Sequence and Series – Arithmetic and Geometric Progressions Chapter 6 Exercise 6A CA foundation maths solutions
CA foundation maths solutions for Chapter 6 Exercise 6A Sequence and Series – Arithmetic and Geometric Progressions are given.
First you should study the CA maths textbook lesson chapter 6 very well.
You must observe and practice all example Problems and solutions which are given in the textbook.
You can observe the solutions given below. You will be try them in your own way.
Sequence and Series – Arithmetic and Geometric Progressions
CA maths Sequence and Series – Arithmetic and Geometric Progressions solutions Exercise 6A chapter 6
CA foundation maths solutions
Chapter 6
Sequence and Series – Arithmetic and Geometric Progressions
Exercise 6A
1. The nth element of sequence 1,3, 5, 7 ..,. is
2. The nth element of sequence -1,-2, -4, 8 …… is
4. The sum to infinity of the series – 5, 25, – 125, 625 …… can be written as
Q. 5. The first three terms of sequence when nth term
6. Which term of the progression -1 -3 – 5, …… is – 39.
7. The value of x such that 8X + 4, 6x – 2, 2x + 7 will form and AP is
12. The last term of the series 5, 7, 9, ….. to 21 terms is
13. The last term of the AP 0.6, 1.2, 1.8, …., to 13 terms is
Chapter 6 Exercise 6A solutions CA maths solutions
14. The sum of the series 9, 5, 1, …..to 100 terms is
15. The two arithmetic means between – 6 and 14 is
16. The sum of three integers in AP is 15 and their product is 80. The integers are
18. The number of numbers between 74 and 25,556 divisible by 5 is
19. The pth term of an AP is (3p – 1) / 6. The sum of the first n terms of the AP is
20. The Arithmetic mean between 33 and 27 is
21. The 4 arithmetic means between -2 and 23 are
22. The first term of an A.P is 14 and the sums of the first five terms and the first ten terms are equal in magnitude but opposite in sign. The third term of the AP is
23. They sum of a certain number of terms of an AP series – 8, – 6, – 4, ….. is 52. The number of terms is
24. The first and the last term of an AP are – 4 and 146. The sum of the terms is 7171. The number of terms is
25. The sum of the series 3 1 / 2 + 7 + 10 1/2 + 14 + …… to 17 terms is
Note : You can observe the solutions and try them in your own method.