1. The first and last term of an A.P are, a and l respectively, show that the sum of nth term from the beginning and nth term from the end is a + l.
2. If mth term of an A.P be 1/n and nth term be 1/m, then show that its mnth term is 1.
3. The sum of four numbers which are in A.P is 32 and the product of whose extremes is 55. Find the numbers.
4. The angles of a quadrilateral are in A.P whose common difference is 10. Find the angles.
5. If a (1/b+1/c), b(1/c+1/b), c(1/a+1/b) are in A.P. Show that a, b, c are in A.P.
6. Evaluate 1/2 +1/3 +1/6+----------------to 10 terms
7. Find the value of x such that 25+22+19+16+---------------+x=115
8. There are n arithmetic means between 5 and 86. Show that the ratio of the first and the last means is 2:11. Find n
9. If a, b, c are in G.P and ax=by=cz then show that 1/x+1/z=2/y
10. The second third and sixth term of an A.P are consecutive terms of G.P. Find the common ratio of G.P
11. The products of three numbers which are in G.P is 216. If 2,8,6 are added to them , the resulting numbers form an A.P. Find the numbers.
12. If a, b, c are in G.P and x, y are respectively the arithmetic means of a, b and b, c . Show that a/x + c/y=2 and 1/x+1/y=2/b.
13. The sum of first eight terms of G.P is five times the sum of the first four terms. Find the common ratio.
14. The ratio of the sum of first three terms is to that of first b terms of a G.P is 125:152. Find the common ratio.
15. Evaluate Ʃ10 k=1(2k+3k-1).
16. Find the sum of geometric series: (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3)+---------------to n terms.
17. Find the sum of n terms: 5+55+555+------------.
18. How many terms of the series 1+4+16+64+------------ make the sum 5461?
19. Find the two positive numbers whose arithmetic mean is 34 and the geometric mean is 16.
20. Insert 6 geometric mean between 27and 1/81.
21. If (a-1) is the G.M between (a-2) and (a+1). Find a.
22. Find the two positive numbers whose difference is 12 and their A.M exceeds the G.M by 2.
23. If a is the A.M between b and c and g1, g2 are two geometric means between b and c. show that
1. To prove that the sum of the nth term from the beginning and the nth term from the end of an arithmetic progression (A.P) is equal to the sum of the first and last term, we can use the formula for the nth term of an A.P.
The nth term of an A.P is given by the formula:
an = a + (n - 1)d
where a is the first term, d is the common difference, and n is the position of the term.
From the given problem, we have the first and last terms as a and l respectively. So, the position of the first term is 1 and the position of the last term is n.
The nth term from the beginning can be represented as:
an = a + (n - 1)d
The nth term from the end can be represented as:
al = l - (n - 1)d
To find the sum of the nth term from the beginning and the nth term from the end, we can add them together:
an + al = (a + (n - 1)d) + (l - (n - 1)d)
= a + (n - 1)d + l - (n - 1)d
The common differences cancel out, leaving us with:
an + al = a + l
Thus, we have proved that the sum of the nth term from the beginning and the nth term from the end of an A.P is equal to the sum of the first and last term.
2. To prove that the mnth term of an arithmetic progression (A.P) with mth term 1/n and nth term 1/m is 1, we can use the formula for the nth term of an A.P.
The nth term of an A.P is given by the formula:
an = a + (n - 1)d
where a is the first term, d is the common difference, and n is the position of the term.
From the given problem, we have the mth term as 1/n. So, the position of this term is m.
The nth term can be represented as:
an = a + (n - 1)d
Since the mth term is given as 1/n, we can substitute m for n and 1/n for an in the equation:
1/n = a + (m - 1)d
Similarly, we are given the nth term as 1/m, so we can substitute n for m and 1/m for an in the equation:
1/m = a + (n - 1)d
Solving these equations simultaneously, we can find the value of a and d.
1/n = a + (m - 1)d
1/m = a + (n - 1)d
Subtracting these equations, we get:
1/n - 1/m = (a + (m - 1)d) - (a + (n - 1)d)
= ad - ad = 0
Simplifying the equation, we have:
1/n - 1/m = 0
Since the left-hand side of the equation is equal to zero, we can conclude that 1/n - 1/m = 0 is equivalent to 1 = 1.
Hence, we have proved that the mnth term of the A.P is 1.