Progressions (A.P., G.P., H.P.) Questions and Answers

Answer: B

The nth term of an A.P. can be found by \(a_n = S_n - S_{n-1}\). First term \(a_1 = S_1 = 3(1)^2 + 5(1) = 8\). Sum of first two terms \(S_2 = 3(2)^2 + 5(2) = 12 + 10 = 22\). The second term \(a_2 = S_2 - S_1 = 22 - 8 = 14\). The common difference \(d = a_2 - a_1 = 14 - 8 = 6\). Alternatively, for a sum in the form \(An^2+Bn\), the common difference is \(2A\). Here A=3, so \(d=2 \times 3=6\).

Answer: B

Let the first five terms of the G.P. be \(a/r^2, a/r, a, ar, ar^2\). The third term is \(a=4\). The product of these five terms is \((a/r^2) \times (a/r) \times a \times (ar) \times (ar^2) = a^5\). Since a=4, the product is \(4^5 = 1024\).

Answer: A

The nth term of a series is the difference between the sum of the first n terms and the sum of the first n-1 terms. \(a_n = S_n - S_{n-1}\). Given \(S_n = 210\) and \(S_{n-1} = 171\). Therefore, \(a_n = 210 - 171 = 39\).

Answer: A

Let the numbers be a and b. A.M. = \(\frac{a+b}{2} = 10 \Rightarrow a+b=20\). G.M. = \(\sqrt{ab} = 8 \Rightarrow ab=64\). We need two numbers whose sum is 20 and product is 64. The numbers are 4 and 16. We can solve this using the quadratic equation \(x^2 - (sum)x + (product) = 0 \Rightarrow x^2 - 20x + 64 = 0 \Rightarrow (x-4)(x-16)=0\).

Answer: B

The first 20 odd natural numbers form an A.P.: 1, 3, 5, ..., 39. The sum of the first n odd numbers is \(n^2\). So the sum is \(20^2=400\). The arithmetic mean is Sum/Number of terms = \(400/20 = 20\). Alternatively, for an A.P., the mean is the average of the first and last term: \((1+39)/2 = 20\).

Answer: A

Let the A.P. have first term 'a' and common difference 'd'. \(a_p = a+(p-1)d = q\) and \(a_q = a+(q-1)d = p\). Subtracting the two equations gives \((p-1-q+1)d = q-p \Rightarrow (p-q)d = -(p-q) \Rightarrow d = -1\). Substituting d=-1 into the first equation: \(a+(p-1)(-1)=q \Rightarrow a-p+1=q \Rightarrow a = p+q-1\). The nth term is \(a_n = a+(n-1)d = (p+q-1) + (n-1)(-1) = p+q-1-n+1 = p+q-n\).

Answer: A

The two-digit numbers divisible by 3 are 12, 15, ..., 99. This is an A.P. with \(a=12, d=3, a_n=99\). First, find the number of terms: \(a_n = a+(n-1)d \Rightarrow 99 = 12+(n-1)3 \Rightarrow 87 = (n-1)3 \Rightarrow 29 = n-1 \Rightarrow n=30\). The sum is \(S_n = \frac{n}{2}(a+a_n) = \frac{30}{2}(12+99) = 15(111) = 1665\).

Answer: B

Here, \(a=3, r=2, a_n=384\). We use the formula \(a_n = ar^{n-1}\). \(384 = 3 \times 2^{n-1} \Rightarrow 128 = 2^{n-1}\). Since \(128 = 2^7\), we have \(2^7 = 2^{n-1}\). Equating the powers gives \(7=n-1\), so \(n=8\). There are 8 terms.

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.

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, ...