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Q.1. The harmonic mean of two numbers is 4. Their arithmetic mean A and the geometric mean G satisfy the relation. 2A + G^{2 }= 27
Find the two numbers. (1979)
Ans. 3 and 6 or 6 and 3
Sol. Let the two numbers be a and b, then
Also 2A + G^{2} = 27 ⇒ a + b + ab = 27 ....(2)
Putting ab = 27 – (a+b) in eqn. (1), we get
a + b = 9 then ab = 27 – 9 = 18
Solving the two we get a = 6, b = 3 or a = 3, b = 6, which are the required numbers.
Q.2. Th e in ter ior an gles of a polygon ar e in ar ith metic progression. The smallest angle is 120°, and the common difference is 5°, Find the number of sides of the polygon. (1980)
Ans. 9
Sol. Let there be n sides in the polygon.
Then by geometry, sum of all n interior angles of polygon = (n – 2) × 180° Also the angles are in A.P. with the smallest angle = 120° , common difference = 5° ∴ Sum of all interior angles of polygon
Thus we should have
⇒ 5n^{2} + 235n= 360n 720
⇒ 5n^{2}  125n + 720= 0
⇒ n^{2 } 25n + 144= 0
⇒ (n 16) (n  9)= 0
⇒ n = 16, 9
Also if n = 16 then 16th angle = 120 + 15 × 5 = 195° > 180°
∴ not possible. Hence n = 9.
Q.3. Does there exist a geometric progression containing 27, 8 and 12 as three of its terms ? If it exits, how many such progressions are possible ? (1982  3 Marks)
Ans. yes, infinite
Sol. If possible let for a G.P.
T_{p}= 27 = ARp^{–1} ....(1)
T_{q}= 8 = AR^{q–1} ....(2)
T_{r}= 12 = AR^{r–1} ....(3)
From (1) and (2)
....(4)
From (2) and (3):
....(5)
From (4) and (5):
R = 3/2; p – q = 3 ; q – r = – 1 p – 2q + r = 4; p, q, r ∈ N ....(6)
As there can be infinite natural numbers for p, q and r to satisfy equation (6)
∴ There can be infinite G.P’s.
Q.4. Find three numbers a, b, c, between 2 and 18 such that (i) their sum is 25 (ii) the numbers 2, a, b sare consecutive terms of an A.P. and (iii) the numbers b, c, 18 are consecutive terms of a G.P. (1983  2 Marks)
Ans. a = 5, b = 8, c = 12
Sol. 2 < a, b, c < 18 a + b + c = 25 ....(1)
2, a, b are in AP ⇒ 2a = b + 2 ⇒ 2a – b = 2 ....(2)
b, c, 18 are in GP ⇒ c^{2} = 18b ....(3)
From (2)
+
(3) ⇒ c^{2} = 6 (48  2c) ⇒ c^{2 }+ 12c  288= 0
⇒ c = 12, – 24 (rejected) ⇒ a = 5, b = 8, c = 12
Q. 5. If a > 0, b > 0 and c > 0, prove that
(1984  2 Marks)
Ans. Sol. Given that a, b, c > 0
We know for +ve numbers A.M. ≥ G.M.
∴ For +ve numbers a, b, c we get
....(1)
Also for +ve numbers we get
....(2)
Multiplying in eqs (1) and (2) we get
Proved.
Q.6. If n is a natural number such that
an d p_{1}, p_{2}, ....., p_{k} are distinct primes, then show that ln n ≥ k ln2 (1984  2 Marks)
Ans. Sol. Given that ....(1)
Where n∈N and p_{1}, p_{2}, p_{3}, ......p_{k }are distinct prime numbers.
Taking log on both sides of eq. (1),
we get log n = α_{1} log p_{1}+ α_{2} log p_{2} + ....+ a_{k} log p_{k} ....(2)
Since every prime number is such that
p_{i }≥ 2
....(3)
∀ i= 1 (1) k Using (2) and (3)
we get log n ≥ α_{1} log 2 + α_{2 }log 2 + α_{3} log 2 + ....+ α_{k} log 2
⇒ log n ≥ (α_{1}+α_{2}+ ....+ α_{k}) log 2
⇒ log n ≥ k log 2 Proved.
Q.7. Find the sum of the series :
... up to terms]
Ans. Sol. The given series is
Now,
Similarly,
Hence the given series is,
.... to terms
[ Summing the G.P.]
Q.8. Solve for x the following equation : (1987  3 Marks)
log_{(2 x +3)} (6 x^{2} + 23x + 21) = 4  log _{(3x+7) }(4 x^{2} + 12x+ 9)
Ans. 1/4
Sol. The given equation is log_{ (2x + 3)} (6x^{2} + 23 + 21)
= 4 – log_{3x+7} (4x^{2} + 12x + 9)
⇒ log_{(2x+3)} (6x^{2} + 23x + 21) + log_{(3x+7)} (4x^{2} + 12x + 9) = 4
⇒ log_{(2x+3)} (2x + 3) (3x + 7) + log_{(3x+7)} (2x + 3)^{2} x =4
⇒ 1+log_{(2x+3) }(3x + 7) + 2log_{(3x+7)} (2x + 3) =4
⇒ log_{(2x+3)}(3x + 7) +
Let log_{(2x+3)}(3x + 7) = y ....(1)
y^{2} 3y+2=0
⇒ (y – 1) (y – 2) = 0 ⇒ y = 1, 2
Substituting the values of y in (1), we get
⇒ log_{(2x+3)}(3x + 7) = 1 and log_{(2x+3)}(3x + 7) = 2
⇒ 3x + 7 = 2x + 3 an d3x + 7 = (2x + 3)^{2 }
⇒ x = – 4 an d4x^{2} + 9x + 2 = 0
⇒ x = – 4 an d(x + 2) (4x + 1) = 0
⇒ x = – 4 and x = – 2, x = –
As log a x is defined for x > 0 and a > 0 (a ≠ 1), the possible value of x should satisfy all of the following inequalities :
⇒ 2x + 3 > 0 and 3x + 7 > 0
Also 2x + 3 ≠ 1 and 3x + 7 ≠ 1
Out of x = – 4, x = – 2 an d x = – only x = –
satisfies the above inequalities.
So only solution is x = –
Q. 9. If log 3^{ 2} , log_{3} (2^{ x} , 5) , and are in arithmetic progression, determine the value of x. (1990  4 Marks)
Ans. 3
Sol. Given that log_{3 }2, log3 (2^{x }– 5), log3 (2^{x} –7/2) are in A.P.
⇒ 2 log_{3} (2^{x} – 5) = log_{3} ^{2 }+ log_{3} (2x – 7/2)
⇒ (2^{ x}  5)^{ 2}
⇒ (2^{x})^{2} – 10.2^{x }+ 25 – 2.2^{x} + 7 = 0 ⇒ (2^{x})^{2} – 12.2^{x }+ 32 = 0
Let 2^{x} = y, then we get, y^{2} – 12y + 32 = 0
⇒ (y – 4) (y – 8) = 0
⇒ y = 4 or 8
⇒ 2x = 22 or 23 ⇒ x = 2 or 3
But for log_{3 }(2^{x} – 5) and log_{3} (2^{x }– 7/2) to be defined
2^{x} – 5 > 0 and 2^{x} – 7/2 > 0 ⇒ 2^{x} > 5 and 2^{x } > 7/2
⇒ 2^{x }> 5 ⇒ x ≠ 2 and therefore x = 3.
Q.10. Let p be the first of the n arithmetic means between two numbers and q the first of n harmonic means between the same numbers. Show that q does not lie between p and
(1991  4 Marks)
Ans. Sol. Let a and b be two numbers and A_{1}, A_{2}, A_{3}, .... A_{n} be n A.M’s between a and b.
Then a, A_{1}, A_{2}, ..... A_{n}, b are in A.P.
There are (n + 2) terms in the series, so that
a + (n + 1) d = b
....(1)
The first H.M. between a and b, when nHM’s are inserted between a and b can be obtained by replacing a by and b by in eq. (1) and then taking its reciptocal.
Therefore,
...(2)
We have to prove that q cannot lie between p
Now, n + 1>n –1
....(3)
Now to prove the given, we have to show that q is less than p.
For this, let,
⇒
⇒
⇒
⇒
⇒ (provided a and b and hence p and q are +ve) p > q ....(4)
From 3 and (4), we get,
∴ q can not lie between p and p, if a and b are +ve numbers.
Q.11. If S_{1}, S_{2} ,S_{3}, ............... , S_{n} are the sums of infinite geometric series whose first terms are 1, 2, 3, ..............., n and whose common ratios are
respectively,,then find the values of (1991  4 Marks)
Ans. Sol. We have,
....................................................
....................................................
NOTE THIS STEP :
Q.12. The real numbers x_{1}, x_{2}, x_{ 3} s atis fying the equation x^{3 }, x^{2} + βx + γ < 0 are in AP. Find the intervals in which β and γ lie. (1996  3 Marks)
Ans. Sol. Since x_{1}, x_{2}, x_{3} are in A.P.
Therefore, let x_{1 }= a – d, x_{2 }= a and x_{3} = a + d and x_{1}, x_{2}, x_{3} are the roots of x_{3} – x_{2} + βx + γ =0
We have ∑α = a – d + a + a + d = 1 ....(1)
∑αβ = (a – d) a + a ( a + d) + (a – d) (a + d) = b ....(2)
αβγ = (a – d) a (a + d) = – γ ....(3)
From (1), we get, 3a = 1 ⇒ a = 1/3
From (2), we get, 3a^{2} – d^{2} = β
⇒ 3(1/3)^{2} – d^{2} = β ⇒ 1/3 – β = d^{2 }
We know that d^{2 }> 0 ∀ d ∈ R
∵ d 2 ≥ 0
From (3), a (a^{2} – d^{2}) = – γ
Q.13. Let a, b, c, d be real numbers in G.P. If u, v, w, satisfy the system of equations (1999  10 Marks)
then show that the roots of the equation
and 20x^{2 }+ 10 (a  d)^{2} x  9 = 0 are reciprocals of each other.
Ans.
Sol. Solving the system of equations, u + 2v + 3w = 6,
4u + 5v + 6w = 12 and 6u + 9v = 4
we get u = – 1/3, v = 2/3, w = 5/3
Let r be the common ratio of the G.P., a, b, c, d. Then b = ar, c = ar^{2}, d = ar^{3} .
Then the first equation
+ [(b – c)^{2 }+ (c – a)^{2 }+ (d – b)^{2}]x + (u + v + w) = 0 becomes
The second equation is, 20 x^{2} + 10 (a – ar^{3})^{2} x – 9 = 0
i.e., 20 x^{2} + 10 a^{2 } (1 – r^{3})^{2 }x – 9 = 0 ....(2)
Since (2) can be obtained by the substitution x → 1/x , equations (1) and (2) have reciprocal roots.
Q.14. The fourth power of the common difference of an arithmatic progression with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer. (2000  4 Marks)
Ans.
Sol. Let a – 3d, a – d, a + d and a + 3d be any four consecutive terms of an A.P. with common difference 2d.
∵ Terms of A.P. are integers, 2d is also an integer.
Hence P = (2d)^{4} + (a – 3d) (a – d) (a + d) (a + 3d)
=16 d^{ 4} + (a^{ 2} – 9 d^{ 2}) (a^{ 2} – d^{ 2}) = (a^{ 2} – 5 d^{ 2})^{2}
Now, a^{ 2} – 5 d^{ 2} = a^{ 2} – 9 d^{ 2} + 4 d^{ 2} = (a – 3 d) (a + 3 d) + (2 d)^{2} = some integer
Thus, P = square of an integer.
Q.15. Let a_{1}, a_{2}, …, a_{n} be positive real numbers in geometric progression. For each n, let A_{n}, G_{n}, H_{n }be respectively, the arithmetic mean, geometric mean, and harmonic mean of a_{1}, a_{2}, …, a_{n}. Find an expression for the geometric mean of G_{1}, G_{2}, …, G_{n} in terms of A_{1}, A_{2} , …, A_{n}, H_{1}, H_{2}, …, H_{n}. (2001  5 Marks)
Ans.
Sol. Given that a_{1}, a_{2}, ....a_{n} are +ve real no’s in G.P.
A_{n} is A.M. of a_{1}, a_{2}, ....., a_{n}
.... (1) (For r ≠ 1)
G_{n} is G.M. of a_{1}, a_{2}, ...., a_{n}
.... (2) (r ≠ 1)
H_{n }is H.M. of a_{1}, a_{2}, ....., a_{n}
(r ≠ 1) ...(3)
We also observe that
∴ A_{n }H_{n} =G_{n}^{2}....(4)
∴ Now, G.M. of G_{1}, G_{2}, .... G_{n} is
[Using equation (4)
G = (A_{1}A_{2}....A_{n}H_{1}H_{2}....H_{n})1/2n ...(5)
If r = 1 then A_{n} = G_{n} = H_{n} = a
Also A_{n} H_{n }= G_{n}^{2}
∴ For r = 1 also, equation (5) holds.
Hence we get, G = (A_{1}A_{2}....A_{n}H_{1}H_{2}....H_{n})^{1/2n}
Q.16. Let a, b be positive real numbers. If a, A_{1}, A_{2}, b are in arithmetic progression, a, G_{1}, G_{2}, b are in geometric progression and a, H_{1}, H_{2}, b are in harmonic progression,
show that (2002  5 Marks)
Ans.
Sol. Clearly A_{1} + A_{2 }= a + b
Q.17. If a, b, c are in A.P., a^{2}, b^{2}, c^{2} are in H.P., then prove that either a = b = c or a, b, form a G.P.. (2003  4 Marks)
Ans.
Sol. Given that a, b, c are in A.P.
⇒ 2b = a + c ....(1)
and a^{2}, b^{2}, c^{2 }are in H.P.
⇒ ac^{2} + bc^{2 }= a^{2}b + a^{2}c [∵ a – b = b – c]
⇒ ac (c – a) + b (c – a) (c + a) = 0
⇒ (c – a) (ab + bc + ca) = 0
⇒ either c – a = 0 or ab + bc + ca = 0
⇒ either c = a or (a + c) b + ca = 0 and then from (i)
2b^{2} + ca = 0
Either a = b = c
i.e. a, b, – c/2 are in G.P. Hence Proved.
Q.18. If a_{n}and
b_{n} = 1 – a_{n}, then find the least natural number n_{0} such that
(2006  6M)
Ans.
Sol.
⇒ (– 3)^{n +1} < 2^{2n–1} For n to be even, inequality always holds. For n to be odd, it holds for n ≥ 7.
∴ The least natural no., for which it holds is 6
(∵ it holds for every even natural no.)
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