The angles of a quadrilateral are in A.P. with a common difference of 10 degrees. Find the smallest angle.
Answer: B
Let the angles be \(a, a+10, a+20, a+30\). The sum of angles in a quadrilateral is 360°. So, \(a + (a+10) + (a+20) + (a+30) = 360 \Rightarrow 4a + 60 = 360 \Rightarrow 4a = 300 \Rightarrow a = 75\). The smallest angle is 75°.
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The geometric mean between two numbers is 12. If one number is 9, what is the other number?
Answer: B
\[ \text{G.M.} = \sqrt{ab} \quad\Longrightarrow\quad 12 = \sqrt{9b} \]
\[ 144 = 9b \quad\Longrightarrow\quad b = 16 \]
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In an A.P., the sum of first four terms is 20 and the sum of first three terms is 12. Find the fourth term.
Answer: B
The fourth term is the difference of the two sums.
\[ 20 - 12 = 8 \]
No property of the progression is needed - only that the sums are of consecutive terms.
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Find the sum of the series \(1 + 1/3 + 1/9 + 1/27 + ...\) to infinity.
Answer: B
An infinite geometric series with \( |r| < 1 \) sums to
\[ S = \frac{a}{1-r} = \frac{1}{1 - 1/3} = \frac{1}{2/3} = \frac{3}{2} \]
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The fourth term of an A.P. is 11 and the seventh term is 20. Find the 15th term.
Answer: B
We have \(a+3d=11\) and \(a+6d=20\). Subtracting the equations gives \(3d=9\), so \(d=3\). Substituting back, \(a+3(3)=11 \Rightarrow a+9=11 \Rightarrow a=2\). The 15th term is \(a_{15} = a+14d = 2 + 14(3) = 2+42=44\).
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If \(x, 2x+2, 3x+3\) are in G.P., then the 4th term is:
Answer: A
If the terms are in G.P., the ratio of consecutive terms is constant. So, \(\frac{2x+2}{x} = \frac{3x+3}{2x+2}\). \((2x+2)^2 = x(3x+3) \Rightarrow 4x^2+8x+4 = 3x^2+3x \Rightarrow x^2+5x+4=0 \Rightarrow (x+1)(x+4)=0\). So \(x=-1\) or \(x=-4\). If x=-1, terms are -1, 0, 0 (not a G.P.). If x=-4, terms are -4, -6, -9. This is a G.P. with \(r = -6/-4 = 3/2\). The 4th term is \(-9 \times (3/2) = -27/2 = -13.5\).
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The sum of the first and third term of a G.P. is 20 and the sum of the first three terms is 26. Find the common ratio.
Answer: D
Let the terms be \(a, ar, ar^2\). Given \(a+ar^2=20 \Rightarrow a(1+r^2)=20\). And \(a+ar+ar^2=26 \Rightarrow a(1+r+r^2)=26\). Dividing the second eq by the first: \(\frac{1+r+r^2}{1+r^2} = \frac{26}{20} = \frac{13}{10}\). \(10+10r+10r^2 = 13+13r^2 \Rightarrow 3r^2-10r+3=0 \Rightarrow (3r-1)(r-3)=0\). So \(r=3\) or \(r=1/3\).
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How many three-digit numbers are divisible by 7?
Answer: A
The three-digit numbers are from 100 to 999. First three-digit number divisible by 7 is 105. Last is 994. This is an A.P. with \(a=105, d=7, a_n=994\). We use \(a_n = a+(n-1)d\). \(994 = 105+(n-1)7 \Rightarrow 889 = (n-1)7 \Rightarrow n-1 = 127 \Rightarrow n=128\). There are 128 such numbers.
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The product of three numbers in a G.P. is 216 and the sum of their products in pairs is 156. Find the numbers.
Answer: A
Let the numbers be \(a/r, a, ar\). Product is \((a/r)(a)(ar) = a^3 = 216\), so \(a=6\). Sum of products in pairs: \((a/r)(a) + a(ar) + (a/r)(ar) = a^2/r + a^2r + a^2 = 156\). Substitute a=6: \(36/r + 36r + 36 = 156 \Rightarrow 36/r+36r=120\). Divide by 12: \(3/r+3r=10 \Rightarrow 3+3r^2=10r \Rightarrow 3r^2-10r+3=0 \Rightarrow (3r-1)(r-3)=0\). So \(r=3\) or \(r=1/3\). If r=3, numbers are 2, 6, 18. If r=1/3, numbers are 18, 6, 2.
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Find the sum of all integers between 100 and 1000 that are divisible by 9.
Answer: A
The numbers are 108, 117, ..., 999. This is an A.P. with \(a=108, d=9, a_n=999\). Number of terms: \(999=108+(n-1)9 \Rightarrow 891=(n-1)9 \Rightarrow n-1=99 \Rightarrow n=100\). Sum = \(\frac{100}{2}(108+999) = 50(1107) = 55350\).
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