The sum of three numbers in an A.P. is 27 and their product is 504. Find the numbers.
Answer: A
Let the three numbers be \(a-d, a, a+d\). Their sum is \((a-d)+a+(a+d) = 3a = 27\), so \(a=9\). Their product is \((a-d)a(a+d) = a(a^2-d^2) = 504\). Substituting a=9, we get \(9(81-d^2)=504\). \(81-d^2 = 504/9 = 56\). \(d^2 = 81-56 = 25\), so \(d = \pm 5\). If d=5, the numbers are 4, 9, 14. If d=-5, the numbers are 14, 9, 4. The numbers are 4, 9, 14.
Enter details here
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 \(a_{10} = 2 + (10-1) \times 5 = 2 + 9 \times 5 = 2 + 45 = 47\).
Enter details here
The third term of a G.P. is 4. The product of the first five terms is:
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 \(4^5 = 1024\).
Enter details here
If the sum of n terms of an A.P. is 210 and the sum of its first n-1 terms is 171, find its nth term.
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, \(a_n = 210 - 171 = 39\).
Enter details here
The sum of the series \(1 + 2x + 3x^2 + 4x^3 + ...\) for \(|x|<1\) is:
Answer: C
This is an Arithmetico-Geometric Progression (A.G.P.) with the A.P. part being 1, 2, 3, ... and G.P. part being \(1, x, x^2, ...\). The sum of an infinite A.G.P. is given by \(S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}\). Here, the first term of the A.P. is \(a_{AP}=1\), \(d=1\), and \(r=x\). The first term of the A.G.P. is \(a=1\). The sum is \(S = \frac{1}{1-x} + \frac{1 \cdot x}{(1-x)^2} = \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2}\). This is a standard result for the expansion of \((1-x)^{-2}\).
Enter details here
The sum of all natural numbers from 100 to 200, which are divisible by 5, is:
Answer: A
The series is 100, 105, ..., 200. This is an A.P. with \(a=100, l=200, d=5\). Number of terms \(n = \frac{l-a}{d} + 1 = \frac{200-100}{5} + 1 = 20+1=21\). Sum = \(\frac{n}{2}(a+l) = \frac{21}{2}(100+200) = \frac{21}{2}(300) = 21 \times 150 = 3150\).
Enter details here
The sum of first 9 terms of an A.P. is 81 and sum of first 18 terms is 324. Find the sum of first 27 terms.
Answer: A
Given \(S_9=81\) and \(S_{18}=324\). Notice that \(81=9^2 \times 1\) and \(324=18^2 \times 1\). No, that's not right. Let's use the formula. \(\frac{9}{2}(2a+8d)=81 \Rightarrow 2a+8d=18\). \(\frac{18}{2}(2a+17d)=324 \Rightarrow 2a+17d=36\). Subtracting the two equations gives \(9d=18 \Rightarrow d=2\). Then \(2a+16=18 \Rightarrow 2a=2 \Rightarrow a=1\). We need \(S_{27} = \frac{27}{2}[2(1)+(26)2] = \frac{27}{2}[2+52] = \frac{27}{2}(54) = 27 \times 27 = 729\).
Enter details here
The sum of all two-digit numbers divisible by 3 is:
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 \(S_n = \frac{n}{2}(a+a_n) = \frac{30}{2}(12+99) = 15(111) = 1665\).
Enter details here
How many terms are there in the G.P. 3, 6, 12, ..., 384?
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.
Enter details here
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.
Enter details here