If x, y and z are pth, qth and rth terms respectively of an A.P. and also of a G.P., then x^{y} – ^{z} y^{z} – ^{x} z^{x – y} is equal to :
∵ x, y, z are the p^{th}, q^{th} and r^{th} terms of an AP..
∴ x = A + (p – 1)D; y = A + (q – 1) D; z = A + (r – 1) D
⇒ x – y = (p – q)D; y – z = (q – r) D z – x = (r – p) D ....(1)
where A is the first term and D is the common difference.
Also x, y, z are the p^{th}, q^{th}, and r^{th} terms of a GP.
∴ x = AR^{p – 1}, y = AR^{q – 1}, z = AR^{r – 1 }
∴ x^{y – z} y^{z – x} z^{x –y} = (AR^{p –1})^{y–z }(AR^{q–1})^{z–x} (AR^{r–1})^{x–y }
= A^{y–z+z–x+x–y}R^{(p–1)(y–z)+(q–1)(z–x)+(r–1)(x–y) }
= A^{0}r^{(p–1) (q–r) D+(q–1)(r–p)D+(r–1)(p–q)D }= A^{0}R^{0} =1
The third term of a geometric progression is 4. The product of the first five terms is (1982  2 Marks)
ar^{2} = 4 ....(1)
a. ar. ar^{2} .ar^{3} .ar^{4} = a^{5} r^{10} = (ar^{2})^{5} = 45.
The rational number, which equals the number with recurring decimal is (1983  1 Mark)
= 2 + .357 + 0.000357 + ...∞
If a, b, c are in G.P., then the equations ax^{ 2 }+ 2bx + c = 0 and dx^{ 2} + 2ex + f = 0 have a common root if are in –– (1985  2 Marks)
a, b, c are in G.P. b^{2} = ac ....(1)
ax^{2} + 2bx + c = 0 and dx^{2} + 2ex + f = 0 have a common root
Let it be α, then aα^{2 }+ 2bα + c = 0 dα^{2} + 2eα + f = 0
⇒
⇒
Substituting the value of α , we get
⇒ (cd – af) ^{2} = 4 (ae – bd) (bf – ce)
Dividing both sides by a^{2}c^{2} we get
[Using eq. (1)]
⇒
⇒
[Using eq.(1)]
⇒
⇒ are in A.P..
Sum of the first n terms of the series....... is equal to (1988  2 Marks)
Let S n terms
..n terms
= (1 + 1 + 1 + ....n terms)
= = n  1+2^{n}
The number log_{ 2} 7 is (1990  2 Marks)
We know that log_{2} 4 = 2 and log_{2} 8 = 3
∴ log_{2} 7 lies between 2 and 3
∴ log_{2} 7 is either rational or irrational but not integer or prime number.
If possible let log_{2} (a rational number)
⇒ 2^{p/q }= 7
⇒ 2^{p }= 7^{q }
⇒ even number = odd number
∴ We get a contradiction, so assumption is wrong.
Hence log_{2} 7 must be an irrational number.
If ln(a + c), ln (a – c), ln (a – 2b + c) are in A.P., then (1994)
In (a + c), In (a – c), In (a – 2b + c) are in A.P.
⇒ a + c, a – c, a – 2b + c are in G.P.
⇒ (c – a)^{2} = (a + c) (a –2b + c)
⇒ (c – a)^{2} = (a + c)^{2 }– 2b (a + c)
⇒ 2b (a + c) = (a + c)2 – (c – a)^{2}
⇒ 2b (a + c) = 4ac ⇒
⇒ a, b, c are in H.P
Let a_{1}, a_{2}, ..... a_{10 }be in A, P, and h_{1}, h_{2},....h_{10} be in H.P. If a_{1} = h_{1} = 2 and a_{10} = h_{10 }= 3, then a_{4}h_{7} is (1999  2 Marks)
a_{1}= h_{1}= 2, a_{10} = h_{10} = 3
3 = a_{10} = 2 + 9d ⇒ d = 1/9
∴ a_{4} = 2 + 3d = 7/3
The harmonic mean of the roots of the equation is (1999  2 Marks)
∴ H = 4.
Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second term is 3/4, then (2000S)
Sum = 4 and second term = 3/4, it is given that first term is a and common ratio r
⇒ = 4 and ar = 3 /4 ⇒
Therefore,
or a^{2} – 4a + 3 = 0 ⇒ (a – 1) (a – 3) = 0
⇒ a = 1 or 3
When a = 1, r = 3/4 and when a = 3, r = 1/4
Let α, β be the roots of x^{2}  x + p = 0 and γ, δ be the roots of x^{2 } 4x + q = 0. If α, β, γ, δ are in G.P., then the integral values of p and q respectively, are (2001S)
α, β are the roots of x^{2} – x + p=0
∴ α + β = 1 ....(1)
αβ = p ...(2)
g,δ are the roots of x^{2} – 4x + q = 0
∴ γ + δ = 4....(3)
γδ = q ....(4)
α,β, γ,δ are in G.P..
∴ Let α = a; β = ar, γ = ar^{ 2} , δ = ar^{3} .
Substituting these values in equations (1), (2), (3) and (4), we get
a + ar = 1 ....(5)
a^{2}r = p ....(6)
ar^{2} + ar^{3 }= 4 ....(7)
a^{2}r^{5} = q ....(8)
Dividing (7) by (5) we get
⇒
(5) ⇒ or
As p is an integer (given), r is also an integer (2 or – 2).
∴ (6) ⇒ Hence a = 1 and r = 2.
∴ p = (– 1)^{2} × (– 2) = – 2
q = (– 1)^{2} × (– 2)^{5} = – 32
Let the positive numbers a, b, c, d be in A.P. Then abc, abd, acd, bcd are (2001S)
a, b, c, d are in A.P.
∴ d, c, b, a are also in A.P.
⇒ are also in A.P..
⇒ are in A.P..
⇒ abc, abd, acd, bcd are in H.P.
If the sum of the first 2n terms of the A.P. 2, 5, 8, …, is equal to the sum of the first n terms of the A.P. 57, 59, 61, …, then n equals(2001S)
ATQ 2 + 5 + 8 + .... 2n terms = 57 + 59 + 61 + .... n terms
⇒
⇒ 6n + 1= n + 56 ⇒ 5n = 55 ⇒ n = 11
Suppose a, b, c are in A.P. and a^{2}, b^{2}, c^{2 }are in G.P. if a < b < c and a + b + c =, then the value of a is (2002S)
Given that a, b, c are in A.P.
⇒ 2b = a + c
⇒ but given a + b + c = 3/2
⇒ 3b = 3/2
⇒ b = 1/2 and then a + c = 1
Again a^{2}, b^{2}, c^{2}, are in G.P. ⇒ b^{4} = a^{2} c^{2}
⇒
and a + c = 1 ....(1) Considering a + c = 1 and ac = 1/4
⇒ (a – c)^{2} = 1 – 1 = 0 ⇒ a = c
but a ≠ c as given that a < b < c
∴ We consider a + c = 1 and ac = – 1/4
⇒ (a – c)^{2} = 1 + 1= 2 ⇒
but a < c ⇒ ....(2)
Solving (1) and (2) we get
An infinite G.P. has first term ‘x’ and sum ‘5’, then x belongs to (2004S)
Since G.P. contains infinite terms
∴ – 1 < r < 1
⇒ 0 < x< 10.
In the quadratic equation ax^{2} + bx + c = 0, Δ = b^{2} – 4ac and α + β, α^{2} + β^{2}, α^{3 }+ β^{3}, are in G.P. where a, b are the root of ax^{2 }+ bx + c = 0, then (2005S)
In the quadratic equation ax^{2} + bx + c = 0
Δ = b^{2}  4ac and
and
Given are in G.P.
⇒ are in G.P.
⇒
⇒ b^{4} + 4a^{2}c^{2 }– 4ab^{2}c = b^{4} – 3ab^{2}c
⇒ 4a^{2}c^{2 }– ab^{2}c = 0 ⇒ ac Δ = 0
In the sum of first n terms of an A.P. is cn^{2}, then the sum of squares of these n terms is (2009)
Given that for an A.P.S_{n }= cn^{2}
Then T_{n} = S_{n}S_{n1} = cn^{2}  c(n 1)^{2} = (2n 1)c
∴ Sum of squares of n terms of this A.P
Let a_{1}, a_{2}, a_{3}, ..... be in harmonic progression with a_{1} = 5 and a_{20} = 25. The least positive integer n for which an < 0 is (2012 )
∵ a_{1}, a_{2}, a_{3}, ....... are in H.P..
.. are in A.P..
Now
Clearly a_{n} < 0 if
⇒ – 4n < – 99 or
Hence least value of n is 25.
Let b_{i} > 1 for i = 1, 2, ..., 101. Suppose loge b_{1}, loge b_{2}, ...., log_{e} b_{101} are in Arithmetic Progression (A.P.) with the common difference log_{e }2. Suppose a_{1}, a_{2}, ...., a_{101} are in A.P. such that a_{1}= b_{1} and a_{51}= b_{51}. If t= b_{1}+b_{2} + .... + b_{51} and s= a_{1}+a_{2}+ .... + a_{53}, then (JEE Adv. 2016)
log_{c} b_{1}, log_{c }b_{2},, log_{c}b_{101} are in A.P.
⇒ b_{1}, b_{2}, b_{101} are in G.P.
Also a_{1}, a_{2}, a_{101} are in A.P. where a_{1} = b_{1} are a_{51} = b_{51}.
∴ b_{2}, b_{3},,b_{50} an d GM’s and a_{2}, a_{3},, a_{50} are AM’s between b_{1} and b_{51}.
∵ GM < AM ⇒ b_{2} < a_{2}, b_{3 }< a_{3},, b_{50} < a_{50 }
∵ b_{1} + b_{2} ++ b_{51 }< a_{1} + a_{2} ++ a_{51 }
⇒ t < s Also a_{1}, a_{51}, a_{101} are in AP b_{1}, b_{51}, b_{101} are in GP
∵ a_{1} = b_{1 }and a_{51} = b_{51}
∴ b_{101} > a_{101}
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