CBSE I NCERT Arithmetic Progression-X Solved Problems

1. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top .If the top and the bottom rungs are two and a half meter apart, what is the length of the wood required for the rungs?

2. The sum of first n terms of an AP is given by Sn = 3n2 + 5n find the nth term of the AP.

3. How many terms of the ap -6,-11/2 , -5,...... are needed to give the sum -25

4. Find a, b such that 27, a, b - 6 are in A.P.

5. Find the sum of all the odd numbers between 50 and 150 divisible by 7

6. The sum of third and seventh term of an AP is 6 and their product is 8.

7. If Sn = n2p and Sm = m2p, (m not equal ton), is an A.P. prove that Sp = p3.

8.Show that the sum of (m+n)th term and (m-n)thterm of an A.Pis equal to twice the mth term.

9. If the ratio of the sums of n terms of 2 APs is n+1:3n+1, then find the ratio of the 7th terms of the AP.

10. Determine the sum of the first 30 terms of the sequence whose nth term is given by tn=2n+9/3

11. 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.

12.The sum of the first and the last terms of an AP is 60,the sum of n terms of the AP is720.what is n?

13. If pth , qth and rth term of an AP are a,b,c respectively , then show that (a-b)r +(b-c)p + (c-a)q = 0

14. if (b+c)/a, (c+a)/b, (a+b)/c are in A.P. show that bc,ca,ab re in A.P.

15. 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

Class X ARITHMETIC PROGRESSIONS (8) Periods

Motivation for studying AP. Derivation of standard results of finding the nth term and sum of first n terms and their application in solving daily life problems.

www.jsuniltutorial.weebly.com/ (To get solution in pdf) 

CBSE I NCERT 10th Arithmetic Progression Problems with solution

Solved Arithmetic Progression Problems By Jsunil

(1) Determine k so that k+2 , 4k-6 and 3k-2 are the three consecutive terms of an AP.

a₁= k+2

a₂= 4k-6

a₃= 3k-2

a₂-a₁= a₃- a₂

4k-6-(k+2) = 3k-2-(4k-6)

4k -6-k-2 = 3k-2-4k+6

3k-8 = -k+4

3k+k = 4+8

4k = 12

k = 3

(2) if 7 times the 7th term of an AP is equal to 11 times the 11th term , show that the 18th term is zero.

Given: 7 times the 7th term of an AP is equal to 11 times the 11th term

7(a+6d) =11(a+10d)

7a+42d=11a+110d

42d-110d=11a-7a

68d = 4a

a =-17d

Now, the 18th term = a+17d=-17d-17d=0

(3) If the nth term of an A.P is 7n-5. Find 100th term

Given that the n th term of the A.P. is 7n-5

So 100 th term will be 7 (100) -5 =695

(4) if m times the mth term of an AP is equal to n times the nth term . Show that (m+n)th term of the AP is zero

We know :- a_n = a +(n-1)d

a _(m+n) = a + (m+n-1)d (just put m+n in place of n ) ------------------------------(1)

Let the first term and common difference of the A.P. be ‘a’ and ‘d’ respectively.

Then, m ^th term = a + (m – 1) d and n ^th term = a + (n – 1) d

By the given condition,

m x a_m = n x a_n

m [a + (m – 1) d] = n [a + (n – 1) d]

⇒ ma + m (m – 1) d = na + n (n – 1) d

=> ma + (m² -m)d - na - (n² -n)d = 0 ( taking the Left Hand Side to the other side )

=> ma -na + (m² - m)d -( n²-n)d = 0 (re-ordering the terms)

=> a (m-n) + d (m²-n²-m+n) = 0 (taking 'a ' and 'd ' common)

=> a (m-n) + d {(m+n)(m-n)-(m-n)} = 0 (a²-b² identity)

Now divide both sides by (m-n)

=> a (1) + d {(m+n)(1)-(1)} = 0

=>a + d (m+n-1) = 0 ---------------(ii)

From equation number 1 and 2 ,

a _(m+n) = a + (m+n-1)d

And we have shown ,

a + d (m+n-1) = 0

So, a _(m+n) = 0 

(5). Prove that the nth term of an AP cannot be n² + 1. Justify your answer.

Common difference of an A.P. must always be a constant.

∴ d cannot be n – 1. Here, d varies when n takes different values.

For n = 1, d = 1 – 1 = 0

For n = 2, d = 2 – 1 = 1

For n = 3, d = 3 – 1 = 2

∴ d is not constant.

Thus, d cannot be taken as n – 1.

a_n is the n ^th term of an A.P. if a_n – a_n –1 = constant

Given, a_n = n ² + 1

a_n – a_n –1 = (n ² + 1) – [(n – 1)² + 1]

= (n ² + 1) – (n ² – 2n + 2)

= 2n – 1        

∴ a_n – a_n –1 ≠ constant

Thus, a_n = n ² + 1 cannot be the n ^th term of A.P.

(6) Find the sum of the first k terms of a series whose n ^th term is 2an+b 

The n ^th term of the AP is given by 2an+b

a1=2a+b

a2 =4a+b

a3=6a+b

Common difference = d=( 4a + b ) - ( 2a + b ) = 2a

Therefore, sum of first k terms =k/2[(2a+(k-1)d]

 = k/2[(2(2a+b)+(k-1)2a]= k/2   x 2 (2a+b+k-a)=k(a+b+ak)

(7) Which term of the AP,  3,10,17 will be 84 more than its 13th term?

Let the nth term be 84 more than the 13^th term.

Now a/q,

a=3, d=10-3=7

So, 13th term= a+12d =3+12x7=87

Then nth term=84+87=171

171=a + (n-1)d

171=3 + (n-1)x7

171-3/7+1=n

168/7+1=n

24+1=25=n

Therefore 25th term of the ap will be 84 more than 13th term

(8) How many terms of the arithmetic series 24 + 21 + 18 + 15 +g, be taken continuously so that their sum is – 351.

In the given arithmetic series, a = 24, d =- 3.

Let us find n such that Sn = – 351

Now, S_n = n/2[(2a + (n-1)d] 

– 351 = n/2[(48 + (n-1)x(-3)] 

on solving we get,  n² - 17n -234 = 0

Þ (n - 26h)(n + 9) = 0

Þ  n = 26 or n = - 9

Here n, being the number of terms needed, cannot be negative

Thus, 26 terms are needed to get the sum -351.

(9) Find the sum of the first 2n terms of the following series. 1² - 2² + 3² - 4² +  

We want to find 1² - 2² + 3² - 4² +.....  to  2n terms

= 1 - 4 + 9 - 16 + 25 ------------2n terms

= (1 – 4) +(9 – 16)+(25 – 36) + ----------- to n terms. (after grouping)

= -3 +(-7)+(-11)+ -------------- n terms

Now, the above series is in an A.P. with first term a = - 3 and common difference d = - 4 

Now, S_n = n/2[(2a + (n-1)d]  = = n/2[(2 x -3) + (n-1)(-4)]  = -n(2n + 1).

(10)  A circle is completely divided into n sectors in such a way that the angles of the sectors are in arithmetic progression. If the smallest-of these angles is 8° and the largest 72°, calculate n and the angle in the fourth sector.

Let the common difference of the A.P. be x

The smallest angle = 8°

⇒ a = 8

And the largest is 72°

⇒ a_n = 72

⇒ a + (n – 1)d = 72

⇒ 8 + (n – 1)d = 72

⇒ (n – 1) d = 72 – 8 = 64 ... (1)

We know that sum of all the angles of a circle is 360°

S_n = n/2[(2a + (n-1)d]  = 360

Þ S_n = n/2[(2x8 + 64] = 360

Þ n= 9

Putting the value of n in equation (1) we get

(9 – 1) d = 64

d = 8

Now angle in fourth sector = a₄ = a + (4 – 1) d

= a + 3d = 8 + 3 × 8 = 8 + 24 = 32

∴ The value of n = 9 and angle in fourth sector is 32°

10th Maths SA-2 Chapter Links

Quadratic Equations

Arithmetic Progressions

Height and Distance

Co-ordinate Geometry

Circles

Area Related to Circles

Surface Areas and Volumes

Probability

CBSE I NCERT 10th Arithmetic Progressions : Problems for self evaluation.

10th Ch- A. P Problems for self evaluation. Download File

CBSE I NCERT 10th Arithmetic Progressions CCE Test Paper

                     

Q. 1 How many terms of A.P. 22, 20, 18,…… should be taken so that their sum is zero?

Q.2  Find the sum of odd positive integers less than 199.

Q. 3 How many two digits numbers between 3 and 102 are divisible by 6?

Q. 4 If 7 times the 7th term is equal to 11 times the 11th term of an A.P. Find its 18th term.

Q. 5 Which term of A.P. 13, 21, 29, . . . . . . will be 48 less than its 19th term?

Q. 6 Find the A.P. whose 3rd term is –13 and 6th term is +2.

Q. 7 Find the A.P., whose 5th term is 23 and 9th term is 43.

Q. 8 The angles of a triangle are in A.P. If the smallest angle is one fifth the sum of other two angles. Find the angles.

Q. 9 Aditi saved Rs. 500 in the first month of a year and then increased her monthly savings by Rs. 50. If in the nth month, her monthly savings become Rs 1000. Find the value of 'n'.

Q. 10 The sum of first n terms of an A.P. is 2n2 + n . Find nth term and common deference of the A.P.

Q. 11 The sum of 3rd and 7th terms of an A.P. is 14 and the sum of 5th and 9th terms is 34. Find the first term and common difference of the A.P.

Q. 12. Find the sum of the first 30 terms of an A.P., whose nth term is 2–3n. If mth and nth terms of an A.P. are 1/n and1/m respectively, then find the sum of mn terms

Q. 13 If mth , nth and rth terms of an A.P. are x, y and z respectively, then prove that :- m( y – z) + n(z – x) + r (x – y) = 0

Q. 14. If the roots of the equation a(b – c) x2 + b(c – a) x + c (a – b) = 0 are equal, then show that 1/a , 1/b , 1/c are in A.P.

Q. 15. If the sum of m terms of an A.P. is n and the sum of n terms is m, then show that sum of (m +n) terms is – ( m + n).

CBSE/NCERT Class X Arithmetic progression Assignment For SA-II

                           

Q. 1. Determine k so that k + 2, 4k - 6 and 3k - 2 are three consecutive terms of an AP.

Q. 2. If m th term of A.P. is , and nth term is , show that the mn th terms is 1.

Q. 3. The first, second and the last terms of an AP are p, q and 2p respectively. Show that its sum is [3pq] / [2(q-p)].

Q. 4. A circle is completely divided into n sectors in such a way that the angles of the sectors are in arithmetic progression. If the smallest-of these angles is 8° and the largest 72°, calculate n and the angle in the fourth sector.

Q. 5. Which term of AP: 3, 10, 17 ... will be 84 more than its 13th term?

Q. 6. If 9th term of an AP is zero, prove that 29th term is double the 19th term.

Q. 7. Find a, b such that 27, a, b - 6 are in A.P.Q. 8. For what value of n, the nth terms of the sequences 3, 10, 17,... and 63, 65, 67,... are equal.

Q. 9. If m times the m th term of an AP is equal to n times its nth term show that the (m + n)th term of the AP is zero.

Q. 10.Find the sum of all odd integers between 78 and 500 which are divisible by 7.

Q. 11. Find n, if the given value of x is nth term of A.P. 17, 22, 27, 32, ...; x = 267

Q. 12. Find the sum of all the odd numbers between 100 and 200

Q. 13.If 10 times the 10th term of an AP is equal to 15 times its 15th term, show that its 25th term is zero.

Q.14. Find the sum 2+4+6+. . . +202

Q. 15. How many terms are there in the A.P -1,-5/6,-2/3,-1/2……….10/3? Also find its general term?

Q. 16. 3 times the tenth term is equal to 5 times the twentieth term. Find twentieth term.

Q. 17.The 5th term of an AP is 24 and its 15th term is 74. Find the sum of its first 10 terms.

Q. 18. If the first term and last term of an AP are a and l respectively and its sum is S , prove that the common difference of the AP is equal to (l2 – a2) / [2S-(l+a)] .

Q. 19. If the difference between the 21st and 10th terms of an AP is 55, find the difference between the 45thand 40th terms.

Q. 20.Find three numbers in an A.P. whose sum is 15 and product 80

CBSE/NCERT 10 Maths Guess Questions Chapter Arithmetic Progressions

Download Test paper model paper link

1) For what value of p, are 2p-1, 7 and 3p three consecutive terms of an A.P? (P=3)

2) Find the value of k, so that 3k + 7, 2k +5, 2k + 7 are in A.P  (k= -4)

3) Find the 15th term from the end of the A.P:   3, 5, 7,………, 201(173)

4) Find the 11th term from the end of the A.P:    10, 7, 4,……, - 62 (-32)

5) If Sn, the sum of first n terms of an A.P is given by Sn = 3n2 – 4n, then find its nth term

(6n – 7)

6) The sum of n terms of an A.P. is 3n2 + 5n. Find the A.P. Hence, find its 16th term

(6n + 2, 98)

7)  In the following A.P. find the missing term   -, 38, -, -  , - , -,22

8)  Find the sum of all natural numbers less than 100 which are divisible by 6                        (816)

9) Find the sum of 3 digit numbers which are not divisible by 7                  (424214)

10)  Find the sum of all three digit numbers which leave the remainder 3 when divided by 5      (99090)

11) Find the sum of first seven multiples of 5                                        (140)

12) Find the sum of all natural numbers up to 100, which are not divisible by 5           (4000)

13) In an A.P , if the 6th and 13th terms are 35 and 70 respectively, find the sum of its first 20 terms.

14) If the 3rd and 9thterm of an A.P. are 4 and -8 respectively, which term is zero (n = 5)

15) The sum of 4th and 8th terms of an A.P is 24 and sum of 6th and 10th term is 44. Find A.P.

16) The 4th term of an A.P is equal to 3 times the first term and the 7th term exceeds twice the 3rd term by 1. Find the A.P  (3 ,5 ,7, …)

17) Which term of the A.P.?  3, 15, 27, 39, will be 120 more than its 21st term   ( n = 31)

18) In an A.P., the first term is 25, nth term is -17 and sum to first n terms is 60.Find n and d the common difference.                                                              1

19) Which term of the sequence 114, 109, 104,is the first negative term?                (n =24)

20) If the 4th term of an A.P is twice the 8th term, prove that the 10th term is twice the 11th term

21) If 2 + 5 + 8 + …………………………+ x = 155, find x    (n = 10, x = a10=29)

22) For A.P. a1, a2, a3, ………., if a4/a7 = 2/3 , find a6/a8                          (4/5)

23) Find the sum of the following A.P:           1 + 3 + 5 + …….. + 199.        (10000)                   24) Find the common difference of an AP whose first term is 100 and sum of first six terms is 5 times the The sum of the next 6 terms                  (d= - 10)

25) Show that progression 7, 2, -3, -8, … ……..Is an A.P . Find its nth term      (12 – 5n)

26) The angles of a triangle are in A.P, the last being half the greatest. Find the angles.     (40˚, 60˚, 80˚)

27)  Find the sum of n terms of an A.P whose nth term is given by tn = 5 – 6n (2n – 3n2)

28) Find the middle term of A.P:     1, 8, 15,  ……………, 505        (253)

29) Find the number of terms of the A.P, 63, 60, 57,  ……….. So that their sum is 693 (n = 22, 21)

30) The sum of 3 numbers in A.P is 3 and their product is -35. Find the numbers

(7, 1, and -5)

31) How many terms of the sequence 18, 16, 14,  …………, should be taken so that their sum is 0                                                      (n= 19)

32) A sum of Rs 1400 is to be used to give 7 cash prizes to students of a school for their overall academic Performance if each prize is Rs40 less than the preceding price, find the value of each of the prizes.    (320, 280, 240, 200, 160, 120, 80)

33) Verify that a + b, (a + 1) + b, (a + 1) + (b + 1) ……….. Is an A.P. and then write its next term                              (a+2) + (b+1)

34) Determine the A.P whose 3rd term is 16 and 7th term exceeds the 5th term by 12                                                    (4, 6, 10, 16 ,…………)

35) If the nth term of the A.P. 9, 7, 5, ………… is the same as the nth term of the A.P. 15, 12, 9, ………., find n                       (n = 7 )

36) Find the sum of first 22 terms of an A.P. in which d = 7 and 22nd term is 149        (1661)

37) Find the sum of the following A.P:      3, 9/2, 6, 15/2,  ………. To 25 terms              (525)

38) The ratio of the sum to p terms and q terms of an A.P. is p2 : q2. Prove that the common difference of the A.P.is twice The first term

39) In an A.P., if the sum of its 4th and 10th terms is 40, and the sum of its 8th and 16th terms is 70, then find the sum of its  First 20 terms

40) Three consecutive positive integers are taken such that the sum of the square of the first and the product of the other two Is 154. Find the integers                        (3, 5, 7……….).

Download Guess Paper:

1. Quadratic Equations

2. Arithmetic Progressions

3. Height and Distance

4.Co-ordinate Geometry

5. Circles