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If the first and the (2n – 1)st terms of an A.P., a G.P. and an H.P. are equal and their n-th terms are a, b and c respectively, th en (1988 - 2 Marks)
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
Thus from (1), (2) and (3), a, b, c are A.M., G.M. and H.M. respectively of x and y
For 0 < φ < π/2, if then: (1993 - 2 Marks)
We have for
[Using sum of infinite G.P. cos2α being < 1]
Substituting the values of cos2φ and sin2φ in (3), from (1) and (2), we get
Thus (b) and (c) both are correct.
Let n be an odd integer. If for every value of q, then (1998 - 2 Marks)
Putting θ = 0, we get b0= 0
= b1 + b2 sinθ+ b3 sin 2θ + ...... +bn sin n-1θ
Taking limit as θ → 0, we obtain
Let Tr be the rth term of an A.P., for r = 1, 2, 3, .... If for some positive integers m, n we have equals (1998 - 2 Marks)
If x > 1, y > 1, z > 1 are in G.P., then are in (1998 - 2 Marks)
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..
For a positive integer n, let . Then (1999 - 3 Marks)
Thus, a (100) < 100
Thus, a (200) >
i.e. a (200) > 100.
A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle, then (2008)
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.
From equations (2) and (3) we get
∴ (d) is also the correct option.
Let = 1, 2, 3, ............ Then, (2008)
For n = 1 we get
Also = 0.34 × 1.73 = 0.58
Let Then Sn can take value(s) (JEE Adv. 2013)
= 8n2 + 8n2 + 4n = 16n2 + 4n
For n = 8, 16n2 + 4n = 1056
and for n = 9, 16n2 + 4n = 1332