D)1684
(E)None of these

The OA is C.

7.5 + 15.5 + 10.5 +13.0 + 13.5 + 10.5....... x –y + z = u
The LHS is the summation of two arithmetic progression.
The first series has a = 7.5 and d = 3.0 and the second
series has a = 15.5 and d = –2.5.
The last three terms of LHS are x, –y and z.
As x, y and z, all are positive, –y; cannot be a part of
the (7.5 + 10.5 + 13.5.....) series. We can break the
LHS as following:
(7.5 + 10.5 +13.5 + .... x + z) + (15.5 + 13.0 + 10.5 + .....
–y)=u
Let there be n terms in the first bracket. The second
bracket will have (n – 1) terms. We can write:- n/2*[2*7.5+(n-1)*3]+(n-1)/2*[2*15.5+(n-2)*(-2.50)]=u
solving n=102,101
so z=310.5
x=307.5
y=-234.5
2x+3y+5z=1464.

ans C

a bit tough

harsh.champ wrote:Assume the given sum of the series, 7.5 + 15.5 + 10.5 + 13.0 + 13.5 + 10.5....+ x – y + z is
20634, (x, y, z > 0) then what is the value of the expression (2x + 3y + 5z)?

(A)1084
(B)1284
(C)1464
(D)1684
(E)None of these

The OA is C.

The given series can be split into

7.5+ 10.5+ 13.5 +......+x+z (n terms)
15.5+13.0 + 10.5 + .........-y (n-1 terms)

z = 7.5 +(n-1) 3
-y = 15.5 - (n-2) 2.5 = 20.5 -2.5n

Sum of the series = n/2(7.5 +z) + (n-1)/2 (15.5 -y)
= n/2 ( 7.5 + 3n+4.5) + n-1/2 (15.5 +20.5 -2.5n)
=n/2 (3n+12) +(n-1)/2 (36-2.5n)
=1/2( 3n^2+12n + 36n-36 -2.5n^2+ 2.5n)
= 1/2( 0.5n^2+ 50.5n -36)
1/2 (n^2 +101n -72) = 20634*2
(n^2 +101n) =
n(n+101) =82 608

I quit