InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term. |
| Answer» `4(a+3d) = 18 (a+17d) rArr 14a + 294d = 0 rArr a+21d = 0 rArr T_(22) = 0` | |
| 2. |
Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623. |
|
Answer» Let three parts be `a-d, a, a+d`. `:. a-d+a+a+d=207` `rArr 3a=207 rArr a=69` and `(a-d).a=4623` `rArr (69-d)69=4623` `rArr 69-d=67 rArr d=2` `:. a-d=69-2=67` `" " a=69` `" "a+d=69+2=71` `rArr` three parts are 67,69,71 |
|
| 3. |
How many terms are there in the AP 18, `15(1)/(2), 13,…, -47?` |
| Answer» Correct Answer - 27 | |
| 4. |
How many terms are there in the AP 11,18,25,32,39,…..,207? |
| Answer» Correct Answer - 29 | |
| 5. |
How many terms are there in the AP `6, 10, 14, 18, ..., 174?` |
| Answer» Correct Answer - 43 | |
| 6. |
Find the four numbers in A.P. whose sum is 20 and the sum of whosesquares is 120. |
|
Answer» Let the required numbers be (a-3d) , (a-d) (a+d) and (a+3d) Then ( a-3d) + (a -d) + ( a+d) + (a +3d) = 20` Rightarrow 4a = 20 Rightarrow a=5` And ` (a-3d)^(2) + (a-d)^(2) + (a +d)^(2) + ( a+3d)^(2) =120 ` ` Rightarrow 2(a^(2)+9a^(2)) =120` ` Rightarrow 4a^(2) + 20 d^(2) = 120 Rightarrow a^(2) + 5d^(2) = 30 ` ` Rightarrow 25 + 5d^(2) =30Rightarrow 5d^(2) =5 Rightarrow d^(2) =1 Rightarrow d = +-1` Hence, the required numbers are 2,4,6,8 or 8,6,4,2 |
|
| 7. |
How many terms of the AP 18,16,14,12 ,….. Are needed to given the sum 78 ? Explain the double answer. |
| Answer» Correct Answer - 6 and 13, sum of all terms from 7th to 13th is zero. | |
| 8. |
The sum of first n terms of an AP is `(5n - n^(2)).` The nth term of the AP isA. (5-2n)B. (6-2n)C. (2n -5)D. (2n-6) |
|
Answer» Correct Answer - B `T_(n) = (S_(n) - S_(n - 1)) = (5n-n^(2)) - {5(n-1) - (n-1)^(2)} = (5n -n^(2)) - (7n-n^(2)-6) = (6 -2n)` |
|
| 9. |
Find the 28th term from the end of the AP 6,9,12,15,18,…..,102 |
| Answer» Correct Answer - 21 | |
| 10. |
Find four numbers in A.P. whose sum is 28 and the sum of whose squares is 216. |
| Answer» Correct Answer - (4, 6, 8, 10) or (10, 8, 6, 4) | |
| 11. |
Find the (i) nth term and (ii) 16th term of the AP 3, 5, 7, 9, 11, … |
|
Answer» The given AP is 3, 5, 7, 9, 11, … Its first term = 3 and common difference = (5-3) = 2. `therefore` a= 3 and d = (5-3) = 2. (i) Its nth term is given by `T_(n) = a + (n-1)d` ` = 3 +(n-1) xx 2 = (2n +1) " " [because a = 3 "and" d = 2].` `therefore "nth term" = (2n +1)` (ii) 16th term of the given AP is `T_(16) = a + (16 -1)d = (a + 15d)` ` = (3 + 15 xx 2) = 33 " " [because a= 3 "and" d= 2]` `therefore ` 16th term = 33.` |
|
| 12. |
find the 16th term form the end of the AP 7,2,-3,-8,-13,…., -113 |
| Answer» Correct Answer - -38 | |
| 13. |
Find the middle term of the A.P. `6, 13 , 20 , 216.` |
| Answer» Correct Answer - 111 | |
| 14. |
Write first four terms of the sequence given by ` a_(n)= 1/6 ( 2n-3)` and obtain the corresponding series. |
|
Answer» We have , ` a_(n) = 1/6 ( 2n-3)` Putting n= 1,2,3,4,…, successively in ( i) we get ` a_(1) = 1/6( 2xx 1-3) = (-1)/6, a_(2) = 1/6 ( 2xx 2-3) = 1/6 ` ` a_(3) = 1/6 ( 2xx 3-3) = 3/6 = 1/2 , a_(4) = 1/6 ( 2 xx 4-3) = 5/6` Hence, the required sequence is `( -1)/6, 1/6, 1/2 ,5/6 ,.....,` The corresponding series is `- 1/6 + 1/6 + 1/2 + 5/6+ ` .... |
|
| 15. |
For the following A.Ps, write the first term `a` and the common difference `d` `(i) 3, 1, -1, -3,.. ` `(ii) 1/3,5/3,9/3,13/3,...` |
|
Answer» (i) 3, 1, -1, -3, ... Here, first term a = 3 Common difference = 1-3= -2 (ii) `(1)/(3),(5)/(3), (9)/(3), (13)/(3), ...` Here , first term `a=(1)/(3)` Common difference `d=(5)/(3)-(1)/(3)=(4)/(3)`. |
|
| 16. |
A sequence is defined as follows : `a_(1)=3, a_(n)=2a_(n-1)+1`, where `n gt 1`. Where `n gt 1`. Find `(a_(n+1))/(a_(n))` for n = 1, 2, 3. |
|
Answer» `a_(1)=3` `a_(n)=2a_(n-1)+1` where `n gt 1` put n=2, we get `a_(2)=2a_(1)+1=2xx3+1=7` put n=3, we get `a_(3)=2a_(2)+1=2xx7+1=15` put n=4, we get `a_(4)=2a_(3)+1=2xx15+1=31` Now, for n=1 `(a_(n+1))/(a_(n))=(a_(2))/(a_(1))=(7)/(3)` For n=2, `(a_(n+1))/(a_(n))=(a_(3))/(a_(2))=(15)/(7)` For n=3, `(a_(n+1))/(a_(n))=(a_(4))/(a_(3))=(31)/(15)` |
|
| 17. |
Fibonacci sequence is defined as follows : `a_(1)=a_(2)=1` and ` a_(n)=a_(n-2)+a_(n-1)`, where `n gt 2`. Find third, fourth and fifth terms. |
|
Answer» `a_(1)=a_(2)=1` `a_(n)=a_(n-2)+a_(n-1),n gt 2` put n = 3 , we get `a_(3)=a_(1)+a_(2)=1+1=2` put n = 4, we get `a_(4)=a_(2)+a_(3)=1+2=3` put n=5, we get `a_(5)=a_(3)+a_(4)=2+3=5` |
|
| 18. |
Find the sum of all odd integers from 1 to 1001. |
|
Answer» The odd integers from 1 to 1001 are 1,3,5,7,……,999, 1001. This is an AP in which a=1, d= ( 3-1) = 2 and l = 1001 . Let the number of terms be n.Then, ` T_(n) 1001 Rightarrow a+ ( n-1) d= 1001` ` Rightarrow 1+ (n-1)xx2 =1001 Rightarrow n = 501` Now, a=1, l = 1001 and n = 501 `S_(n)=n/2(a+l)=501/2.( 1+1001)=(501xx501) = 251001` Hence, the required sum is 251001 |
|
| 19. |
Find the sum of all odd numbers between 100 and 200. |
|
Answer» The series formed by odd numbers lying between 100 and 200 is `101+103+105+.....+199` Here, a=101, `d=103-101=105-103=2` Let `a_(n)=199` `rArr 101+(n-1)2=199` `rArr (n-1)2=98 rArr n-1=49` ltbtgt `rArr n=50` Now, `S_(50)=(50)/(2)(101+199)=7500` |
|
| 20. |
The sum of first 20 odd natural numbers isA. 100B. 210C. 400D. 420 |
|
Answer» Correct Answer - C `S_(20) = 1+3 +5 +7 +…` up to 20 terms. Here a = 1, d = 2. So, `T_(20) = (a+19d) = (1+19 xx 2) = 358 = l.` `therefore S_(20)= (n)/(2)(a+l) = (20)/(2) (1+39) = 400` |
|
| 21. |
Find the sum of all odd numbers between 0 and 50 |
|
Answer» Correct Answer - 625 Required sum = (1+3+5+7+…+49). `T_(n) = 49 rArr 1+(n-1) xx 2 = 49 rArr n = 25` `"Use" S_(n) = (n)/(2) (a+l).` |
|
| 22. |
Write thesum of first `n`odd naturalnumbers. |
|
Answer» Correct Answer - `n^(2)` given sum =1+3+5+…..to n terms ` n/2xx[2xx 1 + (n-1) xx2] =n^(2)` |
|
| 23. |
Find the 17th from the end of the AP-36,-31,-26,-21,…., 79 |
|
Answer» Here a= 36 m d= [ -31-(-36)} = 5and l=79 17th term from the end = l - (n-1)d ` = 79 -( 17-1) xx 5 = ( 79 -80) = -1` Hence, the 17th term from the end is -1. |
|
| 24. |
Find the sum of first n natural numbers. |
|
Answer» Correct Answer - `(1)/(2)n (n+1)` `S_(n) = 1+2+3+….+ n = (n)/(2) (1+n). [S_(n) = (n)/(2)(a+l)]` |
|
| 25. |
The 17th term of an AP exceeds its 10th term by 21. The common difference of the AP isA. 3B. 2C. `-3`D. `-2` |
|
Answer» Correct Answer - A `(T_(17) - T_(10)) = 21 rArr (a+16d) - (a+9d) = 21 rArr 7d = 21 rArr d = 3.` |
|
| 26. |
Which term of the AP 21, 18, 15,… is zero? |
|
Answer» Correct Answer - 8th Let `T_(n) = 0."Then," 21 + (n-1) xx (-3) = 0 rArr n = 8`. So, 8th term is zero. |
|
| 27. |
if the sum of certain number of terms of the AP 27,24,21,18,…., is -30 . Find the last term. |
|
Answer» Correct Answer - -30 Here,`a=27md=(24-27) = -3 and S_(n) = -30` Then, ` n/2.[2a +(n-1)d] = -30 Rightarrow n/2. [54+(n-1)xx(-3)]=-30` `3n^(2)-57n-60=0Rightarrow n^(2)-19n -20 =0 Rightarrow n=20` Now, ` S-9n)=-30 Rightarrow n/2 (a+l) -30 Rightarrow 20/2 (27+l) =-30 Rightarrow l= -30` |
|
| 28. |
Find the sum of 20 terms of the AP (x+y) , (x-y) , (x -3y),…., |
|
Answer» Correct Answer - 20(x-18y) a=(x+y) and d= ( x-y) -(x +y) =-2y ` S_(20)= 20/2 x [2a +(20-1)d] = 10xx [2(x+y)+190xx(-2y)] = 20 (x-18y)` |
|
| 29. |
A man starts repaying a loan as the frist instalment of ₹10000. if he increases the instalments by ₹ 500 every month,what amount will the pay in 30 th instalment ? |
|
Answer» Correct Answer - ₹ 24500 The instalments form the AP 1000 + 10500 + 11000 + 115000 + ….. Up to 30 terms Here a=1000, d=500 and we have to find ` T_(30)` |
|
| 30. |
The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first term and the common difference of the AP |
|
Answer» Correct Answer - a = 1, d = 2 `(9)/(2)(2a+ 8d) = 81 rArr a +4d = 9 " "…(i)` `(20)/(2)(2a+19d) = 400 rArr 2a +19d = 40 ." "..(ii)` |
|
| 31. |
If the sum of three numbers in A.P. is 24 and their product is 440,find the numbers. |
| Answer» Correct Answer - (5, 8, 11) or (11, 8, 5) | |
| 32. |
Insert five numbers between 11 and 29 such that the resulting sequence is an AP. |
| Answer» Correct Answer - 14,17,20,23,26 | |
| 33. |
Find the sums given below : (i) `7+10 1/2+14+dot dot dot+84`(ii) `34+32+30+dot dot dot+10`(iii) `5+(8)+(-11)+dot dot dot+( 230)` |
|
Answer» Correct Answer - (i) `1046(1)/(2)` (ii) 286 (iii)-8930 (iv) 420 `(i) a = 7,d = ((21)/(2) - 7) = (7)/(2) "and" I = 84.` `T_(n) = 84 rArr a + (n-1)d = 84 rArr 7+(n-1) xx (7)/(2) = 84 rArr n = 23.` `S_(23) = (n)/(2)(a+l) = (23)/(2)(7+84) = 1046(1)/(2).` |
|
| 34. |
is `184` a term of the `AP 3, 7, 11, 15, ...?` |
|
Answer» Correct Answer - No Let `T_(n) = 184. "Then", a + (n-d) = 184` `therefore 3 + (n-1) xx 4 = 184 rArr n-1 = (181)/(4) rArr n = ((181)/(4) +1) = (185)/(4) = 46 (1)/(4)` Thus, n is not a natural number. So, 184 is not a term of the given AP. |
|
| 35. |
Find the 8th term from the end of the A.P. 7,10,13, ..., 184 |
| Answer» Correct Answer - 163 | |
| 36. |
Complete the following activity to find the number of terms in the A.P. 1,3,5,…., 149. |
|
Answer» Here,` a = 1, d = square , t_(n) =149` `t_(n) = a+ ( n-1) d ` ` :. 149 = square ` `:. 149 = 2n -square` `:. n = square ` Activity `:` Here, a= 1 , d = 2, `t_(n) = 149` `t_(n) = a + ( n -1)d` `:. 149 = 1+ ( n -1) xx 2` `:. 149 = 1+ 2n -2` `:. 149 = 2n -1` `:. 2n = 149 +1 = 150` `:. n = ( 150)/(2)` `:. n = 75` |
|
| 37. |
The sum of the first 45 terms of an A.P. is 3195. Complete the following activity to find the 23rd term. |
|
Answer» `S_(n) =S_(450 = 3195` Let the first term of the A.P. be a and the common difference d. `S_(n) = ( n )/(2) [ 2a +square ]` ….(Formula) `:. S_(45) = ( 45)/(2) [ 2a + square ]` …(Substituting the values ) `:. 3195 = ( 45)/(2) [ 2a + 44d]` `:. 3195 = 45 xx square ` `:. a + 22d = square ` `:. a + 22d = 71` ...(1) Now, 23 rd term is `t_(23)`. `t_(n) = square ` .....(Formula) ` :. t_(23) = a+ ( 23-1) d ` `:. t_(23) = a+ 22d ` `:. t_(23) = square ` ....[From (1) ] Activity `: S_(n) = ( n )/(2)[2a + ( n-1) d ]` `:. S_(45) /(2) [2a+ ( 45-1) d]` `:. 3195 = ( 45)/(2) [ 2a+ 44d]` `:. 3195 = 45 xx a + 22d` `:. a + 22d = ( 3195)/(45)` `:. a+ 22d = 71` ...(1) Now, 23rd term is `t_(23)`. `t_(n) = a+(n-1)d ` ...(Formula) `:. t_(23) = a+ (23-1)d` `:. t_(23) =a+22d` `:. t_(23) = 71` ...[From (1)] |
|
| 38. |
Complete the following activity to find the 19th term of the A.P. 10,15,20,…. |
|
Answer» Here a=10, `d= square , t_(19)= ?` `t_(n) = a+ square ` `:. t_(19) = 10+ square` `:. t_(19) = square ` Activity `:` Here , `a= 10 , d= 5, t_(19) = ?` `t_(n) = a+ ( n -1) d ` `:. t_(19) = 10+ ( 19-1) xx 5 ` `:. t_(19) = 100` |
|
| 39. |
Find the 19th term of the A.P. 7,13,19,25,…. |
|
Answer» Correct Answer - The 19th term is 115. Here ` a= t_(1) = 7, t_(2) = 13, t_(3) = 19 , t_(4)= 25,…` `d= t_(2) - t_(1) = 13-7=6, n = 19` `t_(n) = a+ ( n -1) d ` …(Formula) `:. T_(19) = 7 + ( 19-1) xx6` …( Substituting the values ) `= 7 + 18 xx 6` `= 7 + 108` ` = 115` |
|
| 40. |
Find the AM between (i) 13 and 19 (ii) (a-b) and (a+b) |
|
Answer» (i) AM between 13 and 19` =(1)/(2) (13 + 19) = 16` (ii) AM between (a-b) and (a+b) ` =(1)/(2) [(a-b) +(a+b)] =a.` |
|
| 41. |
Tanvy joined her job in a company in the year 2015 on a monthly salary of Rs. 40000 with an annual increment of Rs. 2500. In which year will she get Rs. 65000 as monthly salary? |
|
Answer» Monthly salary received by Tanvy in 2015, 2016, 2017, 2018,… is respectively Rs. 40000, Rs. 42500, Rs. 45000, Rs. 47500, …. This is an AP with a = 40000, d = 2500 and l = 65000. Let the number of terms of this AP be n. Then, `T_(n) = 65000 rArr a + (n-1) d = 65000` `rArr 40000 + (n-1) xx 2500 = 65000` `rArr (n-1) xx 2500 = 65000 - 40000 = 25000` `rArr (n-1) = (25000)/(2500) = 10 rArr n = 11.` Thus, the 11th annual salary received by Tanvy will be Rs. 65000. Thus, after 10 years, i.e., in the year 2025, her annual salary will be Rs. 65000. |
|
| 42. |
Ramkali required Rs 2500 after 12 weeks to send her daughter to school. She saved Rs100 in first week and increased her weekly savings by Rs 20 every week. Find whether she will be able to send her daughter to school after 12 weeks. What value is generated in the above situation? |
|
Answer» Monkey required for Ramkali for admission of her daughter =Rs. 2500 Her savings in 12 weeks (in Rs.) are 100, 120,140,160,... upto 12 terms, which is an A.P. Here, a=100, d=120-100=20 `:.` Sum of savings in 12 weeks `=(12)/(2)[2xx100+(12-1)xx20]` `=Rs. (12xx210)=Rs. 2520` So, she will be able to send her daughter as the saved more than the required money. Value : Small savings can fulfil your big dreams. |
|
| 43. |
A man saved ₹660000 in 20 years. In each succeeding year after the first yea he saves ₹ 2000 more than he saved in the previous year. How much did he save in the first year? |
|
Answer» Here, we have an AP in which ` S_(20) = 660000, d= 2000 ` and we have to find the value of a. Using the formula ` S_(n) = n/2 xx [ 2a +(n-1) xx 2000]` ` Rightarrow 10xx [ 2a + 19 xx 2000] = 660000` ` Rightarrow 20a + 380000 = 660000 rRightarrow 20a = 280000 Rightarrow a= 14000` Hence, the man saved ₹ 14000 in the first year. |
|
| 44. |
If the 2nd term of an AP is 13 and 5th term is 25, what is its 7th term ?A. 30B. 33C. 37D. 38 |
|
Answer» Correct Answer - B Here `a_(2)=13` `rArr a+(2-1)d=13 rArr a+d=13 " " ...(1)` and `a_(5)=25` `rArr a+(5-1)d=25 rArr a+4d=25 " " ...(2)` Subtracting equation (1) from equation (2), we get `{:(""a+4d""=25),(underset(-)" "a underset(-)+d=""underset(-)""13),(bar(" "3d=12" ")):}` `rArr " " d=4` Put d=4 in equation (1), we get `a+4=13 rArr a=9` Now `a_(7)=a+(7-1)d=9+6(4)=33` |
|
| 45. |
The 26th, 11th and the last terms of an AP are, 0, 3 and `-(1)/(5)`,respectively. Find the common difference and the number of terms. |
|
Answer» Let the first term, common difference and the number of terms of an A.P. are a, d and n respectively. Then `T_(26)=0` `rArr a+25d=0 " " ...(1)` `T_(11)=3` `rArr a+10d=3 " " ...(2)` Subtracting equation (2) from equation (1), we get `{:(" "a+25d=""0),(underset(-)" "a underset(-)+10d=""underset(-)3),(bar(" "15d=-3" ")):}` `rArr" " d=-(1)/(5)` From equation (1), we get `a=-25d` `=-25xx(-(1)/(5))=5` Now, the last term `=-(1)/(5)` `rArr a+(n-1)d=-(1)/(5)` `rArr 5+(n-1)(-(1)/(5))=-(1)/(5)` `rArr 25-n+1=-1` `rArr n=27` |
|
| 46. |
If the first term of an AP is -5 and the common difference is 2, then the sum of the first 6 terms is |
|
Answer» Correct Answer - A Here `a=-5, d=2, n=6` `:. S_(n)=(n)/(2)[2a+(n-1)d]` `rArr S_(6)=(6)/(2)[(2(-5)+(6-1)(2)]=3(-10+10)=0` |
|
| 47. |
There are 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. |
|
Answer» Correct Answer - n=5 Let `A_(1),A_(2),….,A_(n)` be the arithmetic between 9 and 27. Then , 9 , ` A_(1),A_(2),…..,A_(n) ` , 27 are in AP. 9+( n+2-1) d = 27` Rightarrow d = 18/(n+1)` ` A_(n)/A_(1)=2/1 Rightarrow T_(n-1)/T_(2) = 2/1 Rightarrow ( 9+(n+1-1)d)/(9 +(2-1) d) =2/1` ` Rightarrow 9+ nd=18+2d Rightarrow d= 9/(n-2)` ` 18/(n+2) = 9/(n-2) Rightarrow 18(n-2) = 9(n+1) Rightarrow n=5` |
|
| 48. |
The taxi fare after each km, when the fare is Rs. 15 for the first km and Rs 8 for each additional km, does not form an AP as the total fare (in Rs.) after each km is `15,8,8,8, … .` Is the statement true? Give reasons. |
|
Answer» No. Fare after each km, (in Rs. )is `15,15+8,15+2xx8,15+3xx8,...=15,23,31,39,...` Given series is 15, 8,8,8,... Here `a_(2)-a_(1)=8-15=-7` `a_(3)-a_(2)=8-8=0` `a_(2)-a_(1)!=A_(3)-a_(2)` `:.` It is not an A.P. |
|
| 49. |
A farmer buys a used tractor for ₹ 180000. He pays ₹ 90000 in cash and agrees to pays the balance in annual instalments of ₹ 9000 plus 12% interset on the unpaid amount. How much did the tractor cost him ? |
|
Answer» Correct Answer - ₹ 239400 Total cost = ₹ 180000. paid in cash = ₹ 90000. Balance = ₹ 90000. Amount of each instalment = ₹ 9000. Number of instalments = ` 90000/9000 = 10` 1st instalment = `₹ ( 9000 + 90000 xx 12/100 xx1) = ₹ 19800` 2nd instalment = `₹ (9000 +81000 xx 12/100 xx 1) = ₹ 18720` 3rd instalment = `₹ ( 9000 + 72000 xx 12/100 xx 1) =₹ 17460` and so on. Total payment = ₹ [ 90000 + ( 19800 + 18720 +17640 + ..... up to 10 terms ] ` = ₹ [ 90000 +10/2 . { 39600 +9 xx (-1080) }]` = ₹ (90000 + 149400 ) = ₹ 239400. |
|
| 50. |
A man accepts a position with an initial salary of ₹ 26000 per month. It is understood that he will receive an automatic increase of ₹ 250 in the very next month and each month thereafter. find[ his (i) salary for the 10th month ,(ii) total earings during the frist year. |
|
Answer» Correct Answer - ₹ 28350 (ii) ₹ 328500 The amount form an AP with a = ₹ 26000 and d= ₹ 250 `T_(10) = (a+9d) =₹ (26000 +9 xx 250) = ₹ 28250` (ii) Total earning during 1 year =` ₹ { 12/2 xx(2 xx 26000 +11 xx 250)} = ₹ 328500` |
|