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Q1. The Fibonacci sequence is defined as  for n > 2. Find a_n+1/a_n for n = 2 and 3. 

Solution

The Fibonacci sequence is defined as  for n > 2. The terms of this sequence are:

Q2. Let ‘a’ be the AM and b & c be two GM’s between two any positive numbers. Then prove that .

Solution

Let x and y be two positive numbers, then a is AM of x and y Also, x, b, c, y is the G.P. with common ratio of

Q3. Find the sum of the series 2² + 4² + 6² + 8² + … to n terms.

Solution

Q4.

Solution

Q5. The sum of n terms of two APs are in the ratio (7n + 1):(4n + 27). Find the ratio of their 13^th term.

Solution

Q6. Find the least value of n for which 1 + 3 + 9 + 27 + … to n terms is greater than 7000. [Use log 140.01 = 2.1461 and log 3 = 0.4771]

Solution

Here, we have   Thus, the least value of n is 9.

Q7. If the sum of n terms of three AP’s are S₁, S₂, S₃. The first term of all the AP’s are unity and the common difference are 1, 2 and 3 respectively. Then prove that .

Solution

We have,

Q8. Between 1 and 31, n numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7th and (n – 1)^th numbers is 5 : 9. Find the value of n.

Solution

Let A₁, A₂, A₃, …, A_n  be n arithmetic means between 1 and 31 and common difference be d. Total number of terms = n + 2 First term = 1, last term = (n + 2)^th term = 31 Thus, the number of inserted numbers is 14.

Q9.

Solution

Q10. Find the first three terms of sequence, if sum of n terms is .

Solution

Let sum of n terms be. Here,    Thus, first three terms of this sequence are 1, 2 and 3.

Q11. Find the numbers whose arithmetic mean is 82 and geometric mean is 18.

Solution

Let numbers be a and b. Then the numbers are 162 & 2 or 2 & 162.

Q12. Find the sum of the series 1.2² + 2.3² + 3.4² + … to n terms.

Solution

Q13. Find the sum of the series 2n² - 3n + 5 to n terms.

Solution

Q14. Find the sum to n terms of the series

Solution

The given series is  

Q15. If S be the sum, P the product and R the sum of the reciprocals of n terms of a G.P., then prove that .

Solution

Let first term and common ratio of this G.P. be ‘a’ and r respectively.

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

Solution

Q17.

Solution

Q18. Find the sum of the series 1 + (1 + 2) + (1 + 2 + 3) + … up to n terms.

Solution

 Here, T_n = (1 + 2 + 3+ … + n) = Sum of n terms

Q19.

Solution

Q20. If x, y, z are in A.P. A₁ is the mean of x & y and A₂ is the mean of y & z. Then prove that the mean of A₁ and A₂ is y.

Solution

Here, x, y, z are in AP, then Now, A₁ = mean of x and y Also, A₂ = mean of y and z Now, we have to prove that the mean of A₁ and A₂ is y. Hence, proved.

Q21. Find the sum of the series 1+2+2²+2³+--- +2ⁿ

Solution

Q22.

Solution

Q23.

Solution

Q24. Write the sum to n terms of the series

Solution

Q25. Find the sum of the following seriesup to 16 terms.

Solution

Let T_n be the nth term of the given series. Here, 

Q26. Insert 6 numbers between – 6 and 29 such that the resulting sequence is an A.P.

Solution

Let A₁, A₂, A₃, …, A₆  be 6 arithmetic means between – 6 and 29 and common difference be d. Total number of terms = 6 + 2 = 8   Thus, the A.P. is – 6, – 1, 4, 9, 14, 19, 24, 29.

Q27. Find two positive numbers whose difference is 12 and whose AM exceeds GM by 2.

Solution

Let a and b be two numbers, such that a – b = 12    …(1) and AM – GM = 12 Thus, the numbers are 16 and 4.

Q28. If the 4^th and 9^th terms of a G.P. are 18 and 4374. Find the G.P. 

Solution

Let first term and common ratio of this G.P. be ‘a’ and r respectively. Dividing equation (2) by (1), we get On putting r = 3 in equation (1), we get Thus, the required G.P. is

Q29. Sum the following series to n terms: 2 + 10 + 30 + 68 + 130 + … 

Solution

Q30. Find the number of arithmetic means between 7 and 71 in such a way that 5^th mean is 27.

Solution

Let A₁, A₂, A₃, …, A_n  be n arithmetic means between 7 and 71 and common difference be d. Total number of terms = n + 2 First term = 7, last term = (n + 2)^th term = 71 Thus, there are 15 AM's inserted between 7 and 71.

Q31. The A. M. between two positive numbers a and b is twice the G. M. between them. Find the ratio of the numbers.

Solution

Here, given that AM of a and b = 2(GM of a and b)

Q32. If the fourth term of a G.P. is 3. Find the product of first 7 terms.

Solution

Let first term and common ratio of this G.P. be ‘a’ and r respectively. Fourth term =   …(1) Product of first seven terms = Thus, the product of first seven terms of this G.P. is 2187.

Q33. Find the first five terms of the sequence .

Solution

Here, For n = 1, 2, 3, 4, and 5, we can get the first five terms of this sequence. Thus, the first five terms of the given sequence are 1, 5, 14, 30 and 55.

Q34.

Solution

Q35. Find the 19^th term of the sequence .

Solution

Here, the sequence <a_n> is defines as The 19^th term of the given sequence is 1/3.

Q36. Find n such that  may be the G.M. between positive numbers a and b.

Solution

The G.M. between a and b is.  

Q37. How many terms are there in the AP. .

Solution

Let there be n terms in this A.P. Here, first term a₁ = 20, common difference  and n^th term a_n = – 10 There are 41 terms in the A.P.

Q38. Find the sum of 50 terms of the sequence 9, 9.9, 99.9, 999.9 …

Solution

The given sequence is 9, 9.9, 99.9, 999.9 ……. Sum = 9 + 9.9 + 99.9 + 999.9+ ……50 terms         = [10 –1] + [10 – 0.1] + [10 – 0.01] +…….50 terms         = [10 +10 +10…..50 terms] – [1+0.1+0.01+0.001+….50 terms.]         = 500 – (sum of G.P. of 50 terms with a= 1, r = 0.1)                     =

Q39. Let <a_n> be a sequence defined by . Find  

Solution

Here,