Test: MCQs (One or More Correct Option): Sequences and Series | JEE Advanced - JEE MCQ

 Let x be the first term and y the (2n–1)th terms of AP, GP and HP whose nth terms are a, b, c  respectively.
For AP, y = x + (2n – 2) d

  ....(1)

For  G.P. 

....(2)

For H.P.

....(3)
Thus from (1), (2) and (3), a, b, c are A.M., G.M. and H.M. respectively of x and y

 We have for 

                  ....(1)

[Using sum of infinite G.P. cos²α being < 1]

....(2)

...(3)

Substituting the values of cos²φ and sin²φ in (3), from (1) and (2), we  get

Thus (b) and (c) both are correct.

Putting  θ = 0, we get b₀= 0

= b¹ + b² sinθ+ b³ sin ²θ + ...... +b_n sin ^n-1θ

Taking limit as θ → 0, we obtain

If   x, y, z are in G.P. (x,  y,  z  > 1); log x, log y, log z will be in A.P.
⇒ 1 + log x, 1 + log y, 1 + log z will also be in A.P.

will be in H.P..

We have

Thus, a (100) < 100

Also

Thus, a (200) > 

i.e. a (200) > 100.

We know by geometry   PS × ST = QS × SR ...(1)

∵  S is not the centre of circulm circle,

PS ≠ ST
And we know that for two unequal real numbers.
H.M. < G.M

[using eqn (1)]    ...(2)

∴ (b) is the correct option.

Also 

From equations (2) and (3) we get  

∴ (d) is also the correct option.

We have 

and 

For n = 1 we get

Also  = 0.34 × 1.73 = 0.58

= 8n² + 8n² + 4n = 16n² + 4n
For n = 8,  16n² + 4n = 1056
and for n = 9, 16n² + 4n = 1332