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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.
| 151. |
(i)Which term of the A.P. `4, 3(5)/(7), 3(3)/(7), .....` is the first negative term? (ii) Which term of the progression `20, 19(1)/(2), 18(1)/(2), 17(3)/(4),....` is the first negative term? |
| Answer» Correct Answer - 28th | |
| 152. |
The 7th term of an AP is -4 and its 13th term is -16.Find the AP. |
| Answer» Correct Answer - 8, 6, 4, 2, 0… | |
| 153. |
Divide 32 into four parts which are in A.P. such that the ratio of the product of extremes to the product of means is 7:15. |
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Answer» Correct Answer - (2, 6, 10, 14) or (14, 10, 6, 2) Hint: Let these parts be (a-3d), (a-d), (a+d) and (a + 3d). |
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| 154. |
Find three numbers in A.P., whose sum is 15 and product is 80. |
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Answer» Correct Answer - (2, 5, 8) or (8, 5, 2) Hint: Let the numbers be (a-d), a, (a+d). |
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| 155. |
The internal angles of quadrilateral are in A.P. and their common difference is `10^@`. Find them. |
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Answer» Let the angles of the quadrilateral be `a,a+10^(@),a+20^(@),a+30^(@) " " `(`:.` common difference is `10^(@)`) `:. a+(a+10^(@))+(a+20^(@))+(a+30^(@))=360^(@)` `rArr 4a+60^(@)=360^(@)` `rArr 4a=300^(@)` `rArr a=75^(@)` `:. a+10^(@)=75^(@)+10^(@)=85^(@)` `a+20^(@)=75^(@)+20^(@)=95^(@)` `a+30^(@)=75^(@)+30^(@)=105^(@)` Hence, the angles are `75^(@), 85^(@), 95^(@), 105^(@)`. Alternative Method : Let the four angles of a quadrilateral are `a-3d,a-d,a+d and a+3d` `:.` Here common difference is 2d (remember) `:. (a-3d)+(a-d)+(a+d)+(a+3d)=360^(@)` `rArr 4a=360^(@) rArr a=90^(@)` common difference is given to be `10^(@)` `:. "i.e.," 2d=10^(@) rArr d=5^(@)` `:.` Four angles are `a-3d=90^(@)-3(5^(@))=75^(@)`, `a-d=90^(@)-5^(@)=85^(@)`, `a+d=90^(@)+5^(@)=95^(@)`, `a+3d=90^(@)+3(5^(@))=105^(@)`. |
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| 156. |
The angles of quadrilateral are in `AP` whose common difference is `10^@` then find the angles. |
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Answer» Correct Answer - `75^(@), 85^(@), 95^(@), 105^(@)` Hint: Let these angles be `x^(@), (x+ 10)^(@), (x+20)^(@)` and `(x+30)^(@)`. |
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| 157. |
The interior angles of a polygon are in AP. The smalllest angle is `52^(@)` and the common difference is `8^(@)`. Find the number of sides of the polygon. |
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Answer» Correct Answer - 3 Let the number of sides of the polygon be n . Then , Sum of its interior angles = ` ( 2n-9) x x 90^(@) = ( 180n -360)^(@)` ` 52+ 60+68 +…..tp n terms = ( 180n - 360) ` ` Rightarrow n/2 xx [2 xx 52 + (n-1) xxn 8] = 180n -360` `Rightaarrow n^(2) -33n +90 =0 Rightarrow (n-3) (n-30) =0 Rightarrow n =3 or n=30` When n=30 ,then last angle = ` [ 52+ 29xx 8]^(@) = 284^(@)` , which is a reflex angle , So, ` n ne 30` Hence , n=3 |
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| 158. |
The difference between any two consecutive interior angles of a polygon is `5o`. If the smallest angle is `120o`, find the number of the sides of the polygon. |
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Answer» Let the number of sides of the polygon be n . Then, the sum of its interior angle = ( 2n-4) right angles = ` {( n-2)xx 180}^(@)` Since the difference between any two consecutive interior angles of the polygon is constant, its angles are in AP. In this AP, we have a =12-0 and d=5 Now, ` S_(n) = ( n-2) xx 180` ` Rightarrow n/2. { 2 x 120 + ( n-1) xx 5} = (n -2) xx 180` ` Rightarrow ( 235n)/2 + (5n^(2))/2 = 180 n - 360 Rightarrow 5n^(2) =- 125n + 720=0` ` Rightarrow n^(2) -25n + 144=0 Rightarrow ( n-90 (n-16) =0` but , when n=16 , we have last angle =` { 120 + ( 16 -10 xx 5} ^(@) = 195 ^(@)` , which is not possible n=9 Hence, the number of sides of the given polygon =9 |
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| 159. |
A circle is completely divided into to n sectors in such a way that the angles of the sectors are inAP. If the smallest of these angles is `8^(@)`and the largest is `72^(@)`, calculate n and the angle in the fifth sector. |
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Answer» Correct Answer - n=9,5th angle = `40^(@)` Let the number of sectors be n. Then, ` n/2 (a+l) = 360 Rightarrow n(8 +72) = 720 Rightarrow n = 720/80 =9` Now, a=8` T_(n) = 72 and n=9` 8+8d = 72 ` Rightarrow 8d=64 Rightarrow d=8` So, the AP is 8,16,24,32,….., So,5th angle = (a+4d) ` 40^(@)` |
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| 160. |
The nth term of an AP is (3n +5). Find its common difference. |
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Answer» Correct Answer - 3 `T_(n) = (3n + 5) rArr T_(1) = (3 xx 1 +5) = 8, T_(2) = (3 xx 2 +5) = 11` `therefore d = (T_(2) - T_(1)) = 3.` |
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| 161. |
Which termof the arithmetic progression `5, 15 , 25 , dot`will be 130more than its `31 s t`term? |
| Answer» Correct Answer - 44th | |
| 162. |
The nth term of a progression is (3n + 5). Prove that this progression is an arithmetic progression. Also find its 6th term. (b) The nth term of a progression is (3 - 4n). Prove that this progression is an arithmetic progression. Also find its common difference. (c) The nth term of a progression is `(n^(2) - n + 1).` Prove that it is not an A.P. |
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Answer» Here, `a_(n)=3n+5` `rArr a_(n-1)=3(n-1)+5` `=3n-3+5=3n+2` Now, `a_(n)-a_(n-1)=(3n+5)-(3n+2)=3` Which does not depend on n i.e., it is constant. `:.` Given sequence is in A.P. `" "` Hence Proved. |
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| 163. |
Find the number of terms common to the two arithmetic progression 5,9,13,17,…,216 and 3,9,15,21,…..,321 |
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Answer» Correct Answer - 18 The first common term in two Aps is 9. LCm of their common difference = LCM (4,6) =12 New AP of common elements is 9,21,33,45,…., Let is contain n terms. Then, ` T_(n) gt 216 Rightarrow 9 + ( n-1) xx 12 le 216 Rightarrow (n-1) xx 12 ge 207` ` Rightarrow (n-1) xx 12 = 204 Rightarrow n =18` So, there are 18 common terms in two APs. |
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| 164. |
How many terms of the AP 21, 18, 15,… must be added to get the sum 0? |
| Answer» Correct Answer - 15 | |
| 165. |
The sum of first 6 terms of an arithmetic progression is 42. The ratio of its 10th term to its 30th term is `1:3`. Calculate the first and 13th term of an AP. |
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Answer» Let a be the first term and d the common difference of the given AP. Then, `T_(10) = a + 9d "and" T_(30) = a +29d` `therefore (T_(10))/(T_(30)) = (1)/(3) rArr (a+9d)/(a +29d) = (1)/(3)` `rArr 3a + 27d = a +29d` `rArr 2a = 2d rArr a =d.` `"Also,"S_(n) = (n)/(2) [2a + (n-1)d]` `rArr S_(6) = (6)/(2)(2a +5d)` ` = (6a + 15d) = (6a + 15a) " " [because d =a]` = 21a But, `S_(6) = 42` (given) `therefore 21a = 42 rArr a = 2.` Thus, a = 2 and d = 2. `therefore "13th term," T_(13) = (a+12d) = (2 +12 xx 2) = 26` Hence, the first term is 2 and the 13th term is 26. |
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| 166. |
Which term of the AP 21, 18, 15, … is -81? |
| Answer» Correct Answer - 35th | |
| 167. |
For an A.P. if `t_4` = 12 and d = -10 then find a. A)-18 B)42 C)-5 D)21A. -18B. 42C. -5D. 21 |
| Answer» Correct Answer - B | |
| 168. |
If 1+4+7+10+…+x = 287, find the value of x. |
| Answer» Correct Answer - x = 40 | |
| 169. |
(i) The sum of the first n terms of an AP is `((5n^(2))/(2) + (3n)/(2))`. Find the nth term and the 20th term of this AP. (ii) The sum of the first n terms of an AP is `((3n^(2))/(2) + (5n)/(2)).` Find its nth term and the 25th term. |
| Answer» Correct Answer - `(i) T_(n) = (5n-1), T_(20) = 99 (ii) T_(n) = (3n +1), T_(25) = 76` | |
| 170. |
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x (Hint: `S_[x-1]=S_[49]-S_x`) |
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Answer» We are given an AP, namely 1, 2, 3,…… (x-1), x, (x +1),…, 49 Such that 1+2+3+…+ (x-1) = (x+1) + (x+2) +….+49. Thus, we have `S_(x-1) = S_(49)-S_(x) " "…(i)` Using the formula, `S_(n) = (n)/(2) (a+1)` in (i), we have `((x-1))/(2) * {1+(x-1)} = (49)/(2) * (1+49) - (x)/(2) * (1+x)` `rArr (x(x-1))/(2) + (x (x +1))/(2) = 1225` `rArr 2x^(2) = 2450 rArr x^(2) = 1225 rArr x = sqrt(1225) = 35` Hence, x = 35. |
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| 171. |
(i) Find the 37th term of the AP 6, `7(3)/(4), 9(1)/(2), 11(1)/(4)`,…. (ii) Find the 25th term of the AP `5, 4(1)/(2), 4, 3(1)/(2), 3,…` |
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Answer» Correct Answer - (i) 69 (ii) -7 `(i) a = 6, d = ((31)/(4) - 6) = (7)/(4)` `(ii) a = 5, d = ((9)/(2) -5) = (-1)/(2)` |
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| 172. |
Find the 25th term of the A.P. 6, 10, 14, ... |
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Answer» Here, a = 6, d = 10 - 6 = 14 - 10 = 4, n = 25 `:. a_(n)=a+(n-1)d` `rArr a_(25)=6+(25-1)xx4=6+96=102` `:.` 25th term of the A.P. is 102. |
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| 173. |
Find the 25th term of the AP `-5, (-5)/(2), 0, (5)/(2),….` |
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Answer» The given AP is `-5, (-5)/(2), 0, (5)/(2), ….` Its first term = -5 and common difference `= ((5)/(2) - 0) = (5)/(2)`. ` therefore` a= -5 and d = `(5)/(2)`. `rArr T_(25) = a + (25-1) d` ` =a +24d = (-5) + (24 xx (5)/(2)) = -5 + 60 = 55.` Hence, 25th term = 55. |
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| 174. |
Show that the progression -11,-7,-3,1,5,…, 161 is an AP. How many tems does it have ? |
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Answer» we have (-7) - (-11) = (-7 +11) = 4, (-3) - (-7) = ( -3+7) =4 1-(-3) = ( 1+3) =4 and ( 5-1) =4 { (-7) -(-11)}= { (-3) - (-7)} = {1-(-3)} = ( 5-1) =4, which is contant . So, the given progression is an AP in which a = -11 and d =4 let it have n terms. then , ` T_(n)= 161 Rightarrow a+ ( n-1) d = 161 ` ` Rightarrow (-11) + ( n-1) xx 4 = 161` ` Rightarrow 4n = 176 Rightarrow n= 44 ` ` T_(44) =161` and hence there are 44 terms in the given AP . |
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| 175. |
The next term of the AP `sqrt(7), sqrt(28), sqrt(63)`,… isA. `sqrt(70)`B. `sqrt(84)`C. `sqrt(98)`D. `sqrt(112)` |
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Answer» Correct Answer - D Given terms are `sqrt(7), sqrt(4 xx 7), sqrt(9 xx 7),…., i.e., sqrt(7), 2sqrt(7), 2sqrt(7), 3sqrt(7),`…. So, the next term is `4sqrt(7) = sqrt(4 xx 4 xx 7) = sqrt(112)` |
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| 176. |
The firstand the last terms of an AP are 7 and 49 respectively. If sum of all itsterms is 420, find its common difference. |
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Answer» Let the given AP contain n terms. Here a = 7, l = 49 and `S_(n) = 420.` `therefore S_(n) = (n)/(2) (a+l)` `therefore (n)/(2)(7+49) = 420 rArr (n)/(2) xx 56 = 420` `rArr 28n = 420 rArr n = (420)/(28) = 15.` Thus, the given AP contains 15 terms and `T_(15) = 49.` Let d be the common difference of the given AP. Then, `T_(15) = 49 rArr a + 14d = 49` `rArr 7 + 14d = 49` `rArr 14d = 42 rArr d = 3.` Hence, the common difference of the given AP is 3. |
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| 177. |
The 14th term of an A.P. is twice its 8th term. If its 6th term is -8, then find the sum of itsfirst 20 terms. |
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Answer» Let a be the first term and d be the common difference of the given AP. Then, `T_(14) = 2 xx T_(8) rArr a +13d = 2(a +7d)` `rArr a +d = 0 " "…(i)` `"Also,"T_(6) = -8 rArr a +5d =-8. " "...(ii)` On solving (i) and (ii), we get a = 2 and d = -2. The sum of first 20 terms is given by `S_(20) = (n)/(2) *[2a + (n-1)d,]"where" n = 20` ` = ((20)/(2)) xx {(2 xx 2 +19 xx (-2)}` `= 10 xx (4-38) = 10 xx (-34) = -340.` Hence, the required sum is -340. |
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| 178. |
Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find is 25th term. |
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Answer» Here a = 10 and let d be the common difference. Then, `S_(14) = 1505 rArr (n)/(2) [2a + (n-1)d] = 1505, "where"n = 14 "and" a = 10` `rArr (14)/(2) * (20 + 13d) = 1505 rArr (20 + 13d) = (1505)/(7) = 215` `rArr 13d = 195 rArr d = 15.` Thus,a = 10 and d = 15. `therefore T_(25) = (a+24d) = (10+24 xx 15) = 370.` Hence, the 25th term is 370. |
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| 179. |
The sum of the first `7` terms of an `AP` is `63` and the sum of its next `7 ` terms is `161.` Find the `28th` term of this `AP.` |
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Answer» Let a be the first term and d be the common difference of the given AP. Then, using `S_(n) = (n)/(2) * [2a + (n-1)d]`, we get `S_(7) = (7)/(2) (2a+6d) rArr 7(a+ 3d) = 63 " " [because S_(7) = 63]` `rArr a + 3d = 9. " "...(i)` Clearly, the sum of first 14 terms = 63 + 161 = 224. `therefore S_(14) = 224 rArr (14)/(2) (2a +13d) = 224` `rArr 7(2a +13d) = 224` `rArr 2a + 13d = 32. " "...(ii)` Multiplying (i) by 2 and subtracting the result from (ii), we get `7d = 14 rArr d = 2.` Putting d = 2 in (i), we get a = 9 -6=3. Thus, a = 3 and d =2. `therefore` 28th term of this AP is given by `T_(28) = (a+27d) = (3 + 27 xx 2) = 57.` Hence, the 28 th term of the given AP is 57. |
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| 180. |
In an AP, the sum of first ten terms is -150 and the sum of its next ten terms is -550 Find the AP |
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Answer» Correct Answer - 3, -1, -5, -9,… `S_(10) = -150 "and" S_(20) = (-150) + (-550) = -700` |
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| 181. |
A child puts one five-rupee coin of her saving in the piggy bank on the first day. She increases her saving by one five-rupee coin daily. If the piggy bank can hold 190 coins of five-rupees in all, find the number of days she can continue to put the five-rupee coins into it and find the total money she saved. Write your views on the habit of saving. |
| Answer» Correct Answer - 19, ₹ 950 | |
| 182. |
The first term of an A.P. is `a`and the sum of first `p`terms is zero, show tht the sum of its next `q`terms is `(a(p+q)q)/(p-1)dot` |
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Answer» Let the common difference of the given AP be d. then , ` S_(p) = 0 Rightarrow p/2 xx [ 2a + + ( p-1) d] =0` ` Rightarrow 2a + ( p-1) d=0 Rigtharrow = ( -2a)/( (p-1))` ` S_(p+q) + (( p+q))/2 .[ 2a + ( p+q-1) d]` `= (( p+q))/ 2[ ( 2a + ( p+q-1)xx ((-2a))/((p -1))]` ` =a ( p+q) .[ 1-((p+q -1))/((p-1))] = (-a(=+q)q)/((p-1))` Hence, the requird sum is ` (-a (p+q)q) / (( p-1))` |
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| 183. |
The sum of `n`terms of two arithmetic progressions are in the ratio `(3n+8):(7n+15)dot`Find the ratio of their 12th terms. |
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Answer» Let ` a_(1),a_(2) and d_(1) , d_(2)` be the first terms and common differences of the first and second AP respectively. Then, by the given condition. we have ` (" sum to n terms of first AP")/( " sum to n terms of second AP")= ( 3n +8)/( 7n +15) ` ` Rightarrow (n/2 .[2a_(1) + ( n-1) d_(1)])/ (n/2. [2a_(2) + (n-1) d_(2)]) = ( 3n +8)/( 7n +15)` ` Rightarrow ( 2a_(1) + (n -1) d_(1))/( 2a_(2) + (n-1) d_(2) )= ( 3n+8)/( 7n + 15) ` Now, ( " 12 th term of first AP")/( " 12th term of second AP") = ( a_(1) +11d_(1))/ (a_(2) +11d_(1))` ` (2a_(1) +22d_(1))/ (2a_(2) +22d_(2))= ( 2a_(1) + (23-1)d_(1))/(2a_(2) + (23-1)d_(2))` ` (3xx 23+8)/ 7 xx 23 +15) ` [ putting n=23in (i)] ` 77/ 176 = 7/16` Hence, the ratio of the 12th terms of given APs is 7:16 |
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| 184. |
How manyterms of the A.P. 63, 60, 57, .... must be taken so that their sum is 693? |
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Answer» Correct Answer - n = 21 or n = 22, 22nd term is 0 `n^(2) -43n + 462 = 0 rArr n^(2) - 21n - 22n + 462 = 0` |
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| 185. |
How manyterms of the A.P. `9, 17 , 25 , dot`must betaken so that their sum is 636? |
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Answer» Correct Answer - 12 `4n^(2) + 5n -636 = 0 rArr 4n^(2) + 53n - 48n - 636 =0` |
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| 186. |
Find the value of x if `2+4+6+...+x=650` |
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Answer» Here, a=2, `d=4-2=6-4=2` Let `S_(n)=650` `rArr (n)/(2)[(2(2)+(n-1)2]=650 rArr n(n+1)=650` `rArr n^(2) +n-650=0 rArr (n+26)(n-25)=0` `rArr n=-26 or n=25` n=-26 is not possible as number of terms cannot be negative. `:. n=25` Now, `x=a_(25)=2+24(2)=50` |
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| 187. |
Solve for x : `5+13+21+......+x=2139` |
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Answer» Here, `a=5, d=13-5=8` Let `T_(n)=x` `:. S_(n)=2139` `:. (n)/(2) [2xx5+(n-1)8]=2139` `rArr n(5+4n-4)=2139 rArr 4n^(2)+n-2139=0` `rArr 4n^(2)-92n+93n-2139=0 rArr (n-23)(4n+93)=0` `:. n=23 or n=-(93)/(4)` since `n gt 0`, we have n=23 `:. x = T_(n)=a+(n-1)d=5+(23-1)(8)` `rArr x=181` |
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| 188. |
What is 20th term from the end of the AP 3, 8, 13, …, 253?A. 163B. 158C. 153D. 148 |
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Answer» Correct Answer - B 20th term from the end ` = (253-19 xx 5) = (253-95) = 158. [l - (n-1)d]` |
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| 189. |
`(5+13+21+…+181)=?`A. 2476B. 2337C. 2219D. 2139 |
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Answer» Correct Answer - D Let the number of terms be n. Then, `T_(n) = 181` `therefore 5 + (n-1) xx 8 = 181 rArr (n-1) xx 8 = 176 rArr (n-1) = 22 rArr n = 23.` `therefore "sum" = (n)/(2)(a+l) = (23)/(2)(5+181) = 23 xx 93 = 2139` |
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| 190. |
The sum of first 16 terms of the AP 10, 6, 2 … isA. 320B. `-320`C. `-352`D. `-400` |
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Answer» Correct Answer - B `l = T_(16) = 10 + 15 xx (-4) = -50` `therefore "sum" = (n)/(2)(a+l) = (16)/(2) (10-50) = 8 xx (-40) = -320.` |
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| 191. |
How many terms of the AP 3, 7, 11, 15,…. Will make the sum 406?A. 10B. 12C. 14D. 20 |
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Answer» Correct Answer - C Let `S_(n) = 406. "Then," (n)/(2)[2 xx 3 + (n-1) xx 4] = 406` `rArr (n)/(2)(6+4n-4) = 406 rArr (n)/(2)(4n +2) = 406` `rArr n(2n+1) = 406 rArr 2n^(2) + n - 406 = 0` `therefore n = (-1+-sqrt(1+3248))/(4) = (-1+sqrt(3249))/(4)` [neglecting negative value] `rArr n = (-1+57)/(4) = (56)/(4) = 14` |
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| 192. |
How many terms are there in AP 7, 11, 15, …., 139? |
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Answer» In the given AP, we have a = 7 and d = (11-7) = 4. Suppose there are n terms in the given AP. Then, `T_(n) = 139 rArr a + (n-1)d = 139` `rArr 7 + (n-1) xx 4 = 139` `rArr 4n = 136 rArr n = 34` Hence, there are 34 terms in the given AP. |
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| 193. |
How manythree-digit numbers are divisible by 7? |
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Answer» All 3-digit numbers divisible by 7 are 105, 112, 119,…., 994. Clearly, these numbers form an AP with a = 105, d = (112-105) = 7 and l = 994. Let it contain n terms. Then, `T_(n) = 994 rArr a + (n-1) xx d = 994` `rArr 105 + (n-1) xx 7 = 994` `rArr 98 + 7n = 994 rArr 7n = 896 rArr n = 128.` Hence, there are 128 three-digit numbers divisible by 7.` |
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| 194. |
Find the `11^(th)`from the last term (towards the first term) of the AP : `10 , 7, 4, ... , 62.` |
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Answer» We have `a = 10, d = (7-10) = -3, l = -62 and n = 11`. `therefore` 11th term from the end `= {l - (n-1) xx d}` ` = {-62 -(11-1) xx (-3)}` `= (-62 + 30) = -32.` Hence, the 11th term from the end of the given AP is -32. |
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| 195. |
If mth term of an AP is 1/n and its nth term is 1/m, then show that its (mn)th term is 1 |
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Answer» Let a be the first term and d be the common difference of the given AP. Then, `T_(m) = a + (m-1)d "and" T_(n) = a + (n-1)d.` Now, `T_(m) = (1)/(n) "and" T_(n) = (1)/(m)` (given). `therefore a + (m -1)d = (1)/(n) " "...(i)` `"and" a+ (n-1)d = (1)/(m) " "... (ii)` On subtracting (ii) form (i), we get `(m-n)d = ((1)/(n) - (1)/(m)) = ((m-n)/(mn)) rArr d = (1)/(mn).` Putting `d = (1)/(mn)` in (i), we get `a + ((m-1))/(mn) = (1)/(n) rArr a = {(1)/(n) - ((m-1))/(mn)} = (1)/(mn).` Thus, `a = (1)/(mn) "and" d = (1)/(mn).` `therefore (mn)"th term" = a+ (mn-1)d` ` = {(1)/(mn) + ((mn-1))/(mn)} [because a = (1)/(mn)]` = 1. Hence, the (mn)th term of the given AP is 1. |
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| 196. |
If m times the mth term of an AP is equal to n times its nth term, then show that (m + n)th term of an AP is zero. |
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Answer» Let a be the first term d be the common difference of the given AP. Then, `T_(m) = a + (m-1)d "and" T_(n) = a + (n-1)d.` Now, `(m * T_(m)) = (n * T_(n)) rArr m * {a+ (m-1)d} = n * {a +(n-1)d}` `rArr a * (m-n) + {(m^(2) -n^(2)) - (m-n)} * d = 0` `rArr (m-n) * {a + (m +n -1)}d.` `rArr (m-n) * T_(m+n) = 0` `rArr T_(m+n) = 0 [because (m-n) ne 0].` Hence, the (m+n)th term is zero. |
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| 197. |
The sum of the 4th and 8th terms of an AP is 24 and the sum of its 6th and 10th terms is 44. Find the first terms of the AP. |
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Answer» Let a be the first term and d be the common difference of the given AP. Then, `T_(4) + T_(8) = 24 rArr (a+ 3d) + (a+7d) = 24` `rArr 2a+ 10d = 24` ` rArr a + 5d = 12 " "…(i)` and `T_(6) +T_(10) = 44 rArr (a + 5d) + (a+ 9d) = 44` `rArr 2a +14d = 44` `rArr a + 7d = 22. " "....(ii)` On solving (i) and (ii), we get a = -13 and d = 5. `therefore` first three terms of the given AP are -13, -8 and -3. |
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| 198. |
Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40. |
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Answer» Correct Answer - 3, 5, 7, 9, 11,… `T_(4) = 9 rArr a +3d = 9` `rArr T_(6) + T_(13) = 40 rArr (a+5d) +(a+12d) = 40 rArr 2a +17d = 40` From (i) and (ii), we get a= 3 and d = 2. `therefore AP is 3, 5, 7, 9, 11,… |
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| 199. |
The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16 then the sum of its first ten terms isA. 150B. 175C. 160D. 135 |
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Answer» Correct Answer - B `T_(13) = 4 xx T_(3) rArr a+12d = 4(a+2d) rArr 3a-4d = 0 " "…(i)` `T_(5) = 16 rArr a +4d = 16 " ".. (ii)` On solving (i) and (ii), we get a = 4 and d= 3 `S_(10) = (10)/(2)(2a +9d) = 5(2 xx 4 +9 xx 3) = 175` |
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| 200. |
If a = 6, d= 3 then `S_10` = …… A)192 B)195 C)198 D)201A. 192B. 195C. 198D. 201 |
| Answer» Correct Answer - B | |