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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.
| 51. |
Two cars start together from the same place in same direction. The first goes with unifrom speed of 60 km/hr. The second goes at a speed of 48km/hr in the first hour and increases the speed by 1km each succeding hour. After hour and many hours will the second car overtake the first car if both cars go non-stop? |
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Answer» Correct Answer - 25 hours Suppose the second car overtakes the first car in n hours. Distance covered by the first car in n huours = ( 600n ) km Distance covered by the second car in n hours [ 48+49 +50 +….. To n terms] ` n/2 xx [ 2xx 48+ (n-1) xx1] =n/2 (95+n)` `n/2 ( 95 +n) = 60n Rightarrow 95+n 120 Rightarrow n = 25` hours. |
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| 52. |
Arun buys a scooter for ₹44000. He pays ₹ 8000 in cash and agrees to pay the balance in annual instalments of ₹ 4000 each plus 10% interest on the unpaid amount. How much did he pay for it ? |
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Answer» Correct Answer - ₹ 62000 Total cost = ₹ 44000. parid in cash = ₹ 8000.Balance = ₹ 36000 Amount of each instalment = ₹ 4000 Number of instalment= ` 36000 div 4000 = 9` 1st instalment =` ₹ ( 4000 + 36000 xx 10/100 xx1) = ₹ 7600` 2nd instalment = ` ₹ ( 4000 + 32000 xx 10/100 xx1) = ₹ 7200` 3rd instalment = ` ₹ ( 4000 + 28000 xx 10/100 xx1) = ₹ 6800` ,and so on. Totaol payment = ₹ { 8000 + ( 7600 + 7200 = 6800 + ......to 0 terms)} ` = ₹ { 8000 + 9/2 ( 15200 +8xx (-400)}} = ₹ 62000.` |
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| 53. |
For what value of n are the nth terms of the following two Aps the same 13, 19, 25, … and 69, 68, 67,…? Also, find this term. |
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Answer» Let nth terms of the given progressions be `t_(n) "and" T_(n)` respectively. The first AP is 13, 19, 25, …. Let its first term be a and common difference be d. Then, a = 13 and d = (19-13) = 6. So, its nth term is given by `t_(n) = a + (n-1)d` `rArr t_(n) = 13 + (n-1) xx 6` `rArr t_(n) = 6n + 7 " "...(i)` The second AP is 69, 68, 67, ... Let its first term be A and common difference be D. Then, A = 69 and D = (68-69) =-1. So, it nth term is given by `T_(n) = A + (n-1) xx D` `rArr T_(n) = 69 + (n-1) xx (-1)` `rArr T_(n) = 70-n` Now, `t_(n) = T_(n) rArr 6n + 7 = 70-n` `rArr 7n = 63 rArr n = 9` Hence, the 9th term of each AP is the same This term ` = 70-9 = 61 [ because T_(n) = (70 -n)].` |
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| 54. |
Which term of the AP 72, 68, 64, 60, ..is 88? |
| Answer» Correct Answer - 19th | |
| 55. |
What is the sum of the first five natural numbers divisible by 3 ?A. 54B. 45C. 57D. 75 |
| Answer» Correct Answer - B | |
| 56. |
What is the common difference of an A.P. `- 5 , - (4)/(3), (7)/(3) , 6 , ….` ?A. `(11)/(3)`B. `- (11)/(3)`C. `(10)/(3)`D. `-(10)/(3)` |
| Answer» Correct Answer - A | |
| 57. |
Find `t_(11)` for the A.P. `70,68.5,67,65.5,…`A. 55.5B. 55C. 54.5D. 54 |
| Answer» Correct Answer - B | |
| 58. |
For an A.P., if `S_(10) =150 ` and `S_(9) = 125` , find `t_(10)`A. 20B. 22C. 24D. 26 |
| Answer» Correct Answer - `[S_(n) - S_(n-1) =t_(n)]` | |
| 59. |
If for an A.P. `t_(10) = 25, t_(18) = 45`, what is the value of `t_(14)`? |
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Answer» Correct Answer - The value of `t_(14)` is 35 `t_(14) =[ ( t_(10)+t_(18))/(2)]` |
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| 60. |
If for an A.P.,`d=5,` then `t_(18)-t_(13)=….`A. 5B. 20C. 25D. 30 |
| Answer» Correct Answer - C | |
| 61. |
The sum of first m terms of an AP is `(4m^2-m)` If its nth term is 107, find the value of n. Also find the 21st term of this AP |
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Answer» Correct Answer - n = 14, `T_(21) = 163` `S_(m) = (4m^(2) - m) rArr S_(m-1) = 4(m-1)^(2) - (m-1) = (4m^(2) -9m + 5)` `therefore T_(m) = (S_(m) - S_(m-1)) = (4m^(2)-m) - (4m^(2) -9m + 5) = (8m - 5)` `rArr T_(n) = (8n-5) rArr 8n - 5 = 107 rArr n = 14` `therefore T_(21) = (8 xx 21 - 5) = 163` |
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| 62. |
Which term of the AP 3, 15, 27, 39,… will be 120 more than its 21st term? |
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Answer» Here a= 3 and = (15-3) = 12. `therefore ` 21 st term is given by `T_(21) = a +(21-1)d = a+ 20d = (3 + 20 xx 12) = 243.` Required term = (243 + 120) = 363. Let it be nth term. Then, `T_(n) = 363 rArr a + (n-1)d = 363` `rArr 3 + (n-1) xx 12 = 363` `rArr 12n = 372 rArr n = 31` Hence, 31st term is the required term. |
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| 63. |
If the first two terms of an A.P. Are -3 and 4, then what is the 21st term of this A.P. ?A. -143B. 143C. 137D. 17 |
| Answer» Correct Answer - C | |
| 64. |
If for an A.P., a= 10.5 , d= 0 , dn = 101, then what is the value of `t_(n)` ?A. `0`B. 10.5C. 111.5D. 110.5 |
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Answer» Correct Answer - B `[t_(n) a+ ( n-1) d . d=0 :. (n =1) d= 0 :. t_(n) = a ]` |
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| 65. |
For an A.P. if `a=3.5,d=0,n=101,` then `t_(n)=….` |
| Answer» Correct Answer - B | |
| 66. |
Find the sum of the series 101+99+97+95+…..+43 |
| Answer» Correct Answer - 2160 | |
| 67. |
Find the number of natural numbers between 101 and 999 which aredivisible by both 2 and 5. |
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Answer» Correct Answer - 89 Clearly, each given number is divisible by 10. So, these numbers are 110, 120, 130, …,990 |
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| 68. |
Find the sum of all integers between 101 and 500, which are divisible by 9. |
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Answer» Correct Answer - 13266 Required sum = 108 +117 =126+….+ 495 |
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| 69. |
Find the sum ofintegers from 1 to 100 that are divisible by 2 or 5. |
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Answer» Required sum = ( sim of integers from 1 to 100, dividisble by 2) + ( sum of intergers from 1 to 100 , divisible by 5) - ( sum of intergers from 1 to 100, divisible by 10) ( 2 + 4+ 6+ ……+ 100) + ( 5+ 10+ 15 + ……+ 100) - ( 10+20+30+.......+ 100) ` 50/2 ( 2+100) + 20/2 ( 5 + 105) - 10/2 ( 10+100) ` = ( 2550 + 1050 - 550) = 3050 |
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| 70. |
Find how many integers between 200 and 500 are divisible by 8. |
| Answer» Correct Answer - 37 | |
| 71. |
How many two -digit numbers are divisible by 3? |
| Answer» Correct Answer - 30 | |
| 72. |
How many two -digit numbers are divisible by 3?A. 25B. 30C. 32D. 36 |
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Answer» Correct Answer - B Two-digit numbers divisible by 3 are 12, 15, 18, …, 99 Let `T_(n) = 99. "Then", 12 + (n-1) xx 3 = 99 rArr (n-1) xx 3 = 87 rArr n = 30.` |
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| 73. |
If `a_(n)` denotes the nth term of the AP 3, 18, 13, 18,… then what is the value of `(a_(30)-a_(20))?`A. 40B. 36C. 50D. 56 |
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Answer» Correct Answer - C Here a =3 and d = 5. `therefore (a_(30) -a_(20)) = (3 + 29 xx 5) - (3 +19 xx 5) = 148-98 = 50` |
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| 74. |
How many three-digit numbers are divisible by 9?A. 86B. 90C. 96D. 100 |
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Answer» Correct Answer - D Three-digit numbers divisible by 9 are 108, 117, 126, .., 999 Let `T_(n) = 999."Then," 108 + (n-1) xx 9 = 999` `therefore (n-1) xx 9 = 891 rArr (n-1) = 99 rArr n = 100` |
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| 75. |
What is the common difference of an AP in which `a_(18)-a_(14) = 32?`A. 8B. `-8`C. 4D. `-4` |
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Answer» Correct Answer - A `a_(18) -a_(14) = 32 rArr (a+17d) -(a+13d) = 32 rArr 4d = 32 rArr d= 8` |
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| 76. |
Find the sum of first 51 terms of an AP whose second and third termsare 14 and 18 respectively. |
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Answer» Correct Answer - 5610 Let a be the first term of the given AP. Then, 14-a = 18-14 `rArr` 14-a = 4 `rArr` a = 10 `therefore ` a = 10,d = 4 and n = 51. Now, find `S_(n)`. |
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| 77. |
In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students. Which value is shown in this question? |
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Answer» Correct Answer - 312 Required sum ` = 2 xx [2 + 4 +6 +8 +… + 24] = 2 xx (12)/(2)(2+24) = 312` Planting trees helps in reducing air pollution. |
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| 78. |
There are 25 trees at equal distances of 5 metres n a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener 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 gardener will cover in order to water all the trees. |
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Answer» Correct Answer - 3500 m Total distance covered `= [(2 xx 10) + 2 xx (10 +5 xx 1) +2 xx (10 +5 xx 2) +… + 2 xx (10 +5 xx 24)]m` ` = 2 xx [10 + 15 +20 +….+ 130]m = 2 xx (25)/(2) (10+ 130)m = 3500m.` |
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| 79. |
In a potato race, a bucket is placed at the starting point, which is 5m from the first potato, and the other potatoes are placed 3m apart in a straight line. There are ten potatoes in the line (fig.). A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the nest potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run ? [Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is `2 xx 5 + 2 xx (5 + 3)`] |
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Answer» Correct Answer - 370m Total distance run to pick up the 1st, 2nd, 3rd,.., 10th potato `= [(2 xx 5) + 2 (5+3) +2 xx (5+3 xx 2) +.. +2 xx (5 +3 xx 9)]m` ` = 2 xx [5 + 8 +11 +… +32]m = 2 xx (10)/(2) (+32)m = 370m.` |
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| 80. |
Find the sum of 100 temrs of the AP 0.6,0.61,0.62,0.63 ,….., |
| Answer» Correct Answer - 109.5 | |
| 81. |
Show that the sequence defined by ` T_(n) = 3n+5` is an AP. Find the common difference . |
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Answer» We have ` T_(n) = 3n+5` Replacing n by ( n-1) in (i), we get ` T_(n-1)= 3( n-1) + 5 Rightarrow T_(n-1) = 3n+2` Subtacting (ii) from (i), we get ` ( T_(n) - T_(n-1) = ( 3n + 5) - ( 3n+2) =3`,which is constant Hence, the given sequence in an AP with common differnece 3. |
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| 82. |
if ` (3 +5+7+9+…. " up to 35 terms " )/(5+8+11+ …. " up to n terms ")=7` , find the value of n. |
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Answer» Correct Answer - n=10 `(35/2.{6+34xx2})/(n/2.{10 +(n-1) xx3}) =7 Rightarrow 3n^(2) +7n -370 =0` ` Rightarrow (3n +37) (n-10) =0 Rightarrow n=10` |
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| 83. |
find the value of x such that 1+4+7+10+….+x = 715 |
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Answer» Correct Answer - x=64 Here, a=1and d=3. Let it contain n terms. Then, ` n/2 .[ 3a+(n-1)d] = 715 Rightarrow n/2. [2xx 1 + ( n-1)xx3] = 715` ` Rightarrow n(3n-1) = 1430 Rightarrow 3n^(2)-n-1430 =0 Rightarrow 3n^(2)-66n+65n -1430=0` ` Rightarrow 3n(n-22) + 65 (n-22) =0Rightarrow (n-22) (3n+65) =0 Rightarrow n=22` `x==T_(22) = ( a+21d) = ( 1 + 21xx3) = 64` |
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| 84. |
Show that the sequence `loga ,log((a^2)/b),log((a^3)/(b^2)),log((a^4)/(b^3)), `forms an A.P. |
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Answer» By symmetry , we find that ` T_(n) = log (a^(n)/(b^(n-1))) and T_(n-1) = log(a^(n-1)/(b^(n-2)))` ` ( T_(n)-T_(n-1)) = log(a^(n)/b^(n -1)))- log( a^(n -1)/b^(n-2))` ` log (a ^(n)/ b^(n-1)xx b^(n-2)/a^(n-1)) = log(a/b)` = constant Hence, the given sequence in an AP with common differnce ` log(a/b)` |
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| 85. |
The firstand last terms of an A.P. are 1 and 11. If the sum of its terms is 36, thenthe number of terms will be(a) 5 (b) 6 (c) 7 (d) 8 |
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Answer» Correct Answer - 6 `n/2 (a+l) =36 Rightarrow n/2 xx(1+11) =36 Rightarrow n=6` |
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| 86. |
Show that the progression 7,12,17,22,27, …., is an AP. Find its general term and the 14th term. |
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Answer» we have (12-7) = ( 17-12) = ( 22 -17) = ( 27-22) = 5, which is contant . So , the given progression is an AP in which a=78 and d =5 Its general term, ` T_(n) = ( a+ (n-d)d}= 7 + ( n -1) xx 5 Rightarrow T_(n) = ( 5n +2)` 14th term , ` T_(14) = ( 5xx 14 +2) =72` |
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| 87. |
Find the value of x such that 25+22+19+16 +….+ x=112 |
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Answer» Correct Answer - x=7 `3n^(2)-21n-32n+224=0` |
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| 88. |
The 10th common termbetween the A.P.s 3,7,11,15, and 1,6,11,16,.. is`191`b. `193`c. `211`d. none of these |
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Answer» Correct Answer - 191 The first common term =11 Common differnece of new AP = LCM (4,5) =20 So, the new AP is 11,31,51,….., Its 10th term= ` (11 +9 xx20) =191` |
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| 89. |
The sum of the first ten terms of the A.P.`15,10,5,….is……A. -75B. -125C. 75D. 125 |
| Answer» Correct Answer - A | |
| 90. |
In an A.P.,the first term is 22, `n t h`term is `-11`and the sumto first `n`terms is66. Find `n`and `d`, thecommon difference. |
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Answer» Correct Answer - n = 12, d = -3 `S_(n) = (n)/(2)(a+l) rArr (n)/(2)[22 +(-11)] = 66 rArr n = 12.` `T_(12) =-11 rArr a + 11d = -11 rArr 22 + 11d = -11 rArr d = -3.` |
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| 91. |
The sum of first 7 terms of an A.P. is 10 and that of next 7 terms is167. Find the progression. |
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Answer» Correct Answer - `1+1""1/7,1""2/7,1""3/7`+…. `S_(7)=10 and S_(14) = (17+10) =27` ` 7/2 xx [2a +6a] =10 Rightarrow 7a +21d =10` `and 14/2 xx[2a =13d] =27 Rightarrow 14a +91d =27` Solve for a and d . |
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| 92. |
find the sum of `n` terms of the series `(4-1/n)+(4-2/n)+(4-3/n)+...........` |
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Answer» Correct Answer - `(1)/(2)(7n-1)` Required sum = (4+4+… to n terms) `-((1)/(n) + (2)/(n) + (3)/(n) +… + (n)/(n))` ` = 4n - (n)/(2)((1)/(n) + (n)/(n)) ["sum" = (n)/(2)(a+l)]` ` = 4n -((1+n))/(2) = (1)/(2) (7n-1).` |
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| 93. |
The first and the last terms of an A.P. are 5 and 45 respectively. If thesum of all its terms is 400, find its common difference. |
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Answer» Correct Answer - `d = (8)/(3)` and n = 16 `S_(n) =(n)/(2) (a+l) rArr (n)/(2) (5+45) = 400 rArr n = 16.` `S_(16) = 45 rArr a + 15d = 45 rArr 5 + 15d = 45 rArr d = (8)/(3)` |
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| 94. |
In an AP, it is given that `S_(5) + S_(7) = 167 "and" S_(10) = 235`, then find the AP, where `S_(n)` denotes the sum of its first n terms. |
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Answer» Correct Answer - 1, 6, 11, 16,… `S_(5) + S_(7) = 167 rArr (5)/(2)(2a +4d) + (7)/(2)(2a +6d) = 167` `rArr 12a + 31d = 167 " "…(i)` `S_(10) = 235 rArr (10)/(2) (2a +9d) = 235 rArr 2a + 9d = 47. " "...(ii)` Solve (i) and (ii), we get a = 1 and d = 5. |
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| 95. |
The first and the last terms of an AP are 17 and350 respectively. If the common difference is 9, how many terms are there andwhat is then sum? |
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Answer» Correct Answer - n = 38 and `S_(n) = 6973` Let the number of terms be n. Then, `T_(n) = 350 rArr a +(n-1)d = 350 rArr 17 + (n-1) xx 9 = 350 rArr n = 38.` `S_(n) = (n)/(2)(a+l).` |
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| 96. |
In an A.P., the first term is 2, the last term is 29 and sum of the terms is155. Find the common difference of the A.P. |
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Answer» Correct Answer - d=3 `S_(n) = (n)/(2)(a+l) rArr (n)/(2)(2+29) = 155 rArr n = 10` `T_(10) = 29 rArr 2+ 9d = 29 rArr d =3.` |
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| 97. |
The first term of an AP is p and its common difference is q. Find its 10th term. |
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Answer» Correct Answer - (p+9q) Here a = p and d = q. `therefore T_(10) = a +9d = (p +9q)` |
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| 98. |
If `(4)/(5),` a, 2 are in AP, find the value of a. |
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Answer» Correct Answer - `a = (7)/(5)` We have, `a - (4)/(5) = 2-a rArr 2a = (2 +(4)/(5)) = (14)/(5) rArr a = (7)/(5)` |
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| 99. |
If x, y, z are real numbers satisfying the equation `25(9x^(2)+y^(2))+9z^(2)-15(5xy+yz+3zx)=0` , then prove that x, y and z are in A.P. |
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Answer» We hare, `25(9x^(2)+y^(2))+9z^(2)-15(5xy+yz+3zx)=0` `rArr (15x)^(2)+(5y)^(2)+(3z)^(2)-(15x)(5y)-(5y)(3z)-(3z)(15x)=0` `rArr (1)/(2)[2(15x)^(2)+2(5y)^(2)+2(3z)^(2)-2(15x)(5y)-2(5y)(3z)-2(3z)(15x)]=0` `rArr (1)/(2)[(15x-5y)^(2)+(5y-3z)^(2)+(3z-15x)^(2)]=0` `rArr (15x-5y)^(2)+(5y-3z)^(2)+(3z-15x)^(2)=0` It is possible only when `15x-5y=0 and 5y-3z=0 and 3z-15x=0` `rArr 15x=5y=3z` `rArr (x)/(1)=(y)/(3)=(z)/(5)=k` (say) `:. x=k, y=3k, z=5k` Now, `y-x=3k-k=2k` and `z-y=5k-3k=2k` Since, `y-x=z-y` `:. x,y,z ` are in A.P. `" "` Hence Proved |
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| 100. |
If zeros of `x^3 - 3p x^2 + qx - r` are in A.P., then |
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Answer» Let `alpha-beta, alpha` and `alpha+beta` are the zeroes of given polynomial . `:.` Sum of zeroes `=-((-3p))/(1)=3p` i.e., `alpha-beta+alpha+alpha+beta=3p rArr 3 alpha = 3p` `:. alpha=p " " ...(1)` Now, since `alpha` is one zero of the polynomial. `:.` On putting `x=alpha`, we get the remainder=0. `:. alpha^(3)-3palpha^(2)+qalpha-r=0` `rArr p^(3)-3p(p^(2))+qp-r=0 " "` [from (1)] `rArr p^(3)-3p^(3)+pq-r=0` `rArr 2p^(3)=pq-r`, which is the required relation in p, q and r. |
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