Test: Sum To Infinity Of A GP - JEE MCQ

 6 + 6 + 6.........∞
6^1/2 * 6^1/4 * 6^1/8………..∞
= 6^{1/2 + 1/(2*2)} + 1/(2*2*2)   (sum of infinte G.P.= a/(1−r))
= (1/2)/(6^(1-½))
= (1/2)/(6^1/2)
= 6

 (1 + 1/5²) + (1/2 + 1/5²) + (1/2² + 1/54)+......
= (1 + 1/2 + 1/2² +....) + (1/5² + 1/5⁴ + 1/5⁶ +......)
= r = (1/4)/(1/2) , r = (1/5⁴)/(1/5²)
⇒ r = 1/2 + 1/25
⇒ S = 1 + [(1/2)/(1-1/2)],    S = [(1/25)/(1-1/25)]
⇒ S = 1 + 1,    S = 1/24
Total sum = 2 + 1/24
⇒ 49/24 

 x  = a + a/r  +  a/r² + .....................+ ∞
First term = a  , Common Ratio = 1/r
=> x =  a/(1 - 1/r)  =  ra/(r - 1)
y  = b - b/r  + b/r² + .....................+ ∞
 
First term = b  , Common Ratio = -1/r
=> y =  b/(1 - (-1/r))  =  rb/(r + 1)
z=c+c/r²+c/r⁴+..........................+ ∞
 
First term = c  , Common Ratio = 1/r²
z = c/(1 - (1/r²) = r²c/(r² - 1)
 
LHS = xy/z    
 
=   (ra/(r - 1) )( rb/(r + 1)) / ( r²c/(r² - 1))
= r²ab(r² - 1)/ r²c/(r² - 1)
= ab/c

Therefore

S(infinity) = a/(1-r)
80/9 = a/(1 + 4/5)
80/9 = a/(9/5)
a = 16