Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

Answer:

Answer:x, y,z are in HP

Step-by-step explanation:

Solution is given in attachment

Answer 2

$\displaystyle \tt \: x = \sum_{n = 0}^{ \infty} {a}^{n} \\ \\ \longrightarrow \tt x = {a}^{0} + {a}^{1} + {a}^{2} + {a}^{3}{...{ \infty}}\\ \\ \longrightarrow \tt x = \dfrac{1}{1 - a} \\ \\ \displaystyle \tt \: y = \sum_{n = 0}^{ \infty} {b}^{n}\\ \\ \longrightarrow \tt y = {b}^{0} + {b}^{1} + {b}^{2} + {b}^{3}{...{ \infty}}\\ \\ \longrightarrow \tt y = \dfrac{1}{1 - b} \\ \\ \displaystyle \tt \: z= \sum_{n = 0}^{ \infty} {c}^{n} \\ \\ \longrightarrow \tt z = {c}^{0} + {c}^{1} + {c}^{2} + {c}^{3}{...{ \infty}}\\ \\ \longrightarrow \tt z = \dfrac{1}{1 - c}$

Now , it is given that a,b,c are in A.P.

So (1-a) , (1-b) and (1-c) are also in A.P.

$\tt \therefore \dfrac{1}{1 - a} , \dfrac{1}{1 - b} \: \rm{and } \: \tt{ \dfrac{1}{1 - c}} \:\rm{ are \: in \: H.P}$

Answer :-

Option (a)H.P is correct.