Sequences and Series Exercise 5
Question:
A. G.P. consists of of an even numbers of terms . If the sum of all the terms is 5 times the sum of terms occupying odd places, the find its common ratio.
Answer:

A G.P. is a sequence of numbers in which each term is obtained by multiplying the previous term by a fixed number, called the common ratio.

Since the G.P. consists of an even number of terms, the sum of all the terms will be equal to the sum of the terms occupying the odd places multiplied by 5.

To find the common ratio, divide the term in the G.P. occupying the odd place by the term in the G.P. occupying the even place.

The common ratio can then be found by taking the nth root of the ratio obtained in step 3, where n is the total number of terms in the G.P.
Question:
Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
Answer:
Step 1: List all the integers from 1 to 100 that are divisible by 2 or 5. 2, 4, 5, 6, 8, 10, 11, 12, 14, 15, 16, 18, 20, 22, 24, 25, 26, 28, 30, 32, 33, 34, 36, 38, 40, 41, 42, 44, 45, 46, 48, 50, 52, 54, 55, 56, 58, 60, 62, 64, 65, 66, 68, 70, 72, 74, 75, 76, 78, 80, 82, 84, 85, 86, 88, 90, 92, 94, 95, 96, 98, 100
Step 2: Add all the integers together. 2 + 4 + 5 + 6 + 8 + 10 + 11 + 12 + 14 + 15 + 16 + 18 + 20 + 22 + 24 + 25 + 26 + 28 + 30 + 32 + 33 + 34 + 36 + 38 + 40 + 41 + 42 + 44 + 45 + 46 + 48 + 50 + 52 + 54 + 55 + 56 + 58 + 60 + 62 + 64 + 65 + 66 + 68 + 70 + 72 + 74 + 75 + 76 + 78 + 80 + 82 + 84 + 85 + 86 + 88 + 90 + 92 + 94 + 95 + 96 + 98 + 100 = 2,730
Step 3: The sum of all the integers from 1 to 100 that are divisible by 2 or 5 is 2,730.
Question:
Find the sum of all numbers between 200 and 400 which are divisible by 7.
Answer:
Answer: Step 1: Identify the first number that is divisible by 7 between 200 and 400. Answer: 203
Step 2: Identify the last number that is divisible by 7 between 200 and 400. Answer: 399
Step 3: Subtract the first number from the last number to get the total number of numbers divisible by 7 between 200 and 400. Answer: 399  203 = 196
Step 4: Calculate the sum of all numbers divisible by 7 between 200 and 400. Answer: 203 + 210 + 217 + 224 + 231 + 238 + 245 + 252 + 259 + 266 + 273 + 280 + 287 + 294 + 301 + 308 + 315 + 322 + 329 + 336 + 343 + 350 + 357 + 364 + 371 + 378 + 385 + 392 + 399 = 11,742
Question:
Find the sum of the first n terms of the series :3+7+13+21+31+…
Answer:
Step 1: Determine the value of n, which is the number of terms in the series.
Step 2: Calculate the sum of the first n terms using the formula for the sum of an arithmetic series, S = n/2 (2a + (n1)d), where a is the first term, d is the common difference, and n is the number of terms.
Step 3: Substitute the values for a, d, and n into the formula and calculate the sum.
Question:
The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
Answer:
Given: Sum of first four terms = 56 Sum of last four terms = 112 First term = 11
Step 1: Find the common difference (d) of the A.P.
Let the common difference of the A.P. be d.
Sum of first four terms = 11 + (11 + d) + (11 + 2d) + (11 + 3d)
56 = 11 + 11 + d + 11 + 2d + 11 + 3d
56 = 44 + 4d
4d = 56  44
4d = 12
d = 12/4
d = 3
Step 2: Find the number of terms (n)
Sum of last four terms = (11 + (n  4)d) + (11 + (n  3)d) + (11 + (n  2)d) + (11 + (n  1)d)
112 = 11 + (n  4)d + 11 + (n  3)d + 11 + (n  2)d + 11 + (n  1)d
112 = 44 + 4d + (n  4)d
112  44 = 4d + (n  4)d
68 = 5d
d = 68/5
d = 13.6
n = 13.6 + 4
n = 17.6
Therefore, the number of terms in the A.P. is 17.6.
Question:
Let the sum of n,2n,3n terms of an A.P. be S1,S2 and S3, respectively, show that S3=3(S2−S1).
Answer:
Step 1: Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively.
Step 2: S1 = n + (n + d) + (n + 2d) + … (n + (n1)d)
Step 3: S2 = 2n + (2n + d) + (2n + 2d) + … (2n + (2n1)d)
Step 4: S3 = 3n + (3n + d) + (3n + 2d) + … (3n + (3n1)d)
Step 5: Subtracting S1 from S2, we get S2  S1 = 2n + (2n + d) + (2n + 2d) + … (2n + (2n1)d)  (n + (n + d) + (n + 2d) + … (n + (n1)d))
Step 6: Simplifying the above equation, we get S2  S1 = n + (n + d) + (n + 2d) + … (n + (n1)d)
Step 7: Subtracting S2 from S3, we get S3  S2 = 3n + (3n + d) + (3n + 2d) + … (3n + (3n1)d)  (2n + (2n + d) + (2n + 2d) + … (2n + (2n1)d))
Step 8: Simplifying the above equation, we get S3  S2 = n + (n + d) + (n + 2d) + … (n + (n1)d)
Step 9: Since S2  S1 = S3  S2, we get S3  S2 = S2  S1
Step 10: Therefore, S3 = S2 + (S2  S1)
Step 11: Substituting the values of S2 and S1, we get S3 = 3(S2  S1)
Question:
Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder
Answer:
Step 1: First, we need to identify all twodigit numbers that when divided by 4, yield 1 as a remainder.
Step 2: We can use the following equation to determine the twodigit numbers that yield 1 as a remainder when divided by 4:
4n + 1 = twodigit number
Step 3: We can use this equation to find the twodigit numbers that yield 1 as a remainder when divided by 4. For example, when n = 0, 4(0) + 1 = 1, which is a twodigit number. Similarly, when n = 1, 4(1) + 1 = 5, which is also a twodigit number.
Step 4: We can continue to use this equation to find all the twodigit numbers that yield 1 as a remainder when divided by 4. For example, when n = 2, 4(2) + 1 = 9, which is a twodigit number.
Step 5: We can add all the twodigit numbers that yield 1 as a remainder when divided by 4 to find the sum. For example, the sum of 1, 5, and 9 is 15.
Question:
If a,b,c are in A.P.; b,c,d are in G.P. and 1/c,1/d,1/e are in A.P. prove that a,c,e are in G.P.
Answer:
Given: a, b, c are in A.P. b, c, d are in G.P. 1/c, 1/d, 1/e are in A.P.
To Prove: a, c, e are in G.P.
Proof:
Step 1: We know that a, b, c are in A.P., so a + c = 2b
Step 2: We know that b, c, d are in G.P., so c2 = bd
Step 3: We know that 1/c, 1/d, 1/e are in A.P., so 1/d  1/c = 1/e  1/d
Step 4: Substitute c2 = bd in 1/d  1/c = 1/e  1/d, we get b/d  b/c = b/e  b/d
Step 5: Simplify both sides, we get b(1/d  1/c) = b(1/e  1/d)
Step 6: Substitute a + c = 2b in b(1/d  1/c) = b(1/e  1/d), we get (a + c)(1/d  1/c) = (a + c)(1/e  1/d)
Step 7: Simplify both sides, we get (a/d  a/c) = (a/e  a/d)
Step 8: Divide both sides by a, we get (1/d  1/c) = (1/e  1/d)
Step 9: Since (1/d  1/c) = (1/e  1/d), we can conclude that a, c, e are in G.P.
Question:
If a,b,c are in G.P. prove that (a^n+b^n),(b^n+c^n),(c^n+d^n) are in G.P.
Answer:
Given: a, b, c are in G.P.
To Prove: (a^n+b^n), (b^n+c^n), (c^n+d^n) are in G.P.
Proof:
Since a, b, c are in G.P., then it can be written as a, ar, ar^2, ar^3, …
Therefore,
(a^n+b^n) = (ar^n + ar^(n1)) = ar^(n1)(r+1)
(b^n+c^n) = (ar^(n1) + ar^(n2)) = ar^(n2)(r+1)
(c^n+d^n) = (ar^(n2) + ar^(n3)) = ar^(n3)(r+1)
From the above equations, it can be seen that (a^n+b^n), (b^n+c^n), (c^n+d^n) are in G.P. with common ratio r+1.
Hence, the statement is true.
Question:
The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P if it is positive.
Answer:
Step 1: G.P. stands for Geometric Progression. It is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed nonzero number called the common ratio.
Step 2: The first term of the G.P. is 1.
Step 3: The sum of the third term and fifth term is 90.
Step 4: To find the common ratio of the G.P., we need to calculate the third and fifth terms.
Step 5: The third term is obtained by multiplying the first term (1) by the common ratio twice.
Step 6: The fifth term is obtained by multiplying the first term (1) by the common ratio four times.
Step 7: We can set up an equation to solve for the common ratio: 1 x r2 + 1 x r4 = 90
Step 8: We can solve for the common ratio by dividing both sides of the equation by 1: r2 + r4 = 90
Step 9: We can use the quadratic formula to solve for r: r = ±√(90)
Step 10: Since the common ratio must be positive, the solution is r = √(90).
Question:
Show that the sum of (m+n)th and (m−n)th terms of an A.P. is equal to twice the mth term.
Answer:
Given: A.P. = a, a + d, a + 2d, a + 3d, ….
Step 1: Find the expression for the mth and (m+n)th terms of the A.P.
mth term = a + (m  1)d
(m + n)th term = a + (m + n  1)d
Step 2: Find the expression for the (m−n)th term of the A.P.
(m  n)th term = a + (m  n  1)d
Step 3: Now, add the expressions for the (m+n)th and (m−n)th terms of the A.P.
(m + n)th term + (m  n)th term = a + (m + n  1)d + a + (m  n  1)d
Step 4: Simplify the above expression.
(m + n)th term + (m  n)th term = 2a + (m + n  1)d + (m  n  1)d
Step 5: Simplify further.
(m + n)th term + (m  n)th term = 2a + 2md
Step 6: Now, compare the above expression with the expression for the mth term.
mth term = a + (m  1)d
(m + n)th term + (m  n)th term = 2a + 2md
Step 7: From the above comparison, we can conclude that the sum of (m+n)th and (m−n)th terms of an A.P. is equal to twice the mth term.
Question:
The pth,qth and rth terms of an A.P. are a,b,c respectively. Show that (q−r)a+(r−p)b+(p−q)c=0
Answer:
Step 1: Given, pth, qth and rth terms of an A.P. are a, b, c respectively.
Step 2: We need to prove that (q−r)a+(r−p)b+(p−q)c=0
Step 3: We will use the property of an A.P. which states that the sum of any three consecutive terms of an A.P. is zero.
Step 4: We can rewrite the given equation as (qr)a + (rp)b + (pq)c = (qr)a + (rq+qp)b + (pq)c
Step 5: Simplifying the equation, we get (qr)a + (rp)b + (pq)c = 0a + 0b + 0c
Step 6: Since 0a + 0b + 0c = 0, therefore (q−r)a+(r−p)b+(p−q)c=0
Question:
The ratio of the A.M. and G.M. of two positive numbers a and b such that a> b, is m:n. Show that a:b=(m+√(m^2−n^2)):(m−√(m^2−n^2))
Answer:
Step 1: Given, Ratio of A.M. and G.M. of two positive numbers a and b such that a > b = m : n
Step 2: We have to show that, a : b = (m + √(m2 − n2)) : (m − √(m2 − n2))
Step 3: A.M. = (a + b) / 2 G.M. = √(ab)
Step 4: Since, m : n = (a + b) / 2 : √(ab)
Step 5: We get, (a + b) / 2 : √(ab) = m : n
Step 6: Multiplying both sides by 2√(ab), we get 2(a + b) : 2ab = 2m : 2n
Step 7: Rearranging the terms, we get a + b = m(2ab/2n)
Step 8: Simplifying the above equation, we get a + b = m(ab/n)
Step 9: Substituting a = mx and b = nx in the above equation, we get mx + nx = m(mxnx/n)
Step 10: Simplifying the above equation, we get x(m + n) = mxn
Step 11: Dividing both sides by x, we get m + n = mn
Step 12: Solving the above equation, we get x2 = m2 − n2
Step 13: Taking square root on both sides, we get x = ±√(m2 − n2)
Step 14: Substituting x = ±√(m2 − n2) in the equation a = mx and b = nx, we get a = m ± √(m2 − n2) and b = n ± √(m2 − n2)
Step 15: Since, a > b, we get a = m + √(m2 − n2) and b = n − √(m2 − n2)
Step 16: Hence, a : b = (m + √(m2 − n2)) : (m − √(m2 − n2))
Question:
If S1,S2,S3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9S2/2=S3(1+8S1).
Answer:
Given: S1 = Sum of first n natural numbers S2 = Square of S1 S3 = Cube of S1
To prove: 9S2/2 = S3(1 + 8S1)
Step 1: S1 = n(n+1)/2
Step 2: S2 = n2(n+1)2/4
Step 3: S3 = n3(n+1)3/8
Step 4: Substituting the values of S1, S2 and S3 in the equation, we get
9n2(n+1)2/2 = n3(n+1)3/8(1 + 8n(n+1)/2)
Step 5: Simplifying the above equation, we get
9n2(n+1)2/2 = n3(n+1)3/8 + 8n3(n+1)3/4
Step 6: Comparing the coefficients of n3(n+1)3 on both sides, we get
9/2 = 1/8 + 8/4
Step 7: Simplifying the above equation, we get
9/2 = 8/4
Step 8: Therefore, it is proved that
9S2/2 = S3(1 + 8S1)
Question:
Show that (1×2^2+2×3^2+…+n×(n+1)^2)/(1^2×2+2^2×3+…+n^2×(n+1))=(3n+5)/(3n+1)
Answer:

(1×2^2+2×3^2+…+n×(n+1)^2)/(1^2×2+2^2×3+…+n^2×(n+1))

(1×4+2×9+…+n×(n^2+2n+1))/(2×2+4×3+…+n^2×(n+1))

(n^3+3n^2+2n)/(n^3+3n^2+2n+2)

(n^3+3n^2+2n)/(n^3+3n^2+2n+2) = (n^3+3n^2+2n)/(n^3+3n^2+2n+2) = (n^3+3n^2+2n)/(n^3+3n^2+2n+2) = (3n+5)/(3n+1)
Question:
Find the sum of the following series up to n terms: .6+.66+.666+….
Answer:
Step 1: Define the given series. The given series is an arithmetic progression with common difference of 0.06 and first term equal to 0.6.
Step 2: Write the general formula for the sum of an arithmetic sequence. The general formula for the sum of an arithmetic sequence is Sn = n/2 (a1 + an), where a1 is the first term and an is the nth term.
Step 3: Substitute the values of the given series into the formula. Sn = n/2 (0.6 + an)
Step 4: Calculate the nth term. The nth term can be calculated using the formula an = a1 + (n1)d, where a1 is the first term and d is the common difference.
Substituting the values into the formula, we get: an = 0.6 + (n1)0.06
Step 5: Substitute the value of the nth term into the formula for the sum of the series. Sn = n/2 (0.6 + an)
Sn = n/2 (0.6 + 0.6 + (n1)0.06)
Step 6: Simplify the expression. Sn = n/2 (1.2 + (n1)0.06)
Step 7: Calculate the sum of the series. Sn = n/2 (1.2 + 0.06n  0.06)
Sn = 0.6n + 0.6  0.03n
Therefore, the sum of the series up to n terms is 0.6n + 0.6  0.03n.
Question:
Find the sum of series up to n terms: (1^3)/1+(1^3+2^3)/(1+3)+(1^3+2^3+3^3)/(1+3+5)+….
Answer:
Step 1: Calculate the sum of the first n terms of the series.
S = (1^3)/1 + (1^3 + 2^3)/(1 + 3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) + …. + (1^3 + 2^3 + 3^3 + … + n^3)/(1 + 3 + 5 + … + (2n1))
Step 2: Simplify the expression by factorizing the denominator.
S = (1^3)/1 + (1^3 + 2^3)/(1 + 3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) + …. + (1^3 + 2^3 + 3^3 + … + n^3)/((n)(n+1))
Step 3: Expand the numerator of the expression to obtain the sum of the series.
S = (1^3)/1 + (1^3 + 2^3)/(1 + 3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) + …. + (1^3 + 2^3 + 3^3 + … + n^3)/((n)(n+1))
S = (1^3)/1 + (3^3 + 3^3)/(4) + (6^3 + 6^3 + 6^3)/(10) + …. + (n^3 + n^3 + n^3 + … + n^3)/((n)(n+1))
S = 1/1 + (3^3 + 3^3)/(4) + (6^3 + 6^3 + 6^3)/(10) + …. + (n^3 + n^3 + n^3 + … + n^3)/((n)(n+1))
Step 4: Use the formula for the sum of the series of cubes to calculate the sum of the series.
S = 1 + 3(n(n+1)/2)^2
Therefore, the sum of the series up to n terms is given by S = 1 + 3(n(n+1)/2)^2.
Question:
If (a+bx)/(a−bx)=(b+cx)/(b−cx)=(c+dx)/(c−dx)(x≠0)., then show that a,b,c and d are in G.P.
Answer:
Step 1: Rewrite the given equation as (a+bx)(b−cx)=(b+cx)(a−bx) and (b+cx)(c−dx)=(c+dx)(b−cx)
Step 2: Multiply the two equations to get (a+bx)(b−cx)(c−dx)=(b+cx)(a−bx)(c−dx)
Step 3: Simplify the equation to get a(b−cx)(c−dx)=b(a−bx)(c−dx)
Step 4: Divide both sides by (b−cx)(c−dx) to get a/b= (a−bx)/(c−dx)
Step 5: Since a, b, c and d are in G.P., we can write a/b= (a/b)(b/c)(c/d) = (a/c)(b/c)(c/d)
Step 6: Therefore, (a−bx)/(c−dx) = (a/c)(b/c)(c/d)
Step 7: Hence, a, b, c and d are in G.P.
Question:
If the sum of three numbers in increasing A.P., is 24 and their product is 440, find the numbers.
Answer:
Let the three numbers be x, (x+d) and (x+2d)
According to the given condition,
x + (x + d) + (x + 2d) = 24
3x + 3d = 24
3d = 24  3x
d = (24  3x)/3
Product of three numbers = x(x + d)(x + 2d)
= 440
x(x + (24  3x)/3)(x + 2(24  3x)/3) = 440
x(x + 8  x)(x + 16  2x) = 440
x(x + 8)(x + 16) = 440
x2 + 24x + 128 = 440
x2 + 24x  312 = 0
x2 + 12x  156x  312 = 0
x(x + 12)  156(x + 12) = 0
(x  12)(x + 13) = 0
x = 12 or x = 13
Since the numbers are in A.P., x cannot be negative.
Therefore, x = 12
d = (24  3x)/3 = (24  36)/3 = 12/3 = 4
The three numbers are 12, 8 and 4.
Question:
A manufacturer reckons that the value of the machine, which cost him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
Answer:
Solution:
Step 1: Calculate the depreciation rate for 5 years. Depreciation rate for 5 years = 20% × 5 = 100%
Step 2: Calculate the estimated value at the end of 5 years. Estimated value at the end of 5 years = 15625 × (1  100%) = 15625 × 0 = 0
Question:
The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms .
Answer:
Given, First term (a) = 5 Common ratio (r) = 2 Sum of some terms of G.P. (S) = 315
To find, Last term (l) and Number of terms (n)
Formula used, S = a + ar + ar2 + ar3 + … + arn1
S = a(1  rn) / (1  r)
315 = 5(1  rn) / (1  2)
315 = 5(1  2n) / 1
315 = 5(1  2n)
315/5 = 1  2n
63 = 1  2n
2n = 62
n = 31
Number of terms (n) = 31
Last term (l) = arn1
l = 5 x 2(31)
l = 320
Question:
Find the sum of the following series up to n terms: 5+55+555+….
Answer:
Step 1: Identify the first term (a) and the common difference (d) of the series.
In this series, a = 5 and d = 50.
Step 2: Calculate the nth term of the series.
The nth term of the series is given by an = a + (n  1)d.
Therefore, the nth term of the series is an = 5 + (n  1)50 = 50n  45.
Step 3: Calculate the sum of the series up to n terms.
The sum of the series up to n terms is given by Sn = [n/2] (a + an).
Therefore, the sum of the series up to n terms is Sn = [n/2] (5 + 50n  45) = 25n2  22.5n.
Question:
Let S be the sum , P the product and R the sum of reciprocals of n terms in a G.P. Prove that P^2R^n=S^n.
Answer:

P^2R^n = (a*r^(n1))^2(1+r+r^2+…+r^(n1))^n

P^2R^n = a^2r^(2(n1))(1+r+r^2+…+r^(n1))^n

P^2R^n = a^2r^(2(n1))(1+r+r^2+…+r^(n1))(1+r+r^2+…+r^(n1))^(n1)

P^2R^n = a^2r^(2(n1))(1^2+2r+3r^2+…+(n1)r^(n2)+nr^(n1))(1+r+r^2+…+r^(n1))^(n1)

P^2R^n = a^2r^(2(n1))(1+2r+3r^2+…+(n1)r^(n2)+nr^(n1)+r^n)(1+r+r^2+…+r^(n1))^(n1)

P^2R^n = a^2r^(2(n1))(S+r^n)(1+r+r^2+…+r^(n1))^(n1)

P^2R^n = a^2r^(2(n1))S^n

Therefore, P^2R^n = S^n.
Question:
The sum of three numbers in G.P. is 56. If we subtract 1,7,21 from these numbers in that order, we obtain an arithmetic progression . Find the numbers
Answer:
Step 1: Let the three numbers be x, y, and z.
Step 2: Since the sum of three numbers in G.P. is 56, we can write the following equation: x + y + z = 56
Step 3: Since we are subtracting 1, 7, and 21 from these numbers in that order, we can write the following equation: (x  1) + (y  7) + (z  21) = 0
Step 4: We can combine the two equations to get: x + y + z  1  7  21 = 56  0
Step 5: Simplifying the equation, we get: x + y + z = 28
Step 6: Substituting the value of x + y + z in the first equation, we get: x + y + z = 56
Step 7: Solving the two equations simultaneously, we get: x = 20, y = 12, and z = 24.
Question:
If f is a function satisfying f(x+y)=f(x)f(y) for all x,y ∈N such that f(1)=3 and ∑n x=1 f(x)=120, find the value of n .
Answer:
Step 1: Given that f(1) = 3 and ∑n x=1 f(x) = 120, we can substitute f(1) into the summation to get 3 + ∑n x=2 f(x) = 120.
Step 2: We can then expand the summation to get 3 + f(2) + f(3) + … + f(n) = 120.
Step 3: We can then rearrange the equation to get f(2) + f(3) + … + f(n) = 117.
Step 4: Since f is a function satisfying f(x+y) = f(x)f(y) for all x,y ∈N, we can substitute f(2) = f(1)f(1) = 33 = 9 and f(3) = f(1)f(2) = 39 = 27.
Step 5: We can then continue this pattern to get f(4) = 327 = 81, f(5) = 381 = 243, and so on.
Step 6: We can then add up all of these terms to get 9 + 27 + 81 + 243 + … + f(n) = 117.
Step 7: We can then solve for n by finding the term in the summation that is equal to 117. In this case, it is n = 6, so the value of n is 6.
Question:
If a(1/b+1/c),b(1/c+1/a),c(1/a+1/b) are in A.P. prove that a,b,c are in A.P.
Answer:
Given, a(1/b+1/c),b(1/c+1/a),c(1/a+1/b) are in A.P.
Step 1: We need to prove that a,b,c are in A.P.
Step 2: Let us consider the given expression, a(1/b+1/c) = b(1/c+1/a) = c(1/a+1/b)
Step 3: On solving the above expression, we get a/b = b/c = c/a
Step 4: Now, dividing the above equation by a, we get 1/b = b/ac = c/a2
Step 5: Again, dividing the above equation by b, we get 1/c = c/ab = a/b2
Step 6: Finally, dividing the above equation by c, we get 1/a = a/bc = b/c2
Step 7: From the above equations, we can conclude that a,b,c are in A.P.
Hence, proved.
Question:
If a and b are the roots of x^2−3x+p=0 and c,d are the roots of x^2−3x+p=0, where a,b,c,d form a G.P. Prove that (q+p):(q−p)=17:15
Answer:
Given, a and b are the roots of x^2−3x+p=0 c and d are the roots of x^2−3x+q=0, where a,b,c,d form a G.P
To prove: (q+p):(q−p)=17:15
Step 1: Let us consider the general equation of a quadratic polynomial
x^2−3x+k=0
where k is any constant.
Step 2: Let a and b be the roots of the equation x^2−3x+p=0
Therefore,
a+b=3 and ab=p
Step 3: Let c and d be the roots of the equation x^2−3x+q=0
Therefore,
c+d=3 and cd=q
Step 4: Since a,b,c,d form a G.P.,
ac=bd
Step 5: From Step 2 and Step 3, we have
ab=p and cd=q
Substituting these values in Step 4, we get
pq=bd
Step 6: From Step 2 and Step 3, we have
a+b=3 and c+d=3
Adding these equations, we get
a+b+c+d=6
Step 7: From Step 2 and Step 3, we have
ab=p and cd=q
Multiplying these equations, we get
pq=bd
Step 8: Substituting the value of pq from Step 5 and bd from Step 7 in the equation (a+b+c+d)=6, we get
pq+bd=6
Step 9: Rearranging the equation in Step 8, we get
(p+q)+(b+d)=6
Step 10: From Step 2 and Step 3, we have
a+b=3 and c+d=3
Adding these equations, we get
(a+b)+(c+d)=6
Step 11: Substituting the value of (a+b) from Step 2 and (c+d) from Step 3 in the equation (a+b)+(c+d)=6, we get
(p+q)+(b+d)=6
Step 12: Comparing the equations in Step 9 and Step 11, we get
(p+q)+(b+d)=(p+q)+(b+d)
Step 13: Therefore,
(p+q)=(b+d)
Step 14: From Step 2 and Step 3, we have
a+b=3 and c+d=3
Subtracting these equations, we get
(a−b)=(c−d)
Step 15: From Step 2 and Step 3, we have
ab=p and cd=q
Dividing these equations, we get
(a−b)=(c−d)
Step 16: Substituting the value of (a−b) from Step 14 and (c−d) from Step 15 in the equation (p+q)=(b+d), we get
(p+q)=(q−p)
Step 17: Therefore,
(q+p):(q−p)=(q+p):(q−p)
Step 18: Since (q+p):(q−p)=(q+p):(q−p)
Therefore,
(q+p):(q−p)=17:15
Question:
Find the 20th term of the series 2×4+4×6+6×8+…+n terms.
Answer:
Step 1: To find the 20th term of the series, we need to know the value of ’n'.
Step 2: The series is in the form of 2×4+4×6+6×8+…+n terms.
Step 3: We can find the value of ’n’ by substituting the values of the first 19 terms.
Step 4: 2×4+4×6+6×8+…+18×20 = 2×4+4×6+6×8+…+18×20 = 19×20 = 380
Step 5: Therefore, the value of ’n’ is 20.
Step 6: The 20th term of the series is 20×22 = 440.
Question:
Shamshad Ali buys a scooter for Rs 22000. He pays 4000 cash and agrees to pay the balance in annual installment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?
Answer:
Shamshad Ali will pay a total of Rs 28000 for the scooter.
Calculation: Initial cost: Rs 22000 Cash payment: Rs 4000 Balance: Rs 18000 Interest on Rs 18000 @ 10% = Rs 1800 Total cost: Rs 18000 + Rs 1800 = Rs 28000
Question:
A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual installment of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him ?
Answer:
Step 1: Calculate the amount of interest on the unpaid amount.
Interest = 12% of Rs 6000 = Rs 720
Step 2: Calculate the total amount to be paid.
Total amount = Rs 6000 + Rs 500 (annual installment) + Rs 720 (interest) = Rs 7220
Step 3: Calculate the total cost of the tractor.
Total cost of the tractor = Rs 12000 + Rs 7220 = Rs 19220
Question:
A man deposited Rs 10000 in a bank with interest rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.
Answer:
 Amount in 15th year = Rs 10000 x (1 + 5/100)^15
 Amount in 15th year = Rs 19,976.28
 Amount after 20 years = Rs 10000 x (1 + 5/100)^20
 Amount after 20 years = Rs 25,937.60
Question:
150 workers were engaged to finish a job in a certain number of days.4 workers dropped out on second day, 4more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Answer:
Answer:
Step 1: Calculate the total number of workers who dropped out from the job.
4 workers dropped out on the second day 4 more workers dropped out on the third day 4 more workers dropped out on the fourth day 4 more workers dropped out on the fifth day 4 more workers dropped out on the sixth day 4 more workers dropped out on the seventh day
Total number of workers who dropped out = 4 + 4 + 4 + 4 + 4 + 4 = 24 workers
Step 2: Calculate the number of workers remaining to finish the job.
Number of workers remaining = 150  24 = 126 workers
Step 3: Calculate the number of days in which the work was completed.
Total number of days taken to complete the work = 8 days
Number of days in which the work was completed = 8 + 7 = 15 days
Question:
A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it cost 50 paise to mail one letter, Find the amount spent on the postage when 8th set of letter is mailed.
Answer:
Answer:
 A person writes a letter to four of his friends.
 Each one of them copies the letter and mails to four different persons.
 The chain is not broken and it cost 50 paise to mail one letter.
 The 8th set of letter is mailed.
 The amount spent on the postage is 8 x 50 paise = Rs. 4.00.