The relationship between Arithmetic Mean (A), Geometric Mean (G) and Harmonic Mean (H) is:
Answer: B
For two positive numbers the three means satisfy
\[ G^{2} = AH \]
Checking with 4 and 6: \( A = 5 \), \( G = \sqrt{24} \), \( H = 4.8 \), and \( 24 = 5 \times 4.8 \). The order is always \( A \geq G \geq H \).
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If the sum of 5 terms of an A.P. is 75, find the 3rd term.
Answer: B
For an A.P. with an odd number of terms, the sum is equal to the number of terms multiplied by the middle term. Here, the sum of 5 terms is 75. The middle term is the 3rd term. So, \(5 \times a_3 = 75 \Rightarrow a_3 = 75/5 = 15\).
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Find the geometric mean of 4 and 64.
Answer: B
\[ \text{G.M.} = \sqrt{4 \times 64} = \sqrt{256} = 16 \]
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The first term of an A.P. is 5, the last term is 45, and the sum is 400. Find the number of terms.
Answer: C
We use the formula for the sum of an A.P. when the first and last terms are known: \(S_n = \frac{n}{2}(a+l)\). Given \(S_n=400, a=5, l=45\). \(400 = \frac{n}{2}(5+45) \Rightarrow 400 = \frac{n}{2}(50) \Rightarrow 400 = 25n \Rightarrow n = 400/25 = 16\).
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If \(1/4, 1/x, 1/10\) are in H.P., what is the value of x?
Answer: B
If the reciprocals are in H.P., the numbers themselves are in A.P.
\[ 4,\ x,\ 10 \text{ in A.P.} \quad\Longrightarrow\quad x = \frac{4 + 10}{2} = 7 \]
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Find the sum of the series \(96 - 48 + 24 - 12 + ...\) up to infinity.
Answer: A
This is an infinite G.P. with first term \(a=96\) and common ratio \(r = -48/96 = -1/2\). Since \(|r|<1\), the sum to infinity exists. \(S_\infty = \frac{a}{1-r} = \frac{96}{1 - (-1/2)} = \frac{96}{1+1/2} = \frac{96}{3/2} = 96 \times \frac{2}{3} = 32 \times 2 = 64\).
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In a G.P., the first term is 7 and the last term is 448. If the sum of the terms is 889, find the common ratio.
Answer: A
We use the formula for the sum of a G.P.: \(S_n = \frac{lr-a}{r-1}\), where l is the last term. Given \(S_n=889, l=448, a=7\). \(889 = \frac{448r-7}{r-1} \Rightarrow 889(r-1) = 448r-7 \Rightarrow 889r - 889 = 448r - 7 \Rightarrow 441r = 882 \Rightarrow r = 882/441 = 2\).
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The sum of the first 'n' terms of an A.P. is 288. If the first term is 2 and the common difference is 4, find 'n'.
Answer: B
Correct answer: 12
Substitute into the sum formula with \(a = 2\) and \(d = 4\).
\[S_n = \frac{n}{2}\left[2a + (n-1)d\right] = \frac{n}{2}\left[4 + 4n - 4\right] = \frac{n}{2}(4n) = 2n^2\]
Setting this equal to 288:
\[2n^2 = 288 \quad\Longrightarrow\quad n^2 = 144 \quad\Longrightarrow\quad n = 12\]
The negative root is discarded because \(n\) counts terms. Checking: the 12th term is \(2 + 11(4) = 46\), and \(\tfrac{12}{2}(2 + 46) = 6 \times 48 = 288\).
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The sum of the first 'n' terms of an A.P. is 288. If the first term is 2 and the common difference is 4, find 'n'.
Answer: B
We use the formula for the sum of an A.P.: \(S_n = \frac{n}{2}[2a + (n-1)d]\).
Given \(S_n=288, a=2, d=4\).
\(288 = \frac{n}{2}[2(2) + (n-1)4]\)
\(288 = \frac{n}{2}[4 + 4n - 4]\)
\(288 = \frac{n}{2}[4n] = 2n^2\)
\(n^2 = 288/2 = 144\)
\(n = \sqrt{144} = 12\).
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Find the sum of the first 10 terms of the H.P. \(1/2, 1/5, 1/8, ...\)
Answer: D
There is no direct formula to find the sum of a Harmonic Progression. We can find the sum of the corresponding A.P. (2, 5, 8, ...), but this does not give the sum of the H.P. The sum of an H.P. cannot be calculated using a simple standard formula.
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