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Install Now16. Find a G.P. for which sum of the first two terms is and the fifth term is 4 times the third term.
Ans. Let be the first term and be the common ratio of given G.P.
Given:
……..(i)
And
Putting in eq. (i), we get
Therefore, required G.P. is
Putting in eq. (i), we get
Therefore, required G.P. is
17. If the 4th, 10th and 16th terms of a G.P. are and respectively. Prove that are in G.P.
Ans. Let be the first term and be the common ratio of given G.P.
……….(i)
……….(i)
……….(i)
From eq. (ii),
[From eq. (i) and (iii)]
are in G.P.
18. Find the sum to terms of the sequences 8, 88, 888, 8888, ……
Ans. Here
=
=
19. Find the sum of the product of the corresponding terms of the sequences 2, 4, 16, 32 and 128, 32, 8, 2,
Ans. Multiplying the corresponding terms of the given sequences 2, 4, 16, 32 and 128, 32, 8, 2,
are in G.P.
Here and
when
=
= 496
20. Show that the products of the corresponding terms of the sequences and form a G.P. and find the common ratio.
Ans. Multiplying the corresponding terms of the given sequences, we have
are in G.P.
Here First term = and common ratio =
21. 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 by 4th by 18.
Ans. Let the four numbers in G.P. be
and
Now,
……….(i)
And
……….(ii)
Dividing eq. (ii) by eq. (i), we have
Putting value of in eq. (i), we get
Therefore, the required numbers are
22. If the and terms of a G.P. are and respectively. Prove that
Ans. Let A be the first term and R be the common ratio of given G.P.
……….(i)
……….(ii)
……….(iii)
Now, L.H.S. =
=
=
= = = 1 = R.H.S.
23. If the first and the term of a G.P. are and respectively and if P is the product of terms, prove that
Ans. Here, first term of G.P. is
And
……….(i)
Given: P =
P =
[Squaring both sides]
[From eq. (i)]
24. Show that the ratio of the sum of first terms of a G.P. to the sum of terms from to term is
Ans. Let be the first term and be the common ratio of given G.P.
Then
=
25. If and are in G.P., show that
Ans. Let be the common ratio of given G.P.
Then and
Now, L.H.S. =
=
= =
R.H.S. =
=
=
= =
Therefore, L.H.S. = R.H.S.
26. Insert two numbers between 3 and 81 so that the resulting sequence us G.P.
Ans. Let be two numbers between 3 and 81 such that are in G.P.
Here and
And
Therefore, the required numbers are 9 and 27.
27. Find the value of so that may be the geometric mean between and .
Ans. Since, G.M. between two numbers and is
According to question,
28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio
Ans. Given:
Applying component and dividend, we get
Again, Applying component and dividend, we get
Squaring both sides,
Therefore, the numbers are in the ratio
29. If A and G be A.M. and G.M. respectively between two positive numbers, prove that the numbers are
Ans. Given: A = and G =
Now,
=
=
=
= =
= and
= and
= and
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 hour?
Ans. Bacteria present in the culture originally = 30
Since the bacteria doubles itself after each hour, then the sequence of bacteria after each hour is a G.P.
Here and
Bacteria at the end of 2nd hour =
And Bacteria at the end of 4th hour =
And Bacteria at the end of hour =
31. What will Rs. 500 amount to 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Ans. Original amount = Rs. 500, Rate of interest = 10% compounded annually
Interest of one year = = Rs. 50
And Amount after one year = 500 + 50 = Rs. 550
Here and
Therefore, amount after 10 years =
32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5 respectively then obtain the quadratic equation.
Ans. Let and be the roots of required quadratic equation.
Then A.M. =
And G.M. =
Now, Quadratic equation
Therefore, required equation is .
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