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32.   If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation

Let roots of the quadratic equation be a and b. According to given condition,                                                                      We know that                                           Thus, the quadratic equation = 

31.  What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Given: Bank pays an annual interest rate of 10%  compounded annually. Rs 500 amounts are deposited in the bank. At the end of the first year, the amount                                                     At the end of the second year, the amount  At the end of the third year, the amount  At the end of 10 years, the amount                                                   Thus, at the end of...

30.   The number of bacteria in a certain culture doubles every hour. If there were 30  bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour ?

 The number of bacteria in a certain culture doubles every hour.It forms GP. Given :      a=30   and  r=2. Thus, bacteria present at the end of the 2nd hour, 4th hour and nth hour are 120,480 and  respectively.

29.   If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A \pm \sqrt{( A+G)(A-G)}

  If A and G be A.M. and G.M., respectively between two positive numbers, Two numbers be a and b. ...................................................................1 ...........................................................................2 We know  Put values from equation 1 and 2, ..................................................................3 From 1 and 3 , we have Put value...

28.   The sum of two numbers is 6 times their geometric mean, show that numbers  are in the ratio 

         ( 3+ 2 \sqrt 2 ) : ( 3 - 2 \sqrt 2 ) 

Let there be two numbers a and b geometric mean  According to the given condition, .............................................................(1) Also,  .......................................................(2) From (1) and (2), we get Putting the value of 'a' in (1), Thus, the ratio is 

27.   Find the value of n so that \frac{a^{n+1}+ b ^{n+1}}{a^n+b^n}  may be the geometric mean between a and b.

M of a and b is  Given :             Squaring both sides ,      

26.  Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Let A, B be two numbers between 3 and 81 such that series    3, A, B,81  forms a GP. Let a=first term and common ratio =r.       For , The, required numbers are 9,27.

25.   If a, b, c and d are in G.P. show that    (a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2 .

 If a, b, c and d are in G.P. To prove :  RHS :                              Using equation (1) and (2),     = LHS Hence proved

24.  Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from
         ( n+1)^{th} \: \: to\: \: (2n)^{th} term is \frac{1}{r^n}

.

Let first term =a  and common ratio = r. Since there are n terms from (n+1) to 2n  term. Sum of terms from (n+1) to 2n. Thus, the required ratio  =                                   Thus,  the common ratio of the sum of first n terms of a G.P. to the sum of terms from   term is  .  

23.  If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that  

        P^2 = ( ab)^n.

Given : First term =a  and n th term = b. Common ratio = r. To prove :  Then ,  P = product of n terms Here,    is a AP.                                                                                  Put in equation (2),  Hence proved .

22. If the p^{th} , q ^{th} , r ^{th} terms of a G.P. are a, b and c, respectively. Prove that  a ^{ q-r } b ^{r- p } C ^{p-q} = 1

To prove :  Let A  be the first term and R be common ratio. According to the given information, we have L.H.S :                                  =RHS Thus, LHS = RHS. Hence proved.

2.1  Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and  the second term is greater than the 4 ^{th} by 18.

Let first term be a and common ratio be r. Given : the third term is greater than the first term by 9, and  the second term is greater than the by 18. Dividing equation 2 by 1 , we get Putting value of r , we get Thus, four terms of GP are 

20.  Show that the products of the corresponding terms of the sequences

    a,ar, ar^2 , ...ar^{n-1} \: \: and\: \: A ,AR, AR^2 ....AR^{n-1} form a G.P, and find the common ratio.

To prove :      is a GP. Thus, the above sequence is a GP with common ratio of rR.    

19.   Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2,1/2

    Here,     is a GP. first term =a=4 common ratio =r                      

18.   Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

 8, 88, 888, 8888… is not a GP. It can be changed in GP by writing terms as   to n terms

17.  If the 4 ^{th} , 10 ^{th} , 16 ^ {th}  terms of a G.P. are x, y and z, respectively. Prove that x,y, z are in G.P.

Let x,y, z are in G.P. Let first term=a and common ratio = r Dividing equation 2 by 1, we have Dividing equation 3 by 2, we have Equating values of  ,  we have Thus, x,y,z are in GP

16.  Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term

Given : sum of the first two terms is – 4 and the fifth term is 4 times the third term Let first term be a and common ratio be r If r=2, then   If r= - 2, then Thus, required GP is       or    

15.   Given a G.P. with a = 729 and 7 ^{th} term 64, determine s_7

Given a G.P. with a = 729 and term 64. (Answer)

14.   The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Let GP be  Given :  The sum of first three terms of a G.P. is 16             Given :  the sum of the next three terms is128.         Dividing equation (2) by (1), we have Putting value of r =2 in equation 1,we have 

13.   How many terms of G.P. 3 , 3 ^ 2 , 3 ^ 3, … are needed to give the sum 120? 

G.P.=    , …............ Sum =120 These terms are GP with a=3 and r=3. Hence, we have value of n as 4 to get sum of 120.
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