Question:

The 5th, 8th and 11th terms of a G.P are p,q and s respectively. Show that q^2=ps

Answer:

Given : 5th term = p, 8th term = q, 11th term = s

Step 1: We know that G.P. is a sequence of numbers such that each term after the first is found by multiplying the preceding one by a fixed number, which is called the common ratio.

Step 2: We can write the G.P. as p, pr, pr^2, pr^3, pr^4, pr^5, pr^6, pr^7, pr^8, pr^9, pr^10

Step 3: From the given information, we can say that p = pr^4, q = pr^7, s = pr^10

Step 4: Substituting the values in the equation q^2 = ps, we get

(pr^7)^2 = pr^4 * pr^10

Step 5: Simplifying the equation, we get

pr^14 = pr^14

Step 6: Hence, we can conclude that q^2 = ps.

Question:

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

Answer:

Given, First term (a) = 3 Common ratio (r) = 3 Sum (S) = 120

Step 1: We need to find the number of terms (n) in the G.P. To do this, we need to use the formula

Sn = a(r^n - 1)/(r - 1)

Step 2: Substitute the given values in the formula

120 = 3(3^n - 1)/(3 - 1)

Step 3: Simplify the equation

120 = 3(3^n - 1)/2

Step 4: Divide both sides of the equation by 3

40 = 3^n - 1

Step 5: Add 1 to both sides of the equation

41 = 3^n

Step 6: Take the log of both sides of the equation

log3 41 = n

Step 7: Solve for n

n = log3 41

Hence, the number of terms needed to give the sum 120 is log3 41.

Question:

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

Answer:

Answer: Step 1: Identify the sequence is an arithmetic sequence. Step 2: Calculate the common difference of the sequence by subtracting the second term (88) from the first term (8). Common difference = 8 - 88 = -80 Step 3: Calculate the nth term of the sequence. nth term = a + (n - 1)d nth term = 8 + (n - 1)(-80) nth term = 8 - 80(n - 1) Step 4: Calculate the sum to n terms of the sequence. Sum = n/2 [2a + (n - 1)d] Sum = n/2 [2(8) + (n - 1)(-80)] Sum = n/2 [16 - 80(n - 1)] Sum = 8n - 40(n - 1)

Question:

Find the 20th and nth terms of the G.P.5/2,5/4,5/8…..

Answer:

Solution:

Given, First term, a = 5/2 Common ratio, r = 5/4

nth term = a x r^(n-1)

20th term = a x r^(20-1) = 5/2 x (5/4)^19 = 5/2 x (625/524288) = 5/8192

Question:

Find the 12th term of a G.P whose 8th term is 192 and the common ratio is 2.

Answer:

Answer:

Given: 8th term = 192 Common ratio = 2

Step 1: Find the common difference (d) Common difference (d) = Common ratio (r) - 1

d = 2 - 1 = 1

Step 2: Find the 12th term 12th term = a₈ + (12 - 8)d

12th term = 192 + (12 - 8)1

12th term = 192 + 4(1)

12th term = 192 + 4

12th term = 196

Question:

Find the sum of 20 terms of the series 0.15,0.015,0.0015,…20

Answer:

Answer: Step 1: Write down the first 20 terms of the series. 0.15, 0.015, 0.0015, 0.00015, 0.000015, 0.0000015, 0.00000015, 0.000000015, 0.0000000015, 0.00000000015, 0.000000000015, 0.0000000000015, 0.00000000000015, 0.000000000000015, 0.0000000000000015, 0.00000000000000015, 0.000000000000000015, 0.0000000000000000015, 0.00000000000000000015, 0.000000000000000000015

Step 2: Add the terms together. 0.15 + 0.015 + 0.0015 + 0.00015 + 0.000015 + 0.0000015 + 0.00000015 + 0.000000015 + 0.0000000015 + 0.00000000015 + 0.000000000015 + 0.0000000000015 + 0.00000000000015 + 0.000000000000015 + 0.0000000000000015 + 0.00000000000000015 + 0.000000000000000015 + 0.0000000000000000015 + 0.00000000000000000015 + 0.000000000000000000015 = 0.1515151515151515

Step 3: Simplify the result. 0.1515151515151515

Question:

Sum of the series, √7,√21,3√7,… n terms.

Answer:

Answer: Step 1: Find the common difference of the series. Common difference = √21 – √7 = √14

Step 2: Find the nth term of the series. nth term = (n-1) * √14 + √7

Step 3: Find the sum of the series. Sum of the series = [n/2] * (2 * √7 + (n-1) * √14)

Question:

The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Find the sum of n terms of the G.P.

Answer:

Step 1: A G.P. is an arithmetic progression in which the common ratio between any two consecutive terms is constant.

Step 2: Let the first term of the G.P. be a and the common ratio be r.

Step 3: Since the sum of the first three terms is 16, the equation for the sum of the first three terms can be written as:

a + ar + ar2 = 16

Step 4: Similarly, the sum of the next three terms can be written as:

ar3 + ar4 + ar5 = 128

Step 5: By solving these two equations, we get

a = 2 and r = 2

Step 6: The sum of n terms of the G.P. can be written as:

Sn = a + ar + ar2 + ar3 + ar4 + … + arn-1

Step 7: Substituting a = 2 and r = 2 in the above equation, we get

Sn = 2 + 22 + 22 + 22 + 22 + … + 2n-1

Step 8: The sum of n terms of the G.P. can be written as:

Sn = 2 (1 + 2 + 22 + 23 + … + 2n-1)

Step 9: The sum can be simplified to:

Sn = 2 (2n - 1)

Step 10: Therefore, the sum of n terms of the G.P. is:

Sn = 2 (2n - 1)

Question:

If the pth,qth and rth terms of a G.P. are a, b and c respectively. Prove that a^(q−r)b^(r−p)c^(p−q)=1.

Answer:

Given, pth, qth and rth terms of a G.P. are a, b and c respectively.

To prove: a^(q−r)b^(r−p)c^(p−q)=1

Step 1: As the given terms are in G.P.,

a/b = b/c

Step 2: Rearranging the above equation,

b^2 = ac

Step 3: Substituting the values of p, q and r,

a^(q−r)b^(r−p)c^(p−q) = (a^(q−r))(b^2)c^(p−q)

Step 4: Using the value of b^2 from Step 2,

a^(q−r)b^(r−p)c^(p−q) = (a^(q−r))(ac)c^(p−q)

Step 5: Simplifying the above equation,

a^(q−r)b^(r−p)c^(p−q) = a^(q−r+p−q)c^(p−q+q−r)

Step 6: Simplifying further,

a^(q−r)b^(r−p)c^(p−q) = a^0c^0

Step 7: As a^0 = 1 and c^0 = 1,

a^(q−r)b^(r−p)c^(p−q) = 1

Hence, proved.

Question:

If the first and the nth term of a G.P are a and b, respectively, and if P is product of n terms, prove that P^2=(ab)^n

Answer:

Given, First term, a Nth term, b Product of n terms, P

To prove, P^2=(ab)^n

Proof:

Step 1: P = a × r × r × r × r × … × r × b

Step 2: P^2 = (a × r × r × r × r × … × r × b) × (a × r × r × r × r × … × r × b)

Step 3: P^2 = a^2 × r^2 × r^2 × r^2 × … × r^2 × b^2

Step 4: P^2 = a^2 × r^2 × r^2 × r^2 × … × r^2 × b^2

Step 5: P^2 = (a × r × r × r × r × … × r)^2 × b^2

Step 6: P^2 = (a × r^(n-1))^2 × b^2

Step 7: P^2 = (ab)^n

Hence, proved.

Question:

Show that the ratio of the sum of first n terms of a G.P. to the sum of terms form (n+1)th to (2n)th term is 1/r^n.

Answer:

Given: A G.P. (Geometric Progression) is a sequence of numbers in which each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

Let a be the first term and r be the common ratio of the G.P.

Then, the G.P. is given by a, ar, ar^2, ar^3, ar^4, ar^5, ar^6, …

Solution:

Let S_n be the sum of the first n terms of the G.P.

Then,

S_n = a + ar + ar^2 + ar^3 + ar^4 + … + ar^n-1

Let S_(2n) be the sum of the terms from (n+1)th to (2n)th term

Then,

S_(2n) = ar^n + ar^(n+1) + ar^(n+2) + ar^(n+3) + … + ar^(2n-1)

Now,

Ratio of S_n to S_(2n) =

= (a + ar + ar^2 + ar^3 + ar^4 + … + ar^n-1) / (ar^n + ar^(n+1) + ar^(n+2) + ar^(n+3) + … + ar^(2n-1))

= a / ar^n (since all the terms in the denominator have the factor ar^n)

= 1 / r^n

Hence, the ratio of the sum of first n terms of a G.P. to the sum of terms form (n+1)th to (2n)th term is 1/r^n.

Question:

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

Answer:

Solution: Step 1: Start with 3 and 81. Step 2: Insert two numbers between 3 and 81. For example, 15 and 45. Step 3: The resulting sequence is 3, 15, 45, 81 which is a Geometric Progression (G.P.).

Question:

What will Rs500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Answer:

Step 1: Calculate the principal amount (P) = Rs500

Step 2: Calculate the annual interest rate (r) = 10%

Step 3: Calculate the number of years (n) = 10

Step 4: Calculate the compound interest (CI) using the formula CI = P(1+r/100)^n

Step 5: Calculate the total amount (A) after 10 years A = P + CI

Step 6: Substitute the values in the formula A = 500(1+10/100)^10

Step 7: Calculate the total amount (A) after 10 years A = Rs1,105.10

Question:

The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

Answer:

Given, Sum of first three terms of a G.P. = 39/10 Product of the first three terms = 1

Step 1: Let the first term of the G.P. be a.

Step 2: Let the common ratio of the G.P. be r.

Step 3: As the G.P. is given, the second term of the G.P. = ar and the third term = ar2.

Step 4: Now, using the given information,

Sum of the first three terms = a + ar + ar2 = 39/10

Step 5: Multiplying both sides by 10,

10a + 10ar + 10ar2 = 39

Step 6: Also, using the given information,

Product of the first three terms = a × ar × ar2 = 1

Step 7: Simplifying,

a × ar2 = 1

Step 8: Substituting the value of ar2 in the equation obtained in Step 5,

10a + 10ar + 10 = 39

Step 9: Rearranging,

10a + 10ar - 10 = 39 - 10

Step 10: Simplifying,

10(a + ar - 1) = 29

Step 11: Dividing both sides by 10,

a + ar - 1 = 29/10

Step 12: Substituting the value of ar2 in the equation obtained in Step 11,

a + ar - 1 = 29/10

ar2 = 1

Step 13: Solving the equation obtained in Step 12,

r2 = 10/29

Step 14: Taking the square root of both sides,

r = √(10/29)

Hence, the common ratio of the G.P. is √(10/29).

Step 15: Substituting the value of r in the equation obtained in Step 11,

a + √(10/29) - 1 = 29/10

Step 16: Solving the equation obtained in Step 15,

a = (29/10) - √(10/29) + 1

Hence, the first term of the G.P. is (29/10) - √(10/29) + 1.

Therefore, the common ratio of the G.P. is √(10/29) and the first three terms are (29/10) - √(10/29) + 1, √(10/29) and 1.

Question:

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

Answer:

Step 1: Given, Sum of first two terms = -4 Fifth term = 4 times the third term

Step 2: Let the first term be a and the common ratio be r.

Step 3: According to the given condition,

a + ar = -4 ………………………..(1)

a + ar + ar² + ar³ + 4ar³ = 4ar³ …………………(2)

Step 4: Solving equation (1) and (2), we get

a = -4, r = -1

Step 5: Therefore, the required G.P. is -4, 4, -4, 4, -4, ….

Question:

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

Answer:

Given that a, b, c and d are in G.P.

Step 1: We can express a, b, c and d in terms of common ratio r as a = ar, b = ar2, c = ar3 and d = ar4.

Step 2: Substitute the values of a, b, c and d in the given equation:

(ar2 + ar4 + ar6)(ar4 + ar6 + ar8) = (ar2 × ar3 + ar3 × ar4 + ar4 × ar5)2

Step 3: Simplifying the equation:

ar2 + ar4 + ar6 + ar8 + 2ar2ar4 + 2ar2ar6 + 2ar4ar6 = (ar2 × ar3 + ar3 × ar4 + ar4 × ar5)2

Step 4: Simplifying further:

ar2 + ar4 + ar6 + ar8 + 2ar2ar4 + 2ar2ar6 + 2ar4ar6 = ar4 + 2ar5 + 2ar6 + 2ar7 + ar8

Step 5: Comparing the coefficients of each power of ‘r’ on both sides of the equation:

ar2 = ar2

ar4 = ar4

ar6 = ar6

ar8 = ar8

2ar2ar4 = 2ar5

2ar2ar6 = 2ar6

2ar4ar6 = 2ar7

Hence, the equation holds true.

Question:

The sum of two numbers is 6 times their geometric mean show that numbers are in the ratio (3+2√2):(3−2√2)

Answer:

1. The sum of two numbers is 6 times their geometric mean.
2. Geometric mean = √(x*y), where x and y are the two numbers.
3. 6√(xy) = x + y
4. 6√(xy) = (3+2√2) + (3-2√2)
5. 6√(xy) = 3 + 2√2 - 2√2 + 3
6. 6√(xy) = 6
7. √(x*y) = 1
8. x*y = 1
9. x/y = (3+2√2)/(3−2√2)
10. Therefore, the numbers are in the ratio (3+2√2):(3−2√2).

Question:

If A and G be A.M and G.M, respectively between two positive numbers. prove that the numbers are A±√(A+G)(A−G).

Answer:

1. Let A and G be two positive numbers.

2. A and G are Arithmetic Mean (A.M) and Geometric Mean (G.M.) respectively between two positive numbers.

3. Since A and G are A.M and G.M. respectively, we can write A = (a + b)/2 and G = √(ab).

4. Substituting A and G in the given equation, we get: A ± √(A + G)(A – G) = (a + b)/2 ± √((a + b)/2 + √ab)((a + b)/2 – √ab)

5. Simplifying the above equation, we get: A ± √(A + G)(A – G) = (a + b)/2 ± √(a + b)2/4 – ab

6. Further simplifying, we get: A ± √(A + G)(A – G) = (a + b)/2 ± (a – b)/2

7. Therefore, the two numbers are A ± √(A + G)(A – G).

Question:

If AM and GM of roots of a quadratic equations are 8 and 5 respectively, then obtain the quadratic equation.

Answer:

Step 1: The equation of a quadratic equation is given by ax^2 + bx + c = 0.

Step 2: The Arithmetic Mean (AM) of the roots of the equation is 8 and the Geometric Mean (GM) of the roots is 5.

Step 3: The product of the roots of the equation is given by the formula a*c = GM^2.

Step 4: Substituting GM = 5, we get a*c = 25.

Step 5: The sum of the roots of the equation is given by the formula b = AM*2.

Step 6: Substituting AM = 8, we get b = 16.

Step 7: Substituting the values of a, b and c in the equation ax^2 + bx + c = 0, we get the required quadratic equation as x^2 + 16x + 25 = 0.

Question:

The 4th term of a G.P is square of its second term, and the first term is −3. Determine its 7th term.

Answer:

Given: First term (a) = -3 Second term (a2) = x

4th term (a4) = x2

7th term (a7) = ?

Solution:

We know that the general form of a G.P is a, ar, ar2, ar3,…

Therefore,

ar2 = a4

x2 = a4

x = √a4

Substituting the value of a4,

x = √(-3)2

x = -3

Now, the 7th term (a7) can be found using the general form of a G.P,

a7 = ar6

a7 = (-3)6

a7 = -729

Question:

Sum of the series, 1,−a,a2,−a3,…nterms(ifa≠±1).

Answer:

Step 1: Identify the terms of the series. The terms of the series are 1, -a, a2, -a3, … n terms (if a ≠ ±1).

Step 2: Find the sum of the first n terms of the series. The sum of the first n terms of the series is S = 1 – a + a2 – a3 + … + (–1)n-1 an-1.

Step 3: Simplify the expression. Using the formula for the sum of a geometric series, the expression can be simplified to S = 1 – (–1)n an / (1 – a).

Question:

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

Answer:

Step 1: A.M and G.M of two numbers are given as 8 and 5, respectively.

Step 2: Let the two numbers be x and y.

Step 3: Then, A.M = (x + y)/2 and G.M = (xy)^1/2

Step 4: Substituting these values in the given equation, we get:

(x + y)/2 = 8 and (xy)^1/2 = 5

Step 5: Solving these two equations, we get x = 12 and y = 4

Step 6: Hence, the required quadratic equation is x2 - 16x + 48 = 0

Question:

Which term of the following sequences : (a) 2,2√2,4,…is 128? (b) √3,3,3√3,… is 729? (c) 1/3,1/9,1/27,… is 1/19683?

Answer:

(a) The 8th term is 128. (b) The 6th term is 729. (c) The 9th term is 1/19683.

Question:

Find the value of x such that −2/7,x,−7/2 are three consecutive terms of a G.P.

Answer:

Given that the three terms are in G.P.

Let a = -2/7, r = x

Therefore, the third term, b = -7/2

We know that in G.P.,

b = ar^2

Therefore, -7/2 = (-2/7)x^2

Multiply both sides by 14

-14 = -2x^2

Divide both sides by -2

7 = x^2

Take square root on both sides

x = √7

Question:

Find the sum of the following GP: x3,x5,x7,… to n terms, x>2 A : x3(x^(2n−1)−1)/(x^2−1) B : x3(x^2n−1)/(x^2−1) C : x^2(x^2n−1)/(x^2−1) D : x^2(x^(2n−1)−1)/(x^2−1)

Answer:

Answer: D

Explanation:

The sum of a geometric progression (GP) can be calculated using the formula:

S = a(1-r^n)/(1-r), where a is the first term of the GP, r is the common ratio and n is the number of terms.

In this case, the first term of the GP is x3, the common ratio is x2, and n is the number of terms.

Therefore, the sum of the GP is:

S = x3(x^(2n−1)−1)/(x^2−1)

Question:

If the 4th,10th and 16th terms of a GP are x, y and z respectively, then x,y,z are in : A : AP B : GP C : AGP D : HP

Answer:

Answer: B : GP

Question:

Find four numbers forming a GP in which the third term is greater than the first term by 9 and the second term is greater then 4th by 18. A : 2 , -6, 12, -24 B : 3 , 6, 12 , 24 C : 3 , -6, 12 , -24 D : None of these

Answer:

Answer: C : 3 , -6, 12 , -24

Question:

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?

Answer:

2nd hour: 60 bacteria 4th hour: 120 bacteria nth hour: 30 x 2^n bacteria

Question:

Find the sum of the products of the corresponding terms of the following sequences :- (i) 2,4,8,16,32
(ii) 128,32,8,2,1/2.

Answer:

(i) 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x 1/2

(ii) 128 + 128 + 64 + 32 + 16

Answer: 256 + 448 + 64 + 32 + 16 = 816

Question:

Show that the products of the corresponding terms of the sequences a,ar,ar^2,ar(n−1),and A,A R,AR2,ARn−1 from a G.P. and find the common ratio.

Answer:

Given: Sequences a,ar,ar2,ar(n−1) and A,AR,AR2,ARn−1

Step 1: Determine the common ratio (r) of the sequence a, ar, ar2, ar(n−1).

Step 2: Determine the common ratio (R) of the sequence A, AR, AR2, ARn−1.

Step 3: Show that the products of the corresponding terms of the two sequences (a,ar,ar2,ar(n−1) and A,AR,AR2,ARn−1) are equal.

Step 4: Multiply the first terms of the two sequences (a and A) and compare the result with the product of the second terms (ar and AR), third terms (ar2 and AR2), and so on.

Step 5: If the products of the corresponding terms are equal, then the common ratio of the two sequences is equal (r = R).

Question:

Find the value of n so that a^(n+1)+b^(n+1)/(a^n+b^n) may be the geometric mean between a and b.

Answer:

1. Let G be the geometric mean between a and b.

2. G can be written as the square root of ab.

3. G^2 = ab

4. a^(n+1)+b^(n+1)/(a^n+b^n) = G

5. a^(n+1)+b^(n+1)/(a^n+b^n) = √(ab)

6. a^(n+1)+b^(n+1) = √(ab)(a^n+b^n)

7. a^(2n+2)+b^(2n+2)=ab(a^n+b^n)

8. a^(2n+2)+b^(2n+2)=a^(n+2)b^(n+2)

9. a^(2n+2)-a^(n+2)b^(n+2) = 0

10. a^(n+2)(a^n-b^n) = 0

11. a^n = b^n

12. n = logb/loga

Question:

Find the sum of the GP. 1−a+a^2−a^3+….. to n terms (a≠1).

Answer:

Solution: Step 1: The given GP is 1−a+a^2−a^3+….. to n terms (a≠1).

Step 2: The common ratio (r) of the GP is a.

Step 3: To find the sum of the GP, we will use the formula: S_n = a^n-1/a-1

Step 4: Substituting the values in the formula,

S_n = a^n-1/a-1

Step 5: Therefore, the sum of the GP is S_n = a^n-1/a-1.

Question:

If for a given G.P. a=729 and 7th term is 64, determine S7

Answer:

Given, a = 729 7th term = 64

To find, S7 (7th term of the series)

Solution: Step 1: We know that the general form of a GP is a, ar, ar2, ar3, ar4, ar5, ar6, ar7, …

Step 2: Substitute the given values in the general form.

729, 729r, 729r2, 729r3, 729r4, 729r5, 729r6, 64, …

Step 3: Equate 729r6 = 64

729r6 = 64 r6 = 64/729

Step 4: Calculate the value of r.

r = (64/729)1/6

Step 5: Substitute the value of r in the series.

S7 = 729r7

= 729(64/729)7

Step 6: Calculate the value of S7.

S7 = 729(64/729)7

= 64