Question

If I_n = dx, then I₁, I₂, I₃,…..are in –

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

Answer: (A)

A.P.

Answer 2

We have In =dx

\ I_n + I_n+2 – 2I_n+1

=$\int _ { 0 } ^ { \pi }$ $\frac { ( 1 - \sin 2 n x ) + ( 1 - \sin ( 2 n + 4 ) x ) - 2 ( 1 - \sin ( 2 n + 2 ) x ) } { 1 - \cos 2 x }$dx

=$\int _ { 0 } ^ { \pi }$ $\frac { 2 \sin ( 2 n + 2 ) x - ( \sin ( 2 n + 4 ) x + \sin 2 n x ) } { 1 - \cos 2 x }$dx

=$\frac { 2 \sin ( 2 n + 2 ) x - 2 \sin ( 2 n + 2 ) x \cdot \cos 2 x } { 1 - \cos 2 x }$dx

2$\int _ { 0 } ^ { \pi }$sin(2n + 2)x dx = $\frac { - 2 \cos ( 2 n + 2 ) x } { 2 n + 2 }$

= – $\frac { 1 } { \mathrm { n } + 1 }$ [cos (n + 1) 2p – cos 0] = – $\frac { 1 } { \mathrm { n } + 1 }$ [1 – 1] = 0.

\ I_n, I_n+1, I_n+2 are in A.P.

\ I₁, I₂, I₃,…are in A.P.