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Q1. Define sequence and real sequence?

Solution

Sequence: A sequence is a function whose domain is the set of natural numbers.   Real Sequence: A real sequence is a function with domain N and the range a subset of the set of real numbers.

Q2. Find the ratio of n^th term of the sequence 2,4,8,16,32,64,..... to the n^th term of the sequence 3,9,27,81,243,....?

Solution

  The sequence 2,4,8,16,32,64,.... follow the rule x_n=2ⁿ n=1,2,3,.... The sequence 3,9,27,81,243,.... follow the rule x_n=3ⁿ n=1,2,3,.... So the ratio is    

Q3. A person is entitled to receive an annual payment which for each year is less by one tenth of what it was for the year before.If the first payment is 100 find the maximum possible payment which he can receive however long he may live?

Solution

Q4. Find the sum of the series 2² + 4² + 6² + 8² + … to n terms.

Solution

Q5. If m times the m^th term of an AP is n times the n^th term then find the (m+n)^th?

Solution

Q6. Find two positive numbers whose Arithmetic Means increased by 2 than Geometric Means and their difference is 12.

Solution

Let the two numbers be m and n. Then, m-n=12 ........................ (i) It is given that AM-GM=2 ⇒ (m + n)/2-√mn = 2 ⇒ (m + n)-2√mn = 4 ⇒ (√m-√n)² = 4 ⇒ √m-√n = ±2 ........................ (ii) Now, m-n= 12 ⇒ (√m + √n)(√m-√n) = 12 ⇒ (√m + √n)(±2) = 12 ........................ (iii) ⇒ √m + √n = ± 6, [using (ii)] Solving (ii) and (iii), we get m = 16, n = 4 Hence, the required numbers are 16 and 4.

Q7. Find the 10^th term of the series 63, 58, 53, 48, …

Solution

Here, first term a₁ = 63 and common difference = d = 58 – 63 = – 5   10^th term a₁₀ = [a₁ + (10 – 1)d] = 63 + 9(– 5) = 63 – 45 = 18   Thus, the 10^th term of the series is 18.

Q8. Define finite and infinite sequence?

Solution

Finite Sequence:  A sequence containing finite number of terms is called a finite sequence.   Infinite Sequence:  A sequence containing infinite number of terms is called a finite sequence.

Q9. Prove that the Quadratic equation whose roots are a and b is given by x²-2Ax+G²=0 where A =Arithmetic mean   and G is the geometric mean of a and b respectively.

Solution

the quadratic equation with roots a and b is given by x²-(a+b)x+ab=0 the arithmetic mean of a and b is A=(a+b)/2a+b=2A The geometric mean of a and b is G= So the equation becomes x²-2Ax+G²=0

Q10. Find the ratio of n^th term of the sequence 2,4,8,16,32,64,..... to the n^th term of the sequence 3,9,27,81,243,....?

Solution

The sequence 2,4,8,16,32,64,.... follow the rule x_n=2ⁿ n=1,2,3,.... The sequence 3,9,27,81,243,.... follow the rule x_n=3ⁿ n=1,2,3,.... So the ratio is    

Q11. Sum the following series to n terms: 2 + 10 + 30 + 68 + 130 + … 

Solution

Q12. The Arithmetic and Geometric Means of two positive numbers are 15 and 9 respectively. Find the numbers.

Solution

Let the two positive numbers be x and y. Then according to the problem, (x + y)/2 = 15 or, x + y = 30 .................. (i) and √xy = 9 or xy = 81 Now, (x-y)²= (x + y)² - 4xy = (30)²-4(81)  = 576 = (24)² Therefore, x-y=± 24 .................. (ii) Solving (ii) and (iii), we get, 2x = 54 or 2x = 6 x = 27 or x = 3 When x = 27 then y = 30-x = 30-27 = 3 and when x = 27 then y = 30-x = 30-3 = 27 Therefore, the required numbers are 27 and 3.

Q13. If A is the arithmetic mean of the roots of the equation x²-3x+2=0 and G is the Geometric mean of the roots of the equation x²-5x+6=0 the find the value of 4A-G²?

Solution

The roots of the equation x²-3x+2=0 is x=1,2 A=(1+2)/2=3/2 The roots of the equation x²-5x+6=0 is x=2,3 G= 4A-G²=4(3/2)-6=0

Q14. If the sum of 4 terms of an AP is 36,find the arithmetic mean of the AP?

Solution

Q15. Insert 3 arithmetic means between 5 and 21.

Solution

After inserting the 3 arithmetic means between 5 and 21 the resulting sequence will be an AP with common difference d. First AM is 5+4=9 Second AM is 5+2(4)=13 Third AM is 5+3(4)=17 So the resulting sequence is 5,9,13,17,21

Q16. Sum the following series to n terms: 3 + 18 + 57 + 132 + 255 + … 

Solution

Q17. Find the sum of the series 1.2² + 2.3² + 3.4² + … to n terms.

Solution

Q18. Find the A.P after inserting two arithmetic means between two numbers which are arithmetic means of the sequences 1,2,3,4 and 5,6,7,8 respectively?

Solution

The first number is the arithmetic mean of 1,2,3,4 i.e.  The second number is the arithmetic mean of 5,6,7,8 i.e.  we need to insert two AMs between 5/2 and 13/2 after which the resulting sequence will be an AP with common difference given by  The first AM is 5/2+4/3=23/6 The second AM is 5/2+2(4/3)=31/6 The resulting AP is 5/2 , 23/6 , 31/6 , 13/2

Q19. If a,b,c are x^th,y^th,z^th terms of a GP respectively then (y - z) loga + (z - x) logb + (x - y) logc is?

Solution

Q20. If arithmetic mean of an AP is 16 and the AP consists of 5 terms and the common difference is 2 then find the AP?

Solution

Let the AP is a-2d,a-d,a,a+d,a+2d  Arithmetic mean of the AP is  on solving we get a=16 given d=2 So AP is 12,14,16,18,20

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