Solution Soaps are sodium or potassium salts of long-chain fatty acids. Hard water contains calcium and magnesium ions. When soaps are dissolved in hard water, these ions displace sodium or...
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RS aggarwal class 11 Maths Arithmetic progressions
If the sums of n terms of two APs are in ratio (2n + 3) : (3n + 2), find the ratio of their 10th terms.
Given: sums of n terms of two APs are in ratio (2n + 3) : (3n + 2) To find: find the ratio of their 10th terms. For the sum of n terms of two APs is given by
The sum of n terms of an AP is 1/2 an^2 + bn. Find the common difference.
Given: the sum of n terms of an AP is 1/2 (an^2 + bn) To find: common difference. Put n = 1 we get First term = 1/2+ b Put n = 2 we get First term + second term = 2 Γ a + 2 Γ b Second term = 3/2(a +...
Write the sum of first n odd natural numbers.
n odd natural numbers are given by 3,5,7,9,.......
Write the sum of first n even natural numbers.
Even natural numbers are 2, 4, 6, 8.....
In an AP, the pth term is q and (p + q)th term is 0. Show that its qth term is p.
Given: pth term is q and (p + q)th term is 0. To prove: qth term is p. p th term is given by q = a + (p - 1) Γ d......equation1 (p + q)th term is given by 0 = a + (p + q - 1) Γ d 0 = a + (p - 1) Γ d...
The first and last terms of an AP are 1 and 11 respectively. If the sum of its terms is 36, find the number of terms.
Given: the sum of its terms is 36, the first and last terms of an AP are 1 and 11. To find: the number of terms Sum of AP using first and last terms is given by 36 Γ 2 = n (1 + 11) n = 6
What is the 10th common term between the APs 3, 7, 11, 15, 19, … and 1, 6, 11, 16, …?
To find: 10th common term between the APs Common difference of 1st series = 4 Common difference of 2nd series = 5 LCM of common difference will give us a common difference of new series βΉ 5 Γ 4 βΉ 20...
If 7th and 13th terms of an AP be 34 and 64 respectively then find its 18thterm.
Given: 7th term is 34 and 8th term is 64 To find: find its 18th term 34 = a + 6d .............equation1 64 = a + 12d ............ equation2 Subtract equation1 from equation2 we get d = 5 Put in...
How many 2 – digit numbers are divisible by 7?
The first 2 digit number divisible by 7 is 14, and the last 2 digit number divisible by 7 is 98, so it forms AP with common difference 7 14,...,98 98 = 14 + (n - 1) Γ 7 n = 22
Find the 8th term from the end of the AP 7, 9, 11, …., 201.
To find: 8th term from the end d = 9 - 7 d = 2 Also 201 = 7 + n Γ 2 β 2 n = 98 So 8th term from end will be 7 + 90 Γ 2 βΉ 187
How many terms are there in the AP 13, 16, 19, …., 43?
To find: number of terms in AP Also d = 16 β 13 d = 3 Also 43 = 13 + n Γ 3 β 3 So n = 11
The 9th term of an AP is 0. Prove that its 29th term is double the 19th term.
Given :9th term is 0 To prove: 29th term is double the 19th term a + 8d = 0 a = - 8d 29th term is a + 28d βΉ 20d 19th term is a + 18d βΉ 10d Hence proved 29th term is double the 19th term
Find three arithmetic means between 6 and – 6.
let the three AM be x1,x2,x3. So new AP will be 6,x1,x2,x3, - 6 Also - 6 = 6 + 4d d = - 3 x1 = 3 x2 = 0 x3 = - 3
In an AP it is given that Sn = qn2 and Sm = qm2 . Prove that Sq = q^3 .
Given: Sn = qn2 , Sm = qm2 To prove: Sq = q3 Put n = 1 we get a = q ...... equation 1 Put n = 2 2a + d = 4q ......equation 2 Using equation 1 and 2 we get d = 2q
If 9 times the 9th term of an AP is equal to 13 times the 13th term, show that its 22nd term is 0.
Given : 9 Γ (9th term) = 13 Γ (13th term) To prove: 22nd term is 0 9 Γ (a + 8d) = 13 Γ (a + 12d) 9a + 72d = 13a + 156d - 4a = 84d a = - 21d .....Equation 1 Also 22nd term is given by a + 21d Using...
If the sum of n terms of an AP is given by Sn = (2n2 + 3n), then find its common difference.
Given: Sn = (2n2 + 3n) To find: find common difference Put n = 1 we get S1 = 5 OR we can write a = 5 ...equation 1 Similarly put n = 2 we get S2 = 14 OR we can write 2a + d = 14 Using equation 1 we...
If a, b, c are in AP, show that given are also in AP.
Proof: a, b, c are in A.P. If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P. Multiplying the A.P. with (ab + bc + ac) If a constant is subtracted from...
If given are in AP, prove that a2 (b + c), b2 (c + a), c 2 (a + b) are in AP.
If 1/ a , 1/ b , 1/ c are in AP, prove that
If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P. Multiplying the A.P. with ( a + b + c ) If a constant is subtracted from each term of an A.P., the...
If 1/ a , 1/ b , 1/ c are in AP, prove that
If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P. Multiplying the A.P. with (a + b + c ) If a constant is subtracted from each term of an A.P., the...
A farmer buys a used for βΉ180000. He pays βΉ90000 in cash and agrees to pay the balance in annual instalments of βΉ9000 plus 12% interest on the unpaid amount. How much did the tractor cost him?
Given: -The amount that is to be paid to buy a tractor = βΉ180000. An amount that he paid by cash = βΉ90000. Remaining balance = βΉ90000 Annual instalment = βΉ9000 + interest @12% on unpaid amount....
Arun buys a scooter for βΉ44000. He pays βΉ8000 in cash and agrees to pay the balance in annual instalments of βΉ4000 each plus 10% interest on the unpaid amount. How much did he pay for it?
Given: The amount that is to be paid to buy a scooter = 44000 The amount that he paid by cash = βΉ8000 Remaining balance = βΉ36000 Annual instalment = βΉ4000 + interest@10% on the unpaid amount Thus,...
A carpenter was hired to build 192 window frames. The first day he made 5 frames and each day, thereafter he made 2 more frames than he made the day before. How many days did he take to finish the job?
Let the carpenter take n days to finish the job. To Find: n He builds 5 frames on day 1, 7 on day 2, 9 on day 3 and so on. So it forms an AP 5, 7, 9, 11,... and so on. We need to find the number of...
If Sm = m2p and Sn = n2p, where m β n in an AP then prove that Sp = p3 .
Let the first term of the AP be a and the common difference be d Given: Sm = m2p and Sn = n2p To prove: Sp = p3 According to the problem (m - n)d = 2p(m - n) Now m is not equal to n So d = 2p...
If the sum of n terms of an AP is given, where P and Q are constants then find the common difference.
Let the first term be a and common difference be d To Find: d β 2a + (n - 1)d = 2P + (n - 1)Q β 2(a - P) = (n - 1)(Q - d) Put n = 1 to get the first term as sum of 1 term of an AP is the term...
Find the sum of all natural numbers from 1 and 100 which are divisible by 4 or 5.
To Find: The sum of all natural numbers from 1 to 100 which are divisible by 4 or 5. A number divisible by both 4 and 5 should be divisible by 20 which is the LCM of 4 and 5. Sum of numbers...
The sum of first 7 terms of an AP is 10 and that of next 7 terms is 17. Find the AP.
To Find: AP Given: Sum of first 7 terms = 10 Sum of next 7 terms = 17 According to the problem, Sum of first 14 terms of the given AP is 10 + 17 = 27.
Find the sum of all odd integers from 1 to 201.
To Find: The sum of all odd integers from 1 to 201. The odd integers form the following AP series: 1,3,5....201 First term = a = 1 Common difference = d = 2 Last term = 201 Let the number of terms...
How many terms of the AP 20, 19 1 3 , 18 2 3 must be taken to make the sum 300? Explain the double answer.
To Find: Number of terms required to make the sum of the AP 300. Let the first term of the AP be a and the common difference be d β 300 Γ 6 = n[120 - 2(n - 1)] β n[ - 2n + 122] = 6 Γ 300 β n( - n +...
If a, b, c are in AP, show that (b + c β a), (c + a β b), (a + b β c) are in AP.
(b + c β a), (c + a β b), (a + b β c) are in AP. To prove: (b + c β a), (c + a β b), (a + b β c) are in AP. Given: a, b, c are in A.P. Proof: Let d be the common difference for the A.P. a,b,c Since...
If a, b, c are in AP, show that (a + 2b β c)(2b + c β a)(c + a β b) = 4abc.
To prove: (a + 2b β c)(2b + c β a)(c + a β b) = 4abc. Given: a, b, c are in A.P. Proof: Since a, b, c are in A.P. β 2b = a + c ... (i) Taking LHS = (a + 2b β c) (2b + c β a) (c + a β b) Substituting...
If a, b, c are in AP, prove that a3 + c3 + 6abc = 8b3
a3 + c3 + 6abc = 8b3 To prove: a3 + c3 + 6abc = 8b3 Given: a, b, c are in A.P. Formula used: (a+b)3 = a3 + 3ab(a+b) + b3 Proof: Since a, b, c are in A.P. β 2b = a + c ... (i) Cubing both side,
If a, b, c are in AP, prove that a2 + c2 + 4ac = 2(ab + bc + ca)
a2 + c2 + 4ac = 2(ab + bc + ca) To prove: a2 + c2 + 4ac = 2(ab + bc + ca) Given: a, b, c are in A.P. Proof: Since a, b, c are in A.P. β 2b = a + c
If a, b, c are in AP, prove that (a β c)2 = 4(a β b)(b β c)
(a β c)2 = 4(a β b)(b β c) To prove: (a β c)2 = 4(a β b)(b β c) Given: a, b, c are in A.P. Proof: Since a, b, c are in A.P. β c β b = b β a = common difference β b β c = a β b ... (i) And, 2b = a +...
Prove that the ratio of sum of m arithmetic means between the two numbers to the sum of n arithmetic means between them is m:n.
To prove: ratio of sum of m arithmetic means between the two numbers to the sum of n arithmetic means between them is m:n a = First term l = Last term Let the first series of arithmetic mean having...
Insert five numbers between 11 and 29 such that the resulting sequence is an AP.
To find: Five numbers between 11 and 29, which are in A.P. Given: (i) The numbers are 11 and 29 Formula used: (i) An = a + (n-1)d Let the five numbers be A1, A2, A3, A4 and A5 According to question...
Insert arithmetic means between 16 and 65 such that the 5th AM is 51. Find the number of arithmetic means.
To find: The number of arithmetic means Given: (i) The numbers are 16 and 65 (ii) 5th arithmetic mean is 51 Using Formula, An = a + nd First arithmetic mean, A1 = a + d = 16 + 7 = 23 Second...
There is n arithmetic means between 9 and 27. If the ratio of the last mean to the first mean is 2 : 1, find the value of n.
To find: The value of n Given: (i) The numbers are 9 and 27 (ii) The ratio of the last mean to the first mean is 2 : 1 Formula used: (i) , where, d is the common difference n is the number of...
Insert six arithmetic means between 11 and -10.
To find: Six arithmetic means between 11 and -10 Formula used: (i) , where, d is the common difference n is the number of arithmetic means (ii) An = a + nd We have 11 and -10 Using Formula, An = a +...
Insert three arithmetic means between 23 and 7.
To find: Three arithmetic means between 23 and 7 n is the number of arithmetic means (ii) An = a + nd We have 23 and 7 Using Formula, An = a + nd First arithmetic mean, A1 = a + d = 23 + (-4) = 19...
Insert four arithmetic means between 4 and 29.
To find: Four arithmetic means between 4 and 29 Using Formula, An = a + nd First arithmetic mean, A1 = a + d = 4 + 5 = 9 Second arithmetic mean, A2 = a + 2d = 4 + 2(5) = 4 + 10 = 14 Third arithmetic...
Find the arithmetic mean between:-16 and -8
To find: Arithmetic mean between -16 and -8 The formula used: Arithmetic mean between = a and b = a+b/2 We have -16 and -8 A.M.=(-16)+(-8)2=-242=-12
Find the arithmetic mean between: (i) 15 and -7
To find: Arithmetic mean between 15 and -7 The formula used: Arithmetic mean between a and b = a+b/2 We have 15 and -7 A.M.=15+(-7)2=82=4
Find the arithmetic mean between: (i) 9 and 19
(i) 9 and 19 To find: Arithmetic mean between 9 and 19 The formula used: Arithmetic mean between a and b = a+b/2 We have 9 and 19 A.M.=9+192=282=14
A manufacturer of TV sets produced 6000 units in the third year and 7000 units in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the production (i) in the first year, (ii) in the 10th year, (iii) in 7 years.
Hint: - In the question it is mentioned that the production increases by a fixed number every year. So it is an A.P. (a1, a2, a3, a4, ........an - 1, an). Given: - The 3rd year production is 6000...
A man arranges to pay off a debt of βΉ36000 by 40 annual instalments which form an AP. When 30 of the instalments are paid, he dies, leaving one – third of the debt unpaid. Find the value of the first instalment.
Given: - Total debt = Rs.36000 A man pays this debt in 40 annual instalments that forms an A.P. After annual instalments, that man dies leaving one - third of the debt unpaid. So, Within 30...
A man saves βΉ4000 during the first year, βΉ5000 during the second year and in this way he increases his savings by βΉ1000 every year. Find in what time his savings will be βΉ85000.
A Man saves some amount of money every year. In the first year, he saves Rs.4000. In the next year, he saves Rs.5000. Like this, he increases his savings by Rs.1000 ever year. Given a total amount...
150 workers were engaged to finish a piece of work in a certain number of days. Four workers dropped the second day, four more workers dropped the third day, and so on. It takes 8 more days to finish work now. Find the number of days in which the work was completed.
Given: - Initially let the work can be completed in n days when 150 workers work on every day. However every day 4 workers are being dropped from the work so that work took 8 more days to be...
A man saved βΉ660000 in 20 years. In each succeeding year after the first year, he saved βΉ2000 more than what he saved in the previous year. How much did he save in the first year?
Given: - Amount saved by a man in 20 years is Rs.660000. Let the amount saved by him in the first year be . In every succeeding year, he saves Rs.2000 more than what he saved in the previous year....
A man accepts a position with an initial salary of βΉ26000 per month. It is understood that he will receive an automatic increase of βΉ250 in the very next month and each month thereafter. Find this (i) salary for the 10th month, (ii) total earnings during the first year.
Given: - An initial salary that will be given = βΉ26000 There will be an automatic increase of βΉ250 per month from the very next month and thereafter. Hint: - In the given information the salaries he...
Two cars start together from the same place in the same direction. The first go with a uniform speed of 60 km/hr. The second goes at a speed of 48 km/hr in the first hour and increases the speed by 1 km each succeeding hour. After how many hours will the second car overtake the first car if both cars go non – stop?
Given : Two cars start together from the same place and move in the same direction. The first car moves with a uniform speed of 60km/hr. The second car moves with 48km/hr in the first hour and...
There are 30 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A Gardner waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the Gardner will cover in order to water all the trees.
Hint: Distances between trees and well are in A.P. Given: The distance of well from its nearest tree is 10 metres Distance between each tree is 5 metres. So, In A.P The first term is 10 metres and...
A circle is completely divided into n sectors in such a way that the angles of the sectors are in AP. If the smallest of these angles is 80 and the largest is 720 , calculate n and the angle in the fifth sector.
A circle is divided into n sectors. Given, Angles are in A.P Smallest angle = a = 8Β° Largest angle = l = 72Β° Final term of last term of an A.P series is l = a + (n - 1)Γd So, 72Β° = 8Β° + (n - 1) Γd...
The interior angles of a polygon are in AP. The smallest angle is 520, and the common difference is 80 . Find the number of sides of the polygon.
Given: Interior angles of a polygon are in A.P Smallest angle = a = 52Β° Common difference = d = 8Β° Let the number of sides of a polygon = n Angles are in the following order 52Β°, 52Β° + d, 52Β° + 2d,...
If the sum of n terms of an AP is (3n2 + 5n) and its mth term is 164, find the value of m.
To Find: m Given: Sum of n terms, mth term Put n = 1 to get the first term So a1 = 3 + 5 = 8 Put n = 2 to get the sum of first and second term So a1 + a2 = 12 + 10 = 22 So a2 = 14 Common difference...
Find the sum of all integers between 100 and 600, each of which when divided by 5 leaves 2 as remainder.
The integers between 100 and 600 divisible by 5 and leaves remainder 2 are 102, 107, 112, 117,..., 597. To Find: Sum of the above AP Here a = 102, d = 5, l = 597 a + (n - 1)d = 597 β102 + 5(n - 1) =...
Find the sum of all integers between 101 and 500, which are divisible by 9.
To Find: Sum of all integers between 101 and 500 divisible by 9 The integers between 101 and 500 divisible by 9 are 108, 117, 126,..., 495(Add 9 to 108 to get 117, 9 to 117 to get 126 and so on)....
Find the sum of all even integers between 101 and 199.
To Find: The sum of all even integers between 101 and 199. The even integers form the following AP series - 102, 104, ..., 198 It is and AP series with a = 102 and l = 198. 198 = 102 + (n - 1)2 β 96...
If the ratio between the sums of n terms of two arithmetic progressions is (7n + 1) : (4n + 27), find the ratio of their 11th terms.
Given: Ratio of sum of nth terms of 2 APβs To Find: Ratio of their 11th terms Let us consider 2 AP series AP1 and AP2. Putting n = 1, 2, 3... we get AP1 as 8, 15 22... and AP2 as 31, 35, 39.... So,...
The sums of an terms of two arithmetic progressions are in the ratio (7n β 5) : (5n + 17). Show that their 6th terms are equal.
Given: Ratio of sum of n terms of 2 APβs To Prove: 6th terms of both APβS are equal Let us consider 2 AP series AP1 and AP2. Putting n = 1, 2, 3... we get AP1 as 12,19,26... and AP2 as 22,27,32.......
How many terms of the AP 18 16, 14, 12, …. are needed to give the sum 78? Explain the double answer.
Explanation: Since the given AP is a decreasing progression where an - 1>an,it is bound to have negative values in the series. Sn is maximum for n = 9 or n = 10 since T10 is 0(S10 = S9 = Smax =...
How many terms of the AP 26, 21 16, 11, …. are needed to give the sum 11?
If the sum of a certain number of terms of the AP 27, 24, 21, 18, …. is β30, find the last term.
Find the sum of n term of an AP whose rth term is (5r + 1).
Find the rth term of the AP, the sum of whose first n terms is (3n2 + 2n).
Given: The sum of first n terms. To Find: The rth term. Let the first term be a and common difference be d Put n = 1 to get the first term a = S1 = 3 + 2 = 5 Put n = 2 to get a + (a + d)2a + d = 12...
Find the value of x such that 25 + 22 + 19 + 16 + …. + x = 112.
To Find: The value of x, i.e. the last term. Given: The series and its sum. The series can be written as x, (x + 3), ..., 16, 19, 22, 25 Let there be n terms in the series 25 = x + (n - 1)3 3(n β 1)...
Find the sum of the series 1 + 4 + 7 + 10 + …. + x = 715.
Note: The sum of the series is already provided in the question. The solution to find x is given below. Let there be n terms in the series. x = 1 + (n β 1)3 = 3n β 2 Let S be the sum of the series β...
Find the sum of the series 101 + 99 + 97 + 95 + …. + 43.
To Find: The sum of the given series. Sum of the series is given by Where n is the number of terms, a is the first term and l is the last term Here a = 101, l = 43, n = 30 = 15 Γ 144 = 2160 The sum...
Find the sum of the series 2 + 5 + 8 + 11 + …. + 191.
To Find: The sum of the given series. The nth term of an AP series is given by tn = a + (n β 1)d β191 = 2 + (n - 1)3 β3(n β 1) = 189 β n β 1 = 63 β n = 64 Therefore, The sum of the series is...
Find the sum of n terms of the given AP.
To Find: The sum of n terms of the given AP. Sum of n terms of an AP with first term a and common difference d is given by Here a = x - y, d = 2x - y
Find the sum of 20 terms of the AP (x + y), (x β y), (x β 3y), ….
To Find: The sum of 20 terms of the given AP. Sum of n terms of an AP with first term a and common difference d is given by Here a = x + y, n = 20, d = - 2y β S = 10[2x + 2y + 19( - 2y)] = 10[2x +...
Find the sum of 100 term of the AP 0.6, 0.61, 0.62, 0.63, ….
To Find: The sum of 100 terms of the given AP series. Sum of n terms of an AP with first term a and common difference d is given by Here a = 0.6, n = 100, d = 0.01 Sum of the series is...
Find the sum of 25 terms of the AP β2, 2β2, 3β2, 4β2, ….
To Find: The sum of 25 terms of the given AP series. Sum of n terms of an AP with first term a and common difference d is given by Here, Sum of 25 terms is 325β2.
Find the sum of 16 terms of the AP 6, 5 1 3 , 4 2 3 , 4, ….
To find: Sum of 16 terms of the AP Given: First term = 6 Common difference=-23 The sum of first 16 terms of the series is 16
Find the sum of 23 terms of the AP 17, 12, 7, 2, β3, ….
To Find: The sum of 25 terms of the given AP series. Sum of n terms of an AP with first term a and common difference d is given by Here, a = 17, n = 23 and d = - 5 Sum of 23 terms of the AP IS β...
A man starts repaying a loan as the first instalment of 10000. If he increases the instalment by 500 every month, what amount will he pay in 30thinstalment?
To Find: what amount will he pay in the 30th instalment. Given: first instalment =10000 and it increases the instalment by 500 every month. β΄ So it form an AP with first term is 10000, common...
A side of an equilateral triangle is 24 cm long. A second equilateral triangle is inscribed in it by joining the midpoints of the sides of the first triangle; the process is continued. Find the perimeter of the sixth inscribed equilateral triangle.
To Find: The perimeter of the sixth inscribed equilateral triangle. Given: Side of an equilateral triangle is 24 cm long. As 2nd triangle is formed by joining the midpoints of the sides of the first...
We know that the sum of the interior angles of a triangle is 180Β°. Show that the sum of the interior angles of polygons with 3, 4, 5, 6, …. sides form an arithmetic progression. Find the sum of the interior angles for a 21 – sided polygon.
Show that: the sum of the interior angles of polygons with 3, 4, 5, 6, .... sides form an arithmetic progression. To Find: The sum of the interior angles for a 21 - sided polygon. Given: That the...
Find the number of terms common to the two arithmetic progressions 5, 9, 13, 17, …., 217 and 3, 9, 15, 21, …., 321.
To Find: The number of terms common to both AP Given: The 2 APβs are 5, 9, 13, 17, ...., 217 and 3, 9, 15, 21, ...., 321 As we find that first common term of both AP is 9 and the second common term...
The digits of a 3 – digit number are in AP, and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
To Find: The number Given: The digits of a 3 β digit number are in AP, and their sum is 15. Let required digit of 3 - digit number be (a - d), (a), (a + d). Then, (a - d) + (a) + (a + d)=15 β 3a =...
The angles of a quadrilateral are in AP whose common difference is 10Β°. Find the angles.
To Find: The angles of a quadrilateral. Given: Angles of a quadrilateral are in AP with common difference = 10Β°. Let the required angles be a, (a + 10Β°), (a + 20Β°) and (a + 30Β°). Then, a + (a + 10Β°)...
The sum of three consecutive terms of an AP is 21, and the sum of the squares of these terms is 165. Find these terms
To Find: The three numbers which are in AP. Given: Sum and sum of the squares of three numbers are 21 and 165 respectively. Let required number be (a - d), (a), (a + d). Then, (a β d) + a + (a + d)...
Three numbers are in AP. If their sum is 27 and their product is 648, find the numbers.
To Find: The three numbers which are in AP. Given: Sum and product of three numbers are 27 and 648 respectively. Let required number be (a - d), (a), (a + d). Then, (a - d) + a + (a + d) = 27 β 3a =...
In an AP, it is being given that T4/ T7 = 2 3 . Find T7/ T10 .
(Where Tn is nth term and d is common difference of given AP) Formula Used: Tn = a + (n - 1)d
If ΞΈ1, ΞΈ2, ΞΈ3, …., ΞΈn are in AP whose common difference is d, show that
How many 2 – digit numbers are divisible by 3?
To Find : 2 - digit numbers divisible by 3. First 2 - digit number divisible by 3 is 12 Second 2 - digit number divisible by 3 is 15 and Last 2 - digit number divisible by is 99. Given: The AP is...
How many 3 – digit numbers are divisible by 7?
To Find : 3 - digit numbers divisible by 7. First 3 - digit number divisible by 7 is 105 Second 3 - digit number divisible by 7 is 112 and Last 3 - digit number divisible by 7 is 994. Given: The AP...
Find the 16th term from the end of the AP 7, 2, β3, β8, β13, …., β113
To Find : 28th term from the end of the AP. Given: The AP is 7, 2, β3, β8, β13, ...., β113 a1 = 7, a2 = 2, d = 2β7 = β5 and l = β113 Formula Used: nth term from the end = lβ (nβ1)d (Where l is last...
Find the 28th term from the end of the AP 6, 9, 12, 15, 18, …., 102.
To Find 28th term from the end of the AP. Given: The AP is 6, 9, 12, 15, 18, ...., 102 a1 = 6, a2 = 9, d = 9β6 = 3 and l = 102 Formula Used: nth term from the end = lβ (nβ1)d (Where lis last term...
If 7 times the 7th term of an AP is equal to 11 times its 11th term, show that the 18th term of the AP is zero.
Show that: 18th term of the AP is zero. Given: 7a7= 11a11 (Where a7 is Seventh term, a11 is Eleventh term, an is nth term and d is common difference of given AP) Formula Used: an = a + (n - 1)d 7(a...
The 4th term of an AP is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.
To Find: First term (a) and common difference (d) Given: a4= 3a1 and a7 = 2a3 + 1 (Where a=a1 is first term, an is nth term and d is common difference of given AP) Formula Used: an = a + (n - 1)d...
If the 9th term of an AP is 0, prove that its 29th term is double the 19thterm.
Prove that: 29th term is double the 19th term (i.e. a29 = 2a19) Given: a9= 0 (Where a=a1 is first term, a2 is second term, an is nth term and d is common difference of given AP) Formula Used: an = a...
The 2nd, 31st and the last terms of an AP are 7 3/ 4 , 1/ 2 and β6 1/ 2 respectively. Find the first term and the number of terms.
To Find: First term and number of terms. On solving both equation, we get a = 8 and d = β0.25 On solving the above equation, we get N = 59 So the First term is equal to 8 and the number of terms is...
The 5th and 13th terms of an AP are 5 and β3 respectively. Find the AP and its 30th term.
To Find: AP and its 30th term (i.e. a30=?) Given: a5 = 5 and a13 = β3 Formula Used: an = a + (n - 1)d (Where a=a1 is first term, a2 is second term, an is nth term and d is common difference of given...
Is – 47 a term of the AP 5, 2, β1, β4, β7, ….?
To Find: β47 is a term of the AP or not. Given: The series is 5, 2, β1, β4, β7, .... a1=5, a2= 2, and d=2β5 = β3 (Let suppose an = β47) NOTE: n is a natural number. (Where a=a1 is first term, a2 is...
How many terms are there in the AP
To Find: we need to find number of terms in the given AP. (Where a=a1 is first term, a2 is second term, an is nth term and d is common difference of given AP) Formula Used: an = a + (n - 1)d 27 =...
How many terms are there in the AP 11, 18, 25, 32, 39, …. 207?
To Find: we need to find a number of terms in the given AP. Given: The series is 11, 18, 25, 32, 39, .... 207 a1=11, a2= 18,d=18β11 = 7 and an=207 (Where a=a1 is first term, a2 is second term, an is...
Which term of the AP 9, 14, 19,24, 29, …. is 379?
To Find: we need to find n when an = 379 Given: The series is 9, 14, 19,24, 29, .... and an=379 a1=9, a2= 14 and d=14β9 = 5 (Where a=a1 is first term, a2 is second term, an is nth term and d is...
Find the nth term of the AP
To Find: nth term of the AP So the nth term of AP is equal toΒ 7-n6
Find the nth term of the AP 8, 3, β2, β7, β12, ….
To Find: nth term of the AP Given: The series is 8, 3, β2, β7, β12, .... a1=8, a2= 3 and d= 3β8= β5 (Where a=a1 is first term, a2 is second term, an is nth term and d is common difference of given...
Find the 20th term of the AP
To Find: 20th term of the AP Given: The series is β2, 3β2, 5β2, 7β2, .... a1=β2, a2= 3β2 and d= 3β2ββ2= 2β2 (Where a=a1 is first term, a2 is second term, an is nth term and d is common difference of...
Find the 23rd term of the AP 7, 3, 1, β1, β3, …
To Find: 23rd term of the AP Given: The series is 7, 5, 3, 1, β1, β3, ... a1= 7, a2= 5 and d= 3β5= β2 (Where a=a1 is first term, a2 is second term, an is nth term and d is common difference of given...
Find the first five terms of the sequence, defined by a1 = 1, an = anβ1 + 3 for n β₯ 2.
To Find: First five terms of a given sequence. Condition: n β₯ 2 Given: a1 = 1, an = anβ1 + 3 for n β₯ 2 Put n= 2 in nth term (i.e. an), we have a2 = a2β1 + 3 = a1 + 3 = 1 + 3 = 4 (as a1 = 1) Put n= 3...
Write first 4 terms in each of the sequences:
Given: nth term of series is (β1)nβ1 Γ 2n + 1 Put n=1, 2, 3, 4 in nth term, we get first (a1), Second (a2), Third (a3) & Fourth (a4) term. First four terms of given series are 4, β8 , 16...
Write first 4 terms in each of the sequences:
Given: nth term of series is Put n=1, 2, 3, 4 in nth term, we get first (a1), Second (a2), Third (a3) & Fourth (a4) term. First four terms of given series areΒ -14,14,34,54
Write first 4 terms in each of the sequences: an = (5n + 2)
Given: nth term of series is (5n + 2) Put n=1, 2, 3, 4 in nth term, we get first (a1), Second (a2), Third (a3) & Fourth (a4) term a1 = (5 Γ 1 + 2) = 7 a2 = (5 Γ 2 + 2) = 12 a3 = (5 Γ 3 + 2) = 17...