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प्रश्न
If Sn denote the sum of n terms of an A.P. with first term a and common difference dsuch that \[\frac{Sx}{Skx}\] is independent of x, then
विकल्प
d= a
d = 2a
a = 2d
d = −a
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उत्तर
Here, we are given an A.P. with a as the first term and d as the common difference. The sum of n terms of the A.P. is given by Sn.
We need to find the relation between a and d such that`S_x/S_(kx)` is independent of
So, let us first find the values of Sx and Skx using the following formula for the sum of n terms of an A.P.,
`S_n = n/2 [ 2a + ( n- 1) d ]`
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So, we get,
`S_x = x/2 [ 2a + ( x - 1) d ]`
Similarly,
`S_(kx) = (kx)/2 [ 2a + ( kx - 1) d ]`
So,
`S_x /S_(kx) = (x/2[2a + (x -1)d ] )/((kx)/2 [2a + (kx - 1 ) d])`
`=([2a + ( x -1) d ])/(k[2a + (kx -1) d ]) `
`=(2a + dx - d)/(2ak + k^2 xd -kd)`
Now, to get a term independent of x we have to eliminate the other terms, so we get
2a - d = 0
2a = d
So, if we substitute 2a = d , we get,
`(2a + dx - d)/(2ak + k^2 xd -kd)=(2a + dx -2a)/(2ak + k^2 xd -2ak)`
`=(dx)/(k^2 dx)`
`= 1/(k^2)`
Therefore, 2a = d
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