Solved Problems in Arithmetic Progression 59


Here are some questions on arithmetic progression and their solutions

Question 1: Which term of the A.P. 3,8,13 …is 78?

Solution: Here an = a + (n – 1) d = 78

a= 3, d = 8- 3 = 5

Therefore,

3 + (n -1) (5) = 78

(n-1) * 5 = 78 – 3 = 75

n – 1 = 75/5 = 15

n = 15 + 1 = 16

Hence a 16 (sixteenth term) is 78.

Question 2: Is – 150 a term of the series 11, 8, 5, 2,…?

Solution: Here, a = 11, d = 8-11 = -3. Let an = -150

Therefore,

a + (n-1) d = -150

11+ (n-1) (-3) = -150

-3 (n-1) = -150 – 11 = -161

(n-1) = + 161/3 = 53 2/3 which is not an integral number.

Since number of terms can never be a fraction

Hence, -150 is not a term of the given series.

Question 3: Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.

Solution: Let a be the 1st term and d the common difference.

Here a11 = a + 10d = 38….. (1)

a16 = a + 15d = 73….. (2)

Subtracting (2) from (1), we get

a + 10d – 1 – 15 d = 38 – 73

-5 = – 35

d = 7

Putting d= 7 in (1), we get

a + 10 * 7 = 38

a= 38 – 70 = – 32

a31= a + 30 d

= -32 + 30 * 7

= -32 + 210 = 178

Hence, 31st term is 178.

Question 4:  Which term of the A.P. 3, 15, 27, 39 … will be 132 more than its 54th term?

Solution: Given series is 3, 15, 27, 39…

Here, a =3, d= 15-3 = 12

Since an = ak = (n-k) d

an – a54 = (n-54) 12

132 = 12n – 54 * 12    …..(since an – a 54 = 132 given)

12 n = 132 + 54 * 12 = 12 (11+ 54)

n= 11 + 54 = 65


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59 thoughts on “Solved Problems in Arithmetic Progression

  • Rishabh

    Let there be n problems as that was fed to the comuter.

    Let t be the time it had taken to solve the first question.

    So now aaccording to the problem the time taken to solve the successive questions will be t-t/x, t-2t/x,….., t-(n-1)t/x.

    There are three variables in the problem – n,t,x.

    You need to solve for n.

    There are three conditions in the problem. So now use the three conditions and form equations.

    Then solve for n.

    t-t/x + t-2t/x + t-3t/x + ……. t-(n-1)t/x = 63.5 ———– (1)

    t + t-t/x + t-2t/x +…… + t-(n-2)t/x = 127 ———- (2)

    t-2t/x + t-3t/x + …….. + t-(n-1)t/x = 31.5 ——— (3)

    From the three equations, solve for n.

    (1)-(3):

    t-t/x = 32 ————- (4)

    (2)-(1):

    (n-1)t/x = 63.5 ————(5)

    From (2):

    (n-1)t – (t/x)(n-2)(n-1)/2 = 127

    or (n-1)t*[1 – (n-2)/(2x)] = 127

    63.5x*[1 – (n-2)/(2x)] = 127 ——– from (5) (n-1)t = 63.5x

    or x[1 – (n-2)/(2x)] = 2

    or [2x-n-2] = 4

    or x-1 = (n+4)/2 ———– (6)

    Using (4):

    (1-1/x)[63.5x/(n-1)] = 32

    (x-1)[63.5/(n-1)] = 32 ——– (7)

    So 63.5/(n-1) [n+4] = 64

    or 63.5(n+4) = 64(n-1)

    So 63.5n + 254 = 64n – 64

    or 0.5n = 318.

    or n = 636.

    So 636 Questions would be fed to the computer.

    That solves this question.

    Hope it helps.

    @Himanshu

  • pavan

    first term a And 9th term a+8d sum of these terms is 2a+8d=24 ie a+4d=12 sum of the first 9 numbers is 9/2(2a+8d) ie 9/2(2(a+4d))ie 9/2(2.12)ie 108

  • ADELIZE JUSTINE JANUARIE

    please help me me with this one, i’m new….
    the sum of three numbers in an AP is 21 and their product is 280. find the numbers.

    • Pavan

      Let the nos be a-d , a, a+d
      a-d+a+a+d=21
      3a =21
      a=7

      a-d×a×a+d=280
      a²-d²×a=280
      49-d²×7=280
      49-d²=40
      d²=9
      d=3
      The nos are
      6,7and8

  • abraham

    nikhil :The som ofn terms of two AP are in the ration (2n+3):(6n+5),then the ration of thier 13 th terms is(a) 53:155 (b)29:83 (c) 31:89 (d)27:87

    pls give the procedure to solve the problem

  • steven

    azimah :the maximum speed of an engine is 1200 p.s.m. and minimum speed is 40 p.s.m. the machine should be equipped with a 5-speed. find 3 more speed needed by using the geometric progression.
    please help me solve this problem,.tq.,

    rajeshsaini :nth term of A.P. isa + (n-1) d
    a is the first term{a1}d is the common difference{ a2-a1 = a3-a2 = …… = (an) – (an-1) }

    G.V.SUBRAHMANYAM :The number of terms in an A.P. is even. The sums of odd and even numbered terms are 24 and 30 respectively. If the last term exceeds the first term by 10.t, the number of terms in the A.P. is

    pari :this examples helped me alot

    nikhil :The som ofn terms of two AP are in the ration (2n+3):(6n+5),then the ration of thier 13 th terms is(a) 53:155 (b)29:83 (c) 31:89 (d)27:87

    shubham :the sum of the first 20 terms in an AP is 210.find the sum of the 10th and the 11th terms..
    a.)10.5b.)42c.)21d.)can not be determined

    rajeshsaini :nth term of A.P. isa + (n-1) d
    a is the first term{a1}d is the common difference{ a2-a1 = a3-a2 = …… = (an) – (an-1) }

    @SOMENDRA MISHRA

    azimah :the maximum speed of an engine is 1200 p.s.m. and minimum speed is 40 p.s.m. the machine should be equipped with a 5-speed. find 3 more speed needed by using the geometric progression.
    please help me solve this problem,.tq.,

    azimah :the maximum speed of an engine is 1200 p.s.m. and minimum speed is 40 p.s.m. the machine should be equipped with a 5-speed. find 3 more speed needed by using the geometric progression.
    please help me solve this problem,.tq.,

    azimah :the maximum speed of an engine is 1200 p.s.m. and minimum speed is 40 p.s.m. the machine should be equipped with a 5-speed. find 3 more speed needed by using the geometric progression.
    please help me solve this problem,.tq.,

    @azimah

  • G.V.SUBRAHMANYAM

    The number of terms in an A.P. is even. The sums of odd and even numbered terms are 24 and 30 respectively. If the last term exceeds the first term by 10.t, the number of terms in the A.P. is

  • azimah

    the maximum speed of an engine is 1200 p.s.m. and minimum speed is 40 p.s.m. the machine should be equipped with a 5-speed. find 3 more speed needed by using the geometric progression.

    please help me solve this problem,.tq.,

  • nikhil

    The som ofn terms of two AP are in the ration (2n+3):(6n+5),then the ration of thier 13 th terms is
    (a) 53:155 (b)29:83 (c) 31:89 (d)27:87

  • SOMENDRA MISHRA

    @myza sahira
    Here in question as the AP of the series
    2,7,12…52
    so we use formula is =
    nth = a+(n-1)d
    here nth is = 52, a= 2, d = 5
    52= 2+(n-1)5= 52-2= 5n-5
    = 50 = 5n-5 =50+5 = 5n
    = 55= 5n
    = n= 55/5
    hence n= 11th term
    and we also check this
    the following formula
    nth term = (a+10th d)
    11th = 2+ 10*5
    = 2+50 = 52 is the answer

  • SOMENDRA MISHRA

    @Abhilash
    You can use the formula to calculate the AP term.
    In this question given that series 12,32,42……and so on and find the 20th terms of the series
    formula is that 20th =(a+19d)
    here a = 12, d = 10 that is common difference
    so 20th = (12+19*10)
    = 20th = 12+190 = 202 is the 20th term. ANS .

  • Angela

    In an arithmetic progression, the sum of the first three terms is 18, and the sum of squares of these terms is 126. find the terms.

  • rajendra

    Numbers between 100 and 200 divisible by 5 will be 105, 110, 115,…..195.
    Hence apply formula, nth term = Tn=a+(n-1)d
    Here, Tn=195, a=105, d=5, which makes n=19.

    So, sum of all the terms Sn = n/2(a+l) (l= last term = 195; a = first term=105) which gives 2850.

    @qwerty

  • rajendra

    @myza sahira

    Qn. 1 – check for typo error.

    Ans for Qn 2. You know the n th term of the AP as 52 and common difference d = 5

    Hence, apply the formula Tn = a+(n-1)d

    i.e. 52 = 2+(n-1)5, solving this equation for n gives n=11. Hence, there are 11 terms in the series

  • rajendra

    @Joshua

    n=25; d= 35;
    sum of n terms of an AP Sn =n/2{2a + (n-1)d}

    Here Sn = S25 = 20800= 25/2{2a+(25-1)35}

    Solving this equation gives a= 412 (first month arrear)

    Hope this has helped.

    Last month arrear Sn i.e. S25= a+(n-1)d = 412+24*35 = 1252

  • akash

    the interior angles of convex polygon are in ap, the common difference being 5 degree. if smallest angle is 120 degree then the no of sides is——–

  • myza sahira

    hai..
    i want to ask if we want to find the next term in the progression and the questoin is
    1),-q,0,p+q… how?
    2)find the number of term for the following AP
    2,7,12,…,52
    please give me the solution..
    thankz..

  • Joshua

    Joshua :A company owner paid a sum of 20800$ to an employee as a salary arrears in 25 regular installments such that each month’s payment is 35$ more than the preceeding month. Find the arrear amount of the 1st month and the last month. Please tell me an answer. Thanks

  • Joshua

    A company owner paid a sum of 20800$ to an employee as a salary arrears in 25 regular installments such that each month’s payment is 35$ more than the preceeding month. Find the arrear amount of the 1st month and the last month.

  • shubham

    the sum of the first 20 terms in an AP is 210.
    find the sum of the 10th and the 11th terms..

    a.)10.5
    b.)42
    c.)21
    d.)can not be determined

  • Alagusamy

    Each year a tree grows 5 cm less than it did in the preceding year. If it grew 1 cm in first year in how many years it have ceased growing?

  • Himanshu

    A computer solve several problems in succession. The time it took the computer to solve each successive problem was the same number of times smaller than the time it took to solve the preceding problem. How many problems were suggested to the computer if it spent 63.5 min to solve all the problems except for the first,127 min to solve all the problems except for the last one,and 31.5 min to solve all the problems except for the first two?

  • Abhilash

    1. The sum of the first and the 9th of an arithmetic progression is 24. What is the sum of the first nine
    terms of the progression ?

    2. The sum of 20 terms of the series 12 + 22, 32 + 42, 52 + 62 is?