Question

if in an AP the sum of first x terms is Sx and Sx/Skx is independent of x ,then To proove d=2a

Answer 1

LET a BE THE FIRST TERM AND d BE THE COMMON DIFFERENCE.
 NOW SX/SKX=X/2 (2a + (X-1)d)/KX/2(2a + (KX - 1)d)
                      =(2a + XD- D)/(2AK + K²XD - KD) 
                      =(2a + XD- D)/((2A-D)K + K²XD)

NOW SINCE SX/SKX IS INDEPENDENT OF X THEN,
(2A - D )K=0 D=2A.. HENCE PROVED

Answer 2

Given:

Sum of first x terms of an AP is sx and $\frac {sx}{skx}$

To prove:

$d = 2 \times a$

Answer:

In AP the sum of first x terms is sx and $\frac {sx}{skx}$

$The \ sum \ of \ AP \ formula \ is = \frac { \mathrm { n } ( 2 \mathrm { a } + ( \mathrm { n } - 1 ) \mathrm { d } ) } { 2 }$

$= \frac { \frac { \mathrm { x } } { 2 } ( 2 \mathrm { a } + ( \mathrm { x } - 1 ) \mathrm { d } } { \frac { \mathrm { kx } } { 2 } ( 2 \mathrm { a } + ( \mathrm { kx } - 1 ) \mathrm { d } }$

$= \frac { 2 \mathrm { a } + \mathrm { xd } - \mathrm { d } } { \mathrm { k } ( 2 \mathrm { a } + \mathrm { kxd } - \mathrm { d } ) }$

$= \frac { 2 \mathrm { a } + \mathrm { xd } - \mathrm { d } } { 2 \mathrm { ak } + \mathrm { k } ^ { 2 } \mathrm { xd } - \mathrm { kd } }$

$= \frac { 2 \mathrm { a } + \mathrm { xd } - \mathrm { d } } { ( 2 \mathrm { a } - \mathrm { d } ) \mathrm { k } + \mathrm { k } ^ { 2 } \mathrm { xd } }$

According to the question

sx and $\frac {sx}{skx}$ is independent of x

$2 \times a + 0 - d = 0$

$d = 2 \times a$

Hence proved.