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. |
Determine k, so that `K^(2)+4k+8, 2k^(2)+3k+6` and `3k^(2)+4k+4` are three consecutive terms of an AP. |
|
Answer» Since, `k^(2)+4k+8, 2k^(2)+3k+6` and `3k^(2) +4k+4` are consecutive terms of an AP. ` :. 2k^(2)+3k+6-(k^(2)+4k+8)=3k^(2)+4k+4-(2k^(2)+3k+6)=` Common difference `implies 2k^(2)+3k+6-k^(2)-4k-8=3k^(2)+4k+4-2k^(2)-3k-6` ` implies " " k^(2)-k-2=k^(2)+k-2` `implies " " -k=kimplies2k=0impliesk=0` |
|
| 2. |
Find the 9th term from the ctowards the first term of the A.P. 5,9,13,....,185 |
|
Answer» a=5,d=4,l=185 `T_n=a+(n-1)d` `=185+(9-1)(-4)` `=153`. |
|
| 3. |
A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes. |
|
Answer» Here, we are given, `S_7 = 700, n = 7 and d = -20` We know,`S_n = n/2(2a+(n-1)d)` `700 = 7/2(2a+6(-20))` `200 = 2a-120=>a = 320/2 = 160` So, `a_1 = 160` `a_2 = 160-20 = 140` `a_3 = 140-20 = 120` `a_4 = 120-20 = 100` `a_5 = 100-20 = 80` `a_6 = 80-20 = 60` `a_7 = 60-20 = 40` So, value of each of the prives will be, `160,140,120,100,80,60,40` |
|
| 4. |
A shop keeper buys a number of books for Rs.80. If he had bought 4 more books for the same amount, each book would have cost his Rs.1 less .How many books did he buy? |
|
Answer» Let `x` is the number of books that the shopkeeper purchased. Then, price of each book ` = 80/x` Rs If he had bought `4` more books, then each book would have cost `1` rs less. So, he can buy `x+4` books in the same amount. So the cost of each of the book would be`80/(x+4)`. According to the question, `80/x-80/(x+4)=1` `=>(80x+320-80x)/(x^2 +4x)=1` `=>x^2+4x=320` `=>x^2+4x-320=0` `=>x^2+20x-16x-320=0` `=>x(x+20)-16(x+20)=0` `=>(x+20)(x-16)=0` Therefore, `x= -20, 16` As `x` can not be negative so `x=16`. So he bought `16` books. |
|
| 5. |
The third term and nineth terms of an A.P. are 4 and -8 respectively. Find which term of A.P is equal to 0. |
|
Answer» Let `a` is the first term of `A.P.` and `d` is the common difference of the `A.P.` Then, `a+2d = 4->(1)` `a+8d = -8->(2)` Subtracting (2)-(1), `a+8d-a-2d = -8-4` `d = -2` Therefore, `a+2(-2) = 4 => a = 8` Let `n` th term of this `A.P.` is `0`. `:. 8+(n-1)(-2) = 0` `=> 8-2n+2 = 0` `=>2n = 10` `=> n = 5` `:.5^(th)` term of the `A.P.` will be `0`. |
|
| 6. |
If `a_(n)=3-4n`, then show that `a_(1),a_(2),a_(3), …` form an AP. Also, find `S_(20)`. |
|
Answer» Given that, `n`th term of the series is `a_(n)=3-4n " " `…(i) Put`n=1, " " a_(1)=3-4(1)=3-4= -1` Put`n=2, " " a_(2)=3-4(2)= 3-8=-5` Put`n=3, " " a_(3)=3-4(3)= 3-12=-9` Put`n=4, " " a_(4)=3-4(4)=3-16= -13` So the series becomes `-1,-5,-9,-13, …` We see that, `a_(2)-a_(1)=-5-(-1)=-5+1=-4` `a_(3)-a_(2)=-9-(-5)=-9+5= -4` `a_(4)-a_(3)= -13-(-9)=-13+9= -4` ` "i.e., " a_(2)-a_(1)=a_(3)-a_(2)=a_(4)-a_(3)=...= -4` Since, the each successive term of the series has the same difference. So, it forms an AP. We know that, sun of n terms of an AP, `S_(n)=(n)/(2)[2a+(n-1)d]` ` :. ` Sum of 20 terms of the AP,`S_(20)=(20)/(2)[2(-1)+(20-1)(-4)]` `S_(20)=10(-2+(19)(-4))=10(-2-76)` ` " " =10xx-78= -780` Hence, the required sum of 20 terms i.e., `S_(20)` is `-780`. |
|
| 7. |
The first , second and the last terms of an A.P. are `a ,b , c`respectively. Prove that the sum is `((a+c)(b+c-2a))/(2(b-a))`. |
|
Answer» `1^(st)` term=a `2^(nd)`term=b `3^(rd)`term=c Common diff,d=(b-a) c=a+(n-1)(b-a) c-a=(n-1)(b-a) `n-1=(c-a)/(b-a)` `n=(c-a)/(b-a)+1` `n=(c-a+b-a)/(b-a)=(b+c-2a)/(b-a)` `S=n/2(a+l)` `S=(b+c-2a)/(2(b-a))*(a+c)` `S=((a+c)(b+c-2a))/(2(b-a))`. |
|
| 8. |
Find the `11^(t h)`from the last term (towards the first term) of the AP : `10 , 7, 4, dot dot dot , 62.` |
|
Answer» 10,7,4...,-62 (AP) `a_1`=10 `a_2`=7 d=`a_2-a_1` =7-10 =-3 L=-62 `a_n`=a+(n-1)(d) -62=10+(n-1)(-3) (n-1)=24 n=25 `a_(15)`=a+(15-1)d `a_(15)`=10+(14)(-3) `a_(15)`=-32 |
|
| 9. |
We know that the sum of the interior angles of a triangle is`180^0`. Show that the sums of the interior angles of polygons with `3,4,5,6`sides form an arithmetic progression. Find the sum of the interior angles ofa 21 sided polygon. |
|
Answer» We, know that sum of the interior angles of n-sides=`180*(n-2)` `a_n=180(n-2)` `a_(n-1)=180((n-1)-2)` `a_n-a_(n-1)=180(n-2)-180((n-1)-2)` `a_n-a_(n-1)=180(n-2-n+1+2)` `=180` `d=180` Sum of angles`=180(n-2)` `=180*19` `=3420`. |
|
| 10. |
Check whether ` 150` is a term of the AP : 11, 8, 5, 2... |
|
Answer» given `a=11` b=`8-11=5-8= -3` lets say -150 is `r^(th) term` so`a+(r-1)d= -150` `11+(r-1)(-3)= -150` `(r-1)*3= 161` `r-1=161/3` `r=161/2 +1 cancel(=) ` integer so, -150 is not a term of a given AP |
|
| 11. |
Find the 12th term from the end of the AP`-2, -4, -6, …, -100.` |
|
Answer» Given AP, `-2,-4,-6, …, -100` Here, first term (a)`=-2`, common difference (d)`=-4-(-2)= -2` and the last term (l)`= -100.` We know that, the `n`th term `a_(n)` of an AP from the end is `a_(n)=l-(n-1)d,` where `l` is the last term and d is the common difference. ` :. ` 12th term from the end, `a_(12)= -100-(12-1)(-2)` ` " " = -100+(11)/(2)= -100 +22= -78`. Hence, the 12th term from the end is `-78` |
|
| 12. |
How many terms of the AP: 9, 17, 25, . . . must be taken to give a sum of 636? |
|
Answer» d=8 a=9 `S_n=n/2(2a+(n-1)d)` 636=n/2(2*9+(n-1)8) `4n^2+5n-636` n(4n+53)-12(4n+53)=0 (n-12)(4n+53)=0 n=12,-53/4 n can not be negative so, n=12 |
|
| 13. |
If the 3rd and the `9^(t h)`terms of an AP are 4 and `- 8` respectively, which term of this AP is zero? |
|
Answer» Let `a` is the starting term and `d` is the common difference. Here, we are given, `a_3 = 3=>a+(3-1)d = 4=>a+2d = 4->(1)` `a_9 = -8=>a+(9-1)d = -8=>a+8d = -8->(2)` Subtracting (1) from (2), `8d-2d = -8-4=>d = -2` Putting value of `d` in (1),`a-4 = 4=>a = 8` Let `n^(th)` term of this AP is `0`,then `8+(n-1)(-2) = 0` `n-1 = 4 =>n = 5` So, fifth term of this AP will be `0`. |
|
| 14. |
How many terms of the AP : 24, 21,18,... must be taken so that their sum is 78? |
|
Answer» a=24 and d=21-24=-3 `S_n=n/2(2a+(n-1)d)` `78=n/2(2(24)+(n-1)(-3))` 156=n(48-3n+3) `n^2-51n+156=0` n(n-13)-4(n-13)=0 (n-4)(n-13) n=4,13 n can not be 13 so, n=3 |
|
| 15. |
Write first four terms of the AP, when the first term a and the common difference d are given as follows:(i) a = 10, d= 10 (ii) a = –2, d = 0(iii) a = 4, d = – 3 (iv) a = – 1, d = 1/2(v) a = – 1.25, d = – 0.25 |
|
Answer» First four terms of an `AP` can be given as, `a,a+d,a+2d,a+3d` Here,`a = 10` and `d = 10` So, first four terms are = `10,(10+10),(10+2(10)),(10+3(10)) = 10,20,30,40` |
|
| 16. |
Is 0 a term of the AP 31, 28, 25, …? Justify your answer. |
|
Answer» Let 0 be the nth term fo given AP.i.e., `a_(n)=0`. Given that, first term `a = 31`, common difference, `d=28-31= -3` The nth terms of an Ap, is ` " " a_(n)=a +(n-1)d` ` implies 0=31+(n-1)(-3)` `implies 3(n-1)=31` ` implies n-1=(31)/(3)` ` :. n=(31)/(3)+1=(34)/(3)=11(1)/(3)` Since, n should be positive integer. So, 0 is not a term of the given AP. |
|
| 17. |
Find the sum of the first 15 multiples of 8. |
|
Answer» a=8, d=8 and n=15 `S_15=n/2(2a+(n-1)d)` `S_15=15/2(2(8)+(15-1)(8))` `S+15=960` |
|
| 18. |
In the following APs, find the missing terms in the boxes :(i) `2,square,26` (ii) `square,13,square,3`(iii) `5,square,square,(1)/(2)`(iv) `-4,square,square,square,square,6` (v) `square,38,square,square,square,-22` |
|
Answer» 1)`a_1`=2 `a_3`=26 a+2d=26 2d=24 d=12 `a_2=a+d=2+12=14` 2)`a_4`=3 a+3d=3(1) `a_2`=13 a+d=13(2) subtracting equation 1 from 2 d=-5 a=18 `a_3=a+2d=8` similarly other can be find |
|
| 19. |
Justify whether it is true to say that ` -1, (-3)/(2), -2, (5)/(2), … ` Forms an AP as `a_(2)-a_(1)=a_(3)-a_(2)`. |
|
Answer» False Here, `a_(1)= -1, a_(2)=(-3)/(2), a_(3)= -2` and `a_(4)= (5)/(2)` `a_(2)-a_(1)= (-3)/(2)+1=-(1)/(2)` `a_(3)-a_(2)= -2+(3)/(2)= -(1)/(2)` `a_(4)-a_(3)=(5)/(2)+2=(9)/(2)` Clearly, the difference of successive terms is not same, all though, `a_(2)-a_(1)=a_(3)-a_(2)` but `a_(3)-a_(2) ne a_(4)-a_(3)` therefore it does not form an AP. |
|
| 20. |
The list of number -10, -6, -2, 2, … isA. an AP with d = -16B. an AP with d = 4C. an AP with d = -4D. not an Ap |
|
Answer» Correct Answer - B The given numbers are `-10, -6, -2, 2, ….` Here `a_(1)=-10, a_(2)= -6, a_(3)= -2` and `a_(4)=2, ….` Since, `a_(2) - a_(1)= -6-(-10) = -6 +10=4` `a_(3) - a_(2) = -2-(-6)= -2+6=4` `a_(4) - a_(3)=2-(-2)=2+2=4` … … … … … … … … ... ... ... ... ... ... ... ... ... ... Each successive term of given list has same difference i.e., 4. So, the given list forms an AP with common difference, d = 4. |
|
| 21. |
In an AP, if `a = 3.5`, `d = 0` and `n = 101`, then `a_(n)` will be |
|
Answer» Correct Answer - B For an AP, `a_(n)=a+(n-1)d=3.5+(101-1)xx0 " " ` [by given conditions] ` :. " " =3.5 ` |
|
| 22. |
In an AP, if d = -4, n = 7 and `a_(n)` = 4, then a is equal toA. 6B. 7C. 20D. 28 |
|
Answer» Correct Answer - D In an AP, ` a_(n)=a+(n-1)d` `implies 4=a+(7-1)(-4) " " `[by given conditions] `implies4=a+6(-4)` `implies 4+24=a` `:. " " a=28` |
|
| 23. |
If`(5+9+13+....to n terms)/(7+9+11+....to(n+1) terms)=17/16`, then `n=` |
|
Answer» Let `S_1 = 5+9+13+...to n` terms `=>S_1 = n/2(2*5+(n-1)4) = n/2(6+4n) = n(3+2n)` Let `S_2 = 7+9+11+...to (n+1)` terms `=>S_2 = (n+1)/2(2*7+(n+1-1)2) = (n+1)/2(14+2n)= (n+1)(7+n)` So, `S_1/S_2 = 17/16` `:. (n(3+2n))/((n+1)(n+7)) = 17/16` `=>16n(3+2n) = 17(n+1)(n+7)` `=>48n+32n^2 = 17(n^2+8n+7)` `=>48n+32n^2 = 17n^2+136n+119` `=>15n^2-88n-119 = 0` `=>15n^2-105n+17n+119 = 0` `=>15n(n-7)+17(n-7) = 0` `=>(15n+17)(n-7) = 0` `=>n = -15/17 or n = 7` As `n gt 0`, `:. n = 7.` |
|
| 24. |
If the sum of 3 numbers in AP is 15 and the sum of their square is 93 Find the number |
|
Answer» `a-d,a,a+d` `a-d+a+a+d=15` `3a=15` `a=5` `a-d,a,a+d` `5-d,5,5+d` `(5-d)^2+(5)^2+(5+d)^2=93` `25+d^2-10d+25+25+d^2+10d=93` `75+2d^2=93` `2d^2=93-75` `=18` `d^2=9` `d=3` `5-d,5,5+d` `5-3,5,5+3` `2,5,8` |
|
| 25. |
which term of the A.P. : 120, 116, 112,..... is first negative term? |
|
Answer» In the given `A.P.`, First term, `a = 120` Common difference `d = -4` `:. T_n = a+(n-1)d = 120 - 4(n-1) = 124-4n` For, `T_n` to be negative, `124-4n lt 0` `=> 4n gt 124 => n gt 31` So, for `n = 32`, `T_n` will be negative. So, `32^(nd)` term of the given `A.P.` will be the first negative term. |
|
| 26. |
Which term of the sequence `20,19 1/4, 18 1/2, 17 3/4,...`is the first negative term? |
|
Answer» `a=20,d=-3/4` `a_n=a+(n-1)d` `a_n=20+(n-1)*(-3/4)` `20+(n-1)*(-3/4)<0` `(20*4+(n-1)(-3))/4` `80-3n+3<0` `83<3n` `n>83/3` `n>27.66` `n=28`. |
|
| 27. |
Interior angle of polygon are in A.P.If the smallest angle is `120^@` and the common difference is `5^@`, find the number of sides of polygon. |
|
Answer» Sum of interior angles of a `n`-sided polygon is `(n-2)**180`. Now, we are given, angles are in AP such that , `a = 120 and d = 5` `:. n/2(2a+(n-1)d) = (n-2)180` `=> n/2(240+(n-1)5) = 180n-360` `=>n(235+5n) = 360n-720` `=>5n^2+235n-360n+720 = 0` `=>5n^2-125n+720 = 0` `=>n^2-25n+144 = 0` `=>n^2-16n-9n+144 = 0` `=>n(n-16)-9(n-16) = 0` `=>(n-16)(n-9) = 0` `n=16 and n = 9` So, number of sides in the given polygon can be `16` or `9`. |
|
| 28. |
What least number of terms of the sequence 17, `15 4/5 , 14 3/5 `should be taken so the sum is negative. |
|
Answer» `S_n=n/2(2a+(n-1)d)` `=n/2((a+l)/2)` `s_n<0` `(a+l)/2<0` `2a+(n-1)d<0` `2aL-(n-1)d` `2aL-(n-1)d` `34L-(n-1)-6/5` `170/6<+(n-1)` `170/6+1L.` `29.33`n=30` |
|
| 29. |
The `p^(t h)`term of an A.P. is `a` and `q^(t h)`term is `b`Prove that the sum of its `(p+q)`terms is `(p+q)/2{a+b+(a-b)/(p-q)}dot` |
|
Answer» `a_1+(p-1)d=a-(1)` `a_1+(q-1)d=b-(2)` subtracting equation 2 from 1 `a_1+pd-d-a_!-qd+d=a-b` `(p-q)d=a-b` `d=(a-b)/(p-q)` Sum of p+q terms `S=(p+q)/2[2a+(p+q-1)d]` `S=(p+q)/2[a+b+d]` `S=(p+q)/2[a+b+(a-b)/(p-q)]`. |
|
| 30. |
Find the number of terms in each of the following APs : (i) 7, 13, 19, . . . , 205 (ii) `18 , 15 1/2, 13 , dot dot dot , - 47` |
|
Answer» Let `a` is the first term, `l` is the last term, `n` is number of terms and `d` is the common difference. (i)Here, `a = 7, d = 6, a_n = 205` Then,`205 = 7+(n-1)6` `6(n-1) = 198=>n-1 = 33` `n = 34` So, there are `34` terms in this AP. (ii) Here, `a = 18, d =-5/2, a_n = -47` Then,`-47 = 18+(n-1)(-5/2)` `5/2(n-1) = 65=>n-1 = 26` `n = 27` So, there are `27` terms in this AP. |
|
| 31. |
The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles |
|
Answer» Given that, the angles of a triangle are in AP. Let A, B and C are angles of a `triangle ABC.` ` :. B=(A+C)/(2)` `implies 2B= A+C " " `… (i) We know that, sum of all interior angles of a `triangle ABC=180^(@)` `A+B+C=180^(@)` `implies 2B+B=180^(@) " " ` [from Eq. (i)] `implies " " 3B=180^(@)impliesB=60^(@)` Let the greatest and least angles are A and C respectively. `A=2C " " ` [by condition] ... (ii) Now, put the values of B and A inEq. (i), we get `2xx60=2C+C` `implies120=3CimpliesC=40^(@)` Put the value of C in Eq. (ii), we get `A=2xx40^(@)implies A=80^(@)` Hence, the required angles of triangle are `80^(@), 60^(@)` and `40^(@)`. |
|
| 32. |
Which term of the AP: 21, 18, 15, is 0? |
|
Answer» series of AP is `21,18, 15 .... 0` `a= 21` `d= 18- 21= -3` `T_n = a + (n-1)d = 21 + (n-1)(-3)` `21-3n +3` `= 24 - 3n` `T_n=0` `24- 3n = 0` `24 = 3n` `n= 8` Answer |
|
| 33. |
Which term of the AP : 3, 8, 13, 18, . . . , is 78? |
|
Answer» Here, `a = 3, d = 5 and a_n = 78` We know, `a_n = a+(n-1)d` `=>78 = 3+5(n-1)=>n-1 = 15` `=>n=16` So, `16^(th)` term is the required term. |
|
| 34. |
Find k, if the given value of x is the kth term of given AP. given by ` 2 , -4 , -10 , -16 , -22 .....; x = -448` |
|
Answer» `k^(th)` term in an `AP = a+(k-1)d` Here, `a` is the first term and `d` is the common difference. In the given AP, `a = 2, d = -6` `a+(k-1)d = -448` `=>2+(k-1)(-6) = -448` `=>2-6k+6 = -448` `=>6k = 456` `=> k = 76` |
|
| 35. |
Which of the following list of numbers does form an AP? If they form an AP. write the next two terms :(i) 4, 10, 16. 22, . . .(ii) 1,-1, -3,-5, . . .(iii) -2,2,-2,2, . . .(iv)1,1,1,2,2,2,3,3,3, . . . |
|
Answer» (i) `4,10,16,22...` Here, difference between each term is `6`. So, it is an `AP`. Its next two terms will be `28,36`. (ii)`1,-1,-3,-5...` Here, difference between each term is `-2`. So, it is an `AP`. Its next two terms will be `-7,-9`. (iii)`-2,2,-2,2...` Here, difference between each term is not consistent. Sometimes, it is `4` and sometimes it is `-4`. So, it is not an `AP`. (iv)`1,1,1,2,2,2,3,3,3...` Here, difference between each term is not consistent. Sometimes, it is `0` and sometimes it is `1`. So, it is not an `AP`. |
|
| 36. |
In acertain A.P. the `24 t h`term istwice the `10 t h`term. Provethat the `72 n d`term istwice the `34 t h`term. |
|
Answer» `T24=a+(24-1)d=a+23d` `T10=a+(10-1)d=a+9d` `T24=2T10` `a+23d=2(a+9d)` `a+23d=2a+18d` `a=5d` `T72=a+(72-1)d` `=a+71d=5d+71d` `=76d` `T34=a+(34-1)d` `=a+33d=5d+33d` `T34=38d` `T72=76d` `T34=38d` `T72=2T34` |
|
| 37. |
Which of the following form of an AP ? Justify Your answer. (i) `-1,-1,-1,-1, …` (ii) `0, 2, 0, 2, …` (iii) `1, 1, 2, 2, 3, 3, … ` (iv) `11, 22, 33, … ` (v) `(1)/(2),(1)/(3),(1)/(4), … ` (vi) `2, 2^(2), 2^(3), 2^(4)` (vii) `sqrt(3),sqrt(12),sqrt(27),sqrt(48), ...` |
|
Answer» (i) Here, `t_(1)=1, t_(2)= -1, t_(3)= -1` and `t_(4)= -1` `t_(2)-t_(1)= -1+1=0` `t_(3)-t_(2)= -1+1=0` `t_(4)-t_(3)= -1+1=0` Clearly, the difference of successive terms is same, therefore given list of numbers form an AP. (ii) Here, `t_(1)=0, t_(2)= 2, t_(3)= 0` and `t_(4)= 2` `t_(2)-t_(1)= 2-0=2` `t_(3)-t_(2)= 0-2= -2` `t_(4)-t_(3)= 2-0=2` Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP. (iii) Here, `t_(1)=1, t_(2)= 1, t_(3)= 2` and `t_(4)= 2` `t_(2)-t_(1)= 1-1=0` `t_(3)-t_(2)= 2-1=1` `t_(4)-t_(3)= 2-2=0` Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP. (iv) Here, `t_(1)=11, t_(2)= 22` and `t_(3)= 33` `t_(2)-t_(1)= 22-11=11` `t_(3)-t_(2)= 33-22=11` `t_(4)-t_(3)= 33-22=11` Clearly, the difference of successive terms is same, therefore given list of numbers form an AP. (v) Here, `t_(1)=(1)/(2), t_(2)= (1)/(3)` and `t_(3)= (1)/(4)` `t_(2)-t_(1)= (1)/(3)-(1)/(2)=(2-3)/(6)=(1)/(6)` `t_(3)-t_(2)=(1)/(4)-(1)/(3)=(3-4)/(12)=(1)/(12)` Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP. (vi) `2, 2^(2), 2^(3), 2^(4), ...` i.e.,`2, 4, 8, 16, ...` Here, `t_(1)=2, t_(2)= 4, t_(3)= 8` and `t_(4)= 16` `t_(2)-t_(1)= 4-2=2` `t_(3)-t_(2)= 8-4= 4` `t_(4)-t_(3)= 16-8=8` Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP. (vii) `sqrt(3),sqrt(12),sqrt(27),sqrt(48), ...` i.e., `sqrt(3),2sqrt(3),3sqrt(3),4sqrt(3), ...` Here, `t_(1)=sqrt(3), t_(2)= 2sqrt(3), t_(3)= 3sqrt(3)` and `t_(4)= 4sqrt(3)` `t_(2)-t_(1)= 2sqrt(3)-sqrt(3)=sqrt(3)` `t_(3)-t_(2)= 3sqrt(3)-2sqrt(3)=sqrt(3)` `t_(4)-t_(3)= 4sqrt(3)-3sqrt(3)=sqrt(3)` Clearly, the difference of successive terms is same, therefore given list of numbers form an AP. |
|
| 38. |
A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find :(i) the production in the 1st year(ii) the production in the 10th year(iii) the total production in first 7 years |
|
Answer» `a_3=600` a+2d=600-(1) `a_7=700` a+6d=700-(2) subtracting equation 2 from 1 4d=100 d=25 putting this value in 1 a=550 `a_10`=a+9d =550+9(25) =775 `S_n=n/2(a+l)` =7/2(550+770) =4375 |
|
| 39. |
The famous mathematician associated with finding the sum of the first 100 natural numbers isA. PythagorasB. NewtonC. GaussD. Euclid |
|
Answer» Correct Answer - C Gauss is the famous mathematician associated with finding the sun of the first 100 natural number i.e., ` " " ` 1, 2, 3, ….., 100. |
|
| 40. |
The first four terms of an A.P. whose first term is -2 and the common difference is -2 areA. `-2,0,2,4`B. `-2,4,-8,16`C. `-2,-4,-6,-8`D. `-2,-4,-8,-16` |
|
Answer» Correct Answer - C Let the first four terms of an AP are a, a + d, a + 2d and a + 3d. Given, that first term, a = -2 and common difference, d = -2, then we have an AP as follows -2, -2 -2, -2 + 2(-2), -2 + 3(-2) = -2, -4, -6, -8 |
|
| 41. |
Show that the sequence defined by `a_n=2n^2+1`is not an A.P. |
|
Answer» `a_n=2n^2+1` `a_1=2+1=3` `a_2=2*4+1=9` `a_3=2*9+1=19` `a_3-a_2=19-9-10` `a_2-a_1=9-3=6` `a_3-a_2!=a_2-a_1`. |
|
| 42. |
Which term of the sequence `8-6i ,7-4i ,6-2i , `is (i) purely real (ii) purely imaginary? |
|
Answer» `a=8-6i` `d=7-4i-8+6i` `=-1+2i` `a_n=a+(n-1)d` `a+ib=8-6i+(n-1)(-1+2i)` `a+ib=8-6i+(-1)(n-1)+(n-1)2i` `-6+2(n-1)=0` `2(n-1)=6` `n=4` `a_n=8-6i+(4-1)(-1+2i)` `=8-6i-3+6i=5` `4^(th)` term=5 `8+(-1)(n-1)=0` `8=n-1` `n=9` `a_n=8-6i+(9-1)(-1+2i)` `=8-6i+8(-1+2i)` `=8-6i-8+16i=10i` Purely imaginary `9^(th)` term=10i. |
|
| 43. |
Kanika was given her pocket money on Jan 1st , 2008. She puts Rs. 1 on day 1, Rs. 2 on day 2, Rs. 3 on day 3 and continued doing so till the end of the month, from this money into her piggy bank she also spent Rs. 204 of her pocket money, and found that at the end of the month she still had Rs. 100 with her. How much was her pocket money for the month ? |
|
Answer» Let her pocket money be Rs. `x`. Now, she takes Rs. 1 on day 1, Rs. 2 on day 3 and so on till the end of the month, from this money. i.e., `1+2+3+4+ … +31` which form an AP in which terms are 31 and first term (a) = 1, common difference (d) `= 2-1=1` ` :. ` Sum of first 31 terms `=S_(31)` Sum of n terms, `S_(n)= (n)/(2)[2a+(n-1)d]` `:. S_(n)= (31)/(2)[2xx1+(31-1)xx1]` ` " " =(31)/(2)(2+30)=(31xx32)/(2)` ` " " =31xx16=496` So, Kanika takes Rs. 496 till the end of the month from this money. Also, she spent Rs. 204 of her pocket money and found that at the end of the month she still has Rs. 100 with her. Now, according to the condition, ` (x-496)-204=100` ` implies x-700=100` ` :. x=Rs. 800` Hence, Rs. 800 was her pocket money for the month. |
|
| 44. |
Two APs have the same common difference. The difference between their `100^(t h)`terms is 100, what is the difference between their `1000^(t h)`terms? |
|
Answer» `S_1:a_1,a_2.....d` `S_2:b_1,b_2.....l` `a_100-b_100=100` `a_100=a+(100-1)d` `=a+99d` `b_100=b+(100-1)l` `=b+99l` `a_100-b_100=100` `a-b=100` =`a_1000-b_1000` =`a+999d-b-999l` =`a-b`=100 |
|
| 45. |
Which term of the AP: 3, 15, 27, 39, . . . will be 132 more than its `54^(t h)`term? |
|
Answer» a=3 and d=12 let say `r^(th)` term is 132 more than `54^(th)` term. a+(r-1)d=a+53d+132 (r-1)*12=53*12+132 (r-1)=64 r=65 `65^(th)` term is 132 more than `54^(th)` term |
|
| 46. |
The `17^(th)` term of an AP exceeds its `10^(th)` term by 7. Find the common difference. |
|
Answer» According to Question `a_17=a_10+7` `a_17-a_10=7` (a+16d)-(a+9d)=7 a-a_16d-9d=7 7d=7 d=1 |
|
| 47. |
Find the common difference of an AP whose first term is 1 and the sum of the first four terms is one-third to the sum of the next four terms. |
|
Answer» `1,1+d,1+2d,1+3d,1+4d,....` `1+1+d+1+2d+1+3d=1/3(1+4d+1+5d+1+6d+1+7d)` `4+6d=1/3(4+22d)` `12+18d=4+22d` `d=2`. |
|
| 48. |
How many two–digit numbers are divisible by 3? |
|
Answer» 2 digit no divisible by 3 are `12,15,18,21.....,99` let them be `a_1, a_2 , a_3...... `and so on `a_2-a_1= 15-12=3` `a_3-a_2=18-15=3` so,`a=12 & d=3` and we also know last term of series `=99` now,`a_n= a+(n-1)d` `99=12+(n-1)3` `87=(n-1)3` `n=30` `99 `is the 30th term `:.` total terms are 30 answer |
|
| 49. |
Buses start from a bus terminal with a speed of 40 km/hr, at intervals of 5 minutes What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 4 minutes |
|
Answer» Here, distance travelled by the bus in `5` minutes will be equal to the sum of the distance covered by the bus and the man in `4` minutes. Let speed of the man is `x` km/h.Then, `40**5/60 = 40**4/60+x**4/60` `=>10/3 = 8/3+x/15` `=>2/3 = x/15` `=>x = 10` So, the speed of the man is `10` km/h. |
|
| 50. |
Find the sum of :(i) the first 1000 positive integers(ii) the first n positive integers |
|
Answer» 1)a=1 and d=1 `S_n=n/2(a+l)` `S_n=1000/2(1+1000)` `S_n=500(1001)` `S_n=500500` 2) a=1 and d=1 `S_n=n/2(a+n)` `S_n=n/2(1+n)` `S_n=n/2+n^2/2` |
|