Find the 10th term of the Arithmetic Progression (A.P.) 2, 7, 12, ...
Answer: A
In this A.P., the first term \(a = 2\) and the common difference \(d = 7 - 2 = 5\). The formula for the nth term is \(a_n = a + (n-1)d\). For the 10th term (n=10), we have \(a_{10} = 2 + (10-1) \times 5 = 2 + 9 \times 5 = 2 + 45 = 47\).
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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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Find the sum of the series \(1 + 1/3 + 1/9 + 1/27 + ...\) to infinity.
Answer: B
This is an infinite G.P. with first term \(a=1\) and common ratio \(r = (1/3)/1 = 1/3\). The sum is \(S_\infty = \frac{a}{1-r} = \frac{1}{1-1/3} = \frac{1}{2/3} = 3/2\).
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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 \(a_4\) is the difference between the sum of the first four terms (\(S_4\)) and the sum of the first three terms (\(S_3\)). \(a_4 = S_4 - S_3 = 20 - 12 = 8\).
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The geometric mean between two numbers is 12. If one number is 9, what is the other number?
Answer: B
Let the numbers be a and b. G.M. = \(\sqrt{ab} = 12\), so \(ab = 144\). Given one number a=9, we have \(9b = 144\), so \(b = 144/9 = 16\). The other number is 16.
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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 sum of an infinite geometric series is 15 and the sum of the squares of these terms is 45. Find the series.
Answer: A
Let the series be \(a, ar, ar^2, ...\). Sum \(\frac{a}{1-r} = 15\). The series of squares is \(a^2, a^2r^2, a^2r^4, ...\) which is a G.P. with first term \(a^2\) and common ratio \(r^2\). Its sum is \(\frac{a^2}{1-r^2} = 45\). We can write this as \(\frac{a^2}{(1-r)(1+r)} = 45\). Substituting \(a=15(1-r)\) is complex. Let's use \(\frac{a}{1-r} \times \frac{a}{1+r} = 45 \Rightarrow 15 \times \frac{a}{1+r} = 45 \Rightarrow \frac{a}{1+r} = 3 \Rightarrow a = 3(1+r)\). Now we have \(15(1-r) = 3(1+r) \Rightarrow 5(1-r)=1+r \Rightarrow 5-5r=1+r \Rightarrow 4=6r \Rightarrow r=2/3\). Then \(a = 15(1-2/3) = 15(1/3)=5\). The series is 5, 10/3, 20/9, ...
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If the second term of an H.P. is 1/5 and the 6th term is 1/13, what is the 10th term?
Answer: B
The corresponding A.P. has a 2nd term of 5 and a 6th term of 13. So, \(a+d=5\) and \(a+5d=13\). Subtracting gives \(4d=8\), so \(d=2\). Substituting d=2 gives \(a+2=5\), so \(a=3\). The 10th term of the A.P. is \(a_{10} = a+9d = 3+9(2) = 3+18=21\). The 10th term of the H.P. is the reciprocal, 1/21.
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