The correct option is
A 15LetS∞=1+4x+7x2+10x3+...∞ ...(1)
Now, multiply by x throughout in eqution (1); we get
xS∞=x+4x2+7x3+10x4+...∞ ...(2)
Subtracting (2) from (1); we get
S∞−xS∞=(1+4x+7x2+10x3+...∞)−(x+4x2+7x3+10x4+...∞)
⇒(1−x)S∞=1+4x+7x2+10x3+...∞−x−4x2−7x3−10x4−...∞
⇒(1−x)S∞=1+3x+3x2+3x3+...∞
Notice that the series 3x+3x2+3x3+...∞ is geometric series with the first term a=3x and the common ratio r=x.
Now, use the formula for the sum of an infinite geometric series.
⇒(1−x)S∞=1+3x(1−x), for|x|<1
⇒(1−x)S∞=1+2x(1−x), for|x|<1
Given that, S∞=3516, substitute for S∞ in the above equation; we get
(1−x)3516=1+2x(1−x)
⇒35(1−x)2=16(1+2x)
⇒35(1−2x+x2)=16+32x
⇒35x2−102x+19=0
⇒(7x−19)(5x−1)=0
⇒x=197 or x=15
But x≠197, because for infinity series, |x|<1.
Therefore, x=15.