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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.
| 101. |
If (2p-1), 7, 3p are in AP, find the value of p. |
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Answer» Correct Answer - p =3 We have, 7-(2p-1) = 3p-7 `therefore (2p-1) +3p = 14 rArr 5p = 15 rArr p = 3` |
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| 102. |
Find the value of p for which the numbers 2p-1, 3p+1, 11 are in AP. Hence, find the numbers. |
| Answer» Correct Answer - p = 2; 3, 7, 11 | |
| 103. |
If (2p +1), 13, (5p-3) are in AP, find the value of p. |
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Answer» Correct Answer - p = 4 We have, 13-(2p+1) = (5p -3) - 13. `therefore (5p -3) + (2p +1) = 26 rArr 7p = 28 rArr p = 4.` |
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| 104. |
In a given A.P.. If the `p`th term is `q` and `q`th term is `p` then show that `n` th term is `p+q-n` |
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Answer» Let a be the first term and d be the common difference of the given AP. Then, `T_(p) = a + (p-1)d "and" T_(q) = a + (q-1)d.` Now, `T_(p) = q "and" T_(q) = p` (given). `therefore a + (p-1)d =q " "...(i)` `"and" a+ (q-1)d = p " "...(ii)` On subtracting (i) from (ii), we get `(q-p)d = (p-q) rArr d = -1.` Putting d = -1 in (i), we get a = (p+q-1). Thus, a = (p+q-1) and d = -1. `therefore "nth term" = a + (n-1)d = (p+q-1) + (n-1) xx (-1)` = p+q - n. Hence, nth term = (p+q -n). |
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| 105. |
If pth, qth and rth terms of an A.P. are a, b, c respectively, then show that (i) a(q-r)+b(r-p)+c(p-q)=0 |
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Answer» Let x be the first term and d be the common difference of the given AP. Then, `T_(p) = x + (p-1)d, T_(q) = x +(q-1)d "and" T_(r) = x + (r-1)d.` `therefore x + (p-1)d =a " "…(i)` `x + (p-1)d = b " "...(ii)` `x +(r-1)d = c " "... (iii)` On multiplying (i) (q-r), (ii) by (r-p) and (iii) by (p-q), and adding, we get `a(q-r) +b (r-p) +c(p-q)` ` = x * {(q-r) + (r-p) + (p-q)} + d * {(p-1) (q-r) + (q-1) (r-p) + (r-1) (p-q)}` ` = (x xx 0) + (d xx 0) = 0.` Hence, a(q-r) +b(r-p) +c (p-q) = 0. |
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| 106. |
Find the arithmetic mean between (i) 14 and -6, (ii) (a-b) and (a+b) |
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Answer» (i) Arithmetic mean between 14 and -6 ` ( 14 + (-6))/ 2 = 8/2 =4 ` (ii) Arithmetic mean between 9a-b) and ( a+b) ` ((a -b) + (a+b))/ 2 = (2a)/ 2=a ` |
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| 107. |
If `(a^n+b^n)/(a^(n-1)+b^(n-1))` is the GM between a and b, then the value of n is |
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Answer» We know that the AM between a and b is ` ((a +b))/2` `((a^(n) + b^(n))/( a^(n-1) +b^(n-1)))` is the AM between a and b ` Rightarrow ((a ^(n) +b^(n))/ (a^(n-1) +b^(n -1)) = (a+b)/2` ` Rightarrow 2a^(n) +2b^(n) = a^(n) +a^(n-1) b+b^(n-1) a+b^(n)` ` Rightarrow a^(n) +b^(n) a^(n-1) b+b^(n-1) a` ` Rightarrow a^(n) -a^(n-1) b=b^(n-1) a-b^(n) ` ` Rightarrow a^(n-1) (a-b) =b^(n-1) (a-b)` ` Rightarrow a^(n-1) =b^(n-1)` ` Rightarrow (a/b) ^(n-1) =1=(a/b)^(0) Rightarrow n-1 =0 Rightarrow n=1` Hence, the required value of n is 1. |
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| 108. |
What is the common difference (d) of the A.P. 410, 360, 310….. |
| Answer» `d = 360- 410 = 310 - 360 = - 50` | |
| 109. |
Insert six A.M.s between 15 and -13. |
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Answer» Let ` A_(1) , A_(2),A_(3),A_(4),A_(5),A_(6)` be the six arithmetifc means between 15 and -13. Then , ` 15 , A_(1),A_(2),A_(3),A_(4),A_(5),A_(6),-13` are in AP. Now , ` d = (( -13 -15))/ (( 6+1)) = (-28)/7 = -4 " " [ because d= ( (b-a))/((n+1)) = ((-13-15))/(( 6+1))]` ` A_(1)= (15+d) = ( 15-4) = 11 , A_(2) = (15+2d) = ( 15-8) =7`, ` A_(3)= (15+3d) = ( 15-12) =3, A_(4) = ( 15 + 4d) = (15 -16) =-1` ` A_(5) = ( 15 +5d) = ( 15-20) = -5, A_(6) = ( 15+6d0 = ( 15-24) =-9` Hence, the required six AMs between 15 and -13 are 11,7,2,-5 and -9 |
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| 110. |
If the sixth term of an AP is zero then show that its 33rd term is three times its 15th term. |
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Answer» a +5d =0 `rArr` a = -5d. 33th term = a +32d = -5d +32d = 27d 3(15th term) = 3(a+14d) = 3(-5d+14d) = `3 xx9d` = 27d |
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| 111. |
For an A.P. if a= 100, d = 100, what is `t_(10)` ? |
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Answer» `t_(n) = a+ ( n -1) d ` `:. t_(10) = 10 + ( 10-1) 100 = 100+ 9 xx 100 = 1000`. `t_(10) = 1000`. |
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| 112. |
What is the sum of first six terms of an A.P. whose first term is 3 and the sixth term is 27. |
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Answer» `S_(6)= ( 6)/(2) ( t_(1) + t_(6)) = ( 6)/(2) ( 3+ 27) = 3 xx 30 = 90` The sum is 90 |
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| 113. |
Find the common difference ( d) for an A.P., if `t_(3) = 8 ` and `t_(4) = 12`.A.B.C.D. |
| Answer» `t_(4) - t_(3) = d .` `:. 12 -8= d ` `:. D= 4 ` | |
| 114. |
The first term of an A.P. is 2 and the nth term is 41. What is the value of n, if `S_(n) = 860` ?A. 30B. 31C. 40D. 41 |
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Answer» Correct Answer - C `[ S_(n ) = ( n )/(2) ( a+l)]` |
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| 115. |
How many terms of the A.P. -6, `-(11)/2, -5ddot`are needed to give the sum -? |
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Answer» Let the required number of terms be n. Then , `a=-6,d= {(-11)/2-(-6)}=((-11)/2+6)=1/2and S_(n)= -25` `S_(n)= -25 Rightarrown/2,.[2a+ (n-1)d]=-25` `Rightarrow n/2.[ 2xx (-6)+(n-1).1/2]= -25` `Rightarrown/2.( n/2-25/2) =-25 Rightarrown/2.((n-25))/2 = -25` `Rightarrow n^(2) -25n+100=0 Rightarrow (n-5) (n-20) =0` ` Rightarrow n=5 or n=20` This shows that the sum of first five terms is -25 and that the sum of frist 20 terms is also -25. It means that the sum of all the terms from 6th to 20th is zero. |
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| 116. |
Write `t_(2)` and `t_(3)` for the A.P., if (i) a=11,d=2 (ii) a= -7, d= - 5 |
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Answer» (i) `t_(2) = t_(1) + d = 11+2 = 13 ` `t_(3) = t_(2) + d = 13+2 = 15 ` (ii) `t_(2) = t_(2) + d = - 7+ ( -5)` `= -7 -5 = -12` `t_(3) = t_(2) + d= -12 + ( - 5)` `= - 12 -5= - 17` |
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| 117. |
Write the second term of the sequence `t_(n) = 2n + 1`. |
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Answer» For finding second term,put n = 2. `t_(2) = 2(2) + 1 = 4+1 = 5. ` |
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| 118. |
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)(c-2a))/(2(b-a))`. |
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Answer» Let d be the common difference of the given AP and let it contain n terms. Then d=(b -a) Now, ` T_(n) = c Rightarrow a + (n-1) d=c` ` Rightarrow a+ (n-1) (b-a) =c` ` Rightarrow (n-1) = ((c-a))/((b-a))` `Rightarrown= { ((c-a))/((b-a))+1}=((b+c-2a))/((b-a))` sum of the given `AP= n/2 (a+l)` , where l is the last term `((b+c-2a))/(2(b-a))xx (a+c)` ` [ because n= ((b+c-2a))/(2(b-a))andl=c]`j `((a+c) (b+c-2a))/(2(b-a))` |
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| 119. |
Find the sum of 23 terms of the AP 5,9,13,17,…… |
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Answer» Here,a=5,d= (9-5) =4 and n=23 Now,s ` S_(n) = n/2 xx { 2a + ( n-1) xx d}` ` S_(23) = 23/2 xx { 2 xx 5 + ( 23 -1) xx 4}` ` = ( 23/2 xx 98) = 1127` Hence, the sum of 23 terms of the given AP is 1127. |
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| 120. |
Write the next two terms of the following sequences `:` (i) 3,5,8,12,17 (ii) -25, -23, - 21, -19 |
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Answer» (i) 23, 30 [ The number go on increasing by 2,3,4,5,….] (ii) -17,-15 [ The common difference d= 2 ] |
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| 121. |
क्या संख्याओं की सूची 5, 11, 17, 23,.. का कोई पद 301 है? क्यों ? |
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Answer» Correct Answer - 301 is not a term of the given A.P. Here, `a = t_(1) = 5, t_(2) = 11,` ….. `d = t_(2) - t_(1) = 11-5 = 6` Let 301 be the nth term of the given A.P. `t_(n) = a+ ( n-1) d ` …(Formula) `:. 301 = 5+ ( n -1) xx6` `= 5+ 6n -6` `:. 301 = 6n -1 ` `:. 6n = 301 + 1 = 302 ` `:. n = ( 302)/(6)` `:. n = ( 151)/93)`. But this is not an integer. `:. 301` cannot be a term of the given A.P. |
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| 122. |
How many terms of the A.P.: 9,17,25,..... must be taken to give a sum of 636? |
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Answer» 9, 17, 25, .... Here, a=9, `d=17-9=25-17=8` Let `S_(n)=636` `rArr (n)/(2)[2(9)+(n-1)8]=636` `rArr (n)/(2)(18+8n-8)=636` `rArr n(4n+5)=636` `rArr 4n^(2)+5n-636=0` `rArr (4n+53)(n-12)=0` `rArr n=-(53)/(4) or n=12` `n=-(53)/(4)` is not possible `:. n=12` |
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| 123. |
Find how many two-digit numbers are divisible by 6. |
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Answer» Correct Answer - 15 The given numbers are 12, 18, 24, 30,…,96. Let their number be n. Then, `12 +(n-1) xx 6 = 96`. Find n. |
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| 124. |
The first and the last terms of an A.P. are `a a n d l`respectively. Show that the sum of nth term from the beginning and nthterm from the end is `a+ldot` |
| Answer» Required sum` = {a + (n-1)d} + {l -(n-1)d} = (a+l)` | |
| 125. |
If `9t h`term of anA.P. is zero, prove that its `29 t h`term isdouble the `19 t h`term. |
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Answer» `T_(9)=0Rightarrow a+8d Rightarrow a=-8d ` ` T_(19)=a+8d =-8d+18d =10d and T_(29) =a+28d =-8d +28d =-8d +28d =20d` `T_(29) = 2xx T_(19)` |
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| 126. |
If the sum of nterms of an A.P. in `n P" "+1/2n(n-1)Q ,`where P and Q are constants, find the commondifference. |
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Answer» Correct Answer - Q `S_(n)=nP+1/2n(n-1) Qrightarrow S_(n-1) = (n-1) P+1/2(n-1) (n-2) Q` ` T_(n) = ( S_(n)-S_(n-1)) = P ( n-1) Q Rightarrow T_(n-1) =P +(n-2)Q` ` d= (T_(n) -T_(n-1)=Q` |
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| 127. |
Find the sum of all natural numbers from 1 and 100 which are divisble by 4 or 5. |
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Answer» Correct Answer - 2050 Requrired numbers = ( numbers divisible by 4)+(numbers divisible by 5) - (numbers divisible by 20) = ( 4+8+12 +…..+ 100) +( 5+10+15+…+100) - (20 +40 +60 +….+ 100) ` 25/2 xx (4 +100) + 20/2 xx( 5 +100) -5/2 xx ( 20 +100)` ` 25/2xx(4 +100) +20/2 xx (5+100) -5/2 xx (20 +100)` (1300 +1050 -300) = 2050 |
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| 128. |
Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to `((a+c)(b+c-2a))/(2(b-a))`. |
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Answer» Le the number of terms in A.P. =n common difference d=b-a c=nth term of the A.P. `rArr c=a+(n-1)(b-a)` `rArr c-a(n-1)(b-a)` `rArr n-1=(c-a)/(b-a)` `rArr n=(c-a)/(b-a)+1=(c-a+b-a)/(b-a)=(b+c-2a)/(b-a)` `:.` Sum of n term of A.P. `=(n)/(2)(a+l)` `=((b+c-2a))/(2(b-a))(a+c)=((a+c)(b+c-2a))/(2(b-a))` |
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| 129. |
Sum of the first p, q and r terms of an A.P are a, b and c, respectively.Prove that `a/p(q-r)+b/q(r-p)+c/r(p-q)=0` |
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Answer» Let the first term and common difference of A.P. be x and y respectively. `:. S_(p)=a` `rArr (p)/(2)[2x+(p-1)y]=a` `rArr (a)/(p)=x+(p-1)(y)/(2)` `rArr (a)/(p)(q-r)=x(q-r)+(p-1)(q-r)(y)/(2) " " ...(1)` Similarly, `S_(q)=b` `rArr (q)/(2)[2x+(q-1)y]=b` `rArr (b)/(q)=x+(q-1)(y)/(2)` `rArr (b)/(q)(r-p)=x(r-p)+(q-1)(r-p)(y)/(2) " " ...(2)` and `S_(r)=c` `rArr (r)/(2)[2x+(r-1)y]=c` `rArr (c)/(r)=x+(r-1)(y)/(2)` `rArr (c)/(r)(p-q) = x(p-q)+(r-1)(p-q)(y)/(2) " " ...(3)` Adding eqs. (1), (2) and (3), we get `(a)/(p)(q-1)+(b)/(q)(r-p)+(c)/(r)(p-q)=x{(q-r)+(r-p)+(p-q)}+(y)/(2){(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}` `= 0 +(y)/(2){pq-pr-q+r+qr-pq-r+p+rp-rq-p+q}` `=0 " "` Hence Proved. |
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| 130. |
Find the sum of all multiples of 9 lying between 300 and 700. |
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Answer» Correct Answer - 21978 Required sum = (306 +315 +324 +…+693). |
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| 131. |
Find thesum of all 3 digit natural numbers, which are multiples of 11. |
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Answer» All 3-digit natural numbers, which are multiples of 11 are given as 110, 121, 132,…., 990. This is an AP is which a = 110, d = 11 and l = 990. Let the given AP contain n terms. Then, `T_(n) = 990 rArr a + (n-1)d = 990 rArr 110 + (n-1) xx 11 = 990` `rArr 99 + 11n = 990 rArr 11n = 891 rArr n = 81.` `therefore "required sum" = (n)/(2) (a+l) = (81)/(2) xx (110 + 990)` ` = ((81)/(2) xx 1100) = 44550.` Hence, the required sum is 44550. |
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| 132. |
How manythree digit numbers are divisible by 7? |
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Answer» Correct Answer - 128 These numbers are 105, 112, 119, 126, .., 994. Let `T_(n) = 994.` Then, 105 + (n-1) `xx` 7 = 994 `rArr` n = 128. |
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| 133. |
How many three-digit numbers are divisible by 9? |
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Answer» Correct Answer - 100 These numbers are 108, 117, 126, …, 999. `"Let" T_(n) = 999. "Then," 108 + (n-1) xx 9 = 999 rArr n = 100` |
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| 134. |
The 5th term of an AP is -3 and its common difference is -4. The sum of its first 10 terms isA. 50B. `-50`C. 30D. `-30` |
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Answer» Correct Answer - B `a+4d = -3 "and" d = -4`. So, a = 13 `therefore S_(10) = (10)/(2) [2a +9d] = 5 [2 xx 13 + 9xx (-4)] = -50` |
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| 135. |
If the sum of first m terms of an AP is `(2m^(2) + 3m)` then what is its second term? |
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Answer» Correct Answer - 9 `S_(m) = (2m^(2) +3m) rArr S_(1) = 2+3 = 5 "and" S_(2) = 2 xx 4 +3 xx 2 = 14.` `therefore T_(2) = S_(2) -S_(1) = 14-5 =9.` |
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| 136. |
What is the 5th term from the end of the AP 2, 7, 12, …, 47? |
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Answer» Correct Answer - 27 nth term from the end = l - (n-1)d. `therefore "5th term from the end" = 47-4 xx 5 = 47-20 = 27` |
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| 137. |
What is the sum of first n terms of the AP a, 3a, 5a,…. |
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Answer» Correct Answer - `n^(2)a` Required sum ` = (n)/(2)[2a +(n-1) xx 2a] = n(na) = n^(2)a`. |
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| 138. |
The sequence -10,-6,-2,2,…. IsA. is an A.P., Reason d = -16B. is an A.P., Reason d = 4C. is an A.P., Reason d = -4D. is not an A.P. |
| Answer» Correct Answer - B | |
| 139. |
The 4th term from the end of an AP -11, -8, -5, …, 49 isA.B.C.D. |
| Answer» Correct Answer - `therefore`Fourth term from end given A.P. is 40 | |
| 140. |
The first four terms of an Ap 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 | |
| 141. |
200 logs are stacked in such a way that there are 20 logs in the bottom row, 19 in the next row, 18 in the next row and so on. In how many rows, 200 logs are placed and how many logs are there in the top row? |
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Answer» Let the required number of rows be n. Then, 20 + 19 +18 + … to n terms = 200. This is an arithmetic series in which a = 20, d = (19-20) = -1 and `S_(n) = 200.` We know that `S_(n) = (n)/(2) * {2a + (n-1)d}` `therefore (n)/(2) * {2 xx 20 + (n-1) xx (-1)} = 200` `rArr n(41-n) = 400 rArr n^(2) -41n +400 =0` `rArr n^(2) -25n - 16n +400 = 0 rArr n(n-25) - 16(n-25) = 0` `rArr (n-25) (n-16) = 0 rArr n - 25 = 0 "or" n - 16 =0` `rArr n = 25 "or" n = 16.` Now, `T_(25) = (a+24d) = 20+24 xx (-1) = -4.` This is meaningless as the number of logs cannot be negative. So, we reject the value n = 25. `therefore n = 16.` Thus, there are 16 rows in the whole stack. Now, `T_(16) = (a + 15d) = 20 + 15 xx (-1) = 20-15 =5.` Hence, there are 5 logs in the top row. |
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| 142. |
The 9th term of the A.P. is 499 and the 499th term is 9. Which term of this A.P. is zero ? |
| Answer» 508 term is zero. | |
| 143. |
The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year. Find the production during (i) first year (ii) 8th year (iii) first 6 years. |
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Answer» Let the production during first year be a and let d be the increase in production every year. Then, `T_(6) = 16000 rArr a +5d = 16000 " "…(i)` `"and" T_(9) = 22600 rArr a + 8d = 22600 " "...(ii)` On subtracting (i) from (ii), we get `3d = 6600 rArr d = 2200` Putting d = 2200 in (i), we get `a + 5 xx 2200 = 16000 rArr a + 11000 = 16000` ` rArr a = 16000 -11000 = 5000` Thus, a = 5000 and d = 2200. (i) Production during first year, a = 5000 (ii) Production during 8th year is given by `T_(8) = (a +7d) = (5000 + 7 xx 2200) = (5000 + 15400) = 20400.` (iii) Production during first 6 years is given by `S_(6) = (6)/(2) {2a+5d} = 3(2 xx 5000 + 5 xx 2200)` ` = 3(10000 + 11000) = 63000` |
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| 144. |
Find the sum of all numbers from 150 to 200 which are divisible by 7. |
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Answer» Correct Answer - The sum is 1225 The numbers from 150 to 200 divisible by 7 are 154,161 ,168,…., 196 Here, `a= 154, d= 7 ` and `t_(n) = 196` `t_(n) = a+ ( n -1)d` …(Formula ) `:. 196 = 154 + ( n-1) xx 7 ` …(Substituting the values ) `:. 196 - 154 = ( n-1) xx 7` ` :. ( 42)/(7) = n-1` `:. n -1 = 6` `:. n = 7` Now, we find the sum of 7 numbers. `S_(n) = (n)/(2) [t_(1) + t_(n)]` ...(Formula ) `= ( 7)/(2) [ 154 + 196)` `= ( 7)/(2) xx 350` ` = 7 xx 175 ` `= 1225` |
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| 145. |
In an A.P. the sum of three consecutive terms is 27 and their product is 504. Find the terms. ( Consider the terms to be in ascending order. ) |
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Answer» Correct Answer - The three consecutive terms are 4,9 and 14. Let the three consecutive terms in A.P. be a-d,a and a+d. From the first condition. a-d+a+a+d= 27 `:. 3a = 2` `:. A= 9` …..(1) From the second condition. `(a-d) (a) (a+d) = 504` `:. a(a^(2) -d^(2))= 504` `:. 9((9)^(2) - d^(2)) = 504 ` ...[From (1)] `:. 81 -d^(2) = 56` ...(Dividing both the sides by 9) `:. d^(2) = 81-56 ` `:. d^(2) = 25 ` `:. d +- 5` ......(Taking square root ) But the terms are in ascending order, `:. d = 5` Substituting a= 9 and d=5,a = 9-5=4, a = 9 , ` `a+d = 9+5= 14` |
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| 146. |
If `x ,y ,z`are in A.P. and `A_1`is the A.M. of `xa n dya n dA_2`is the A.M. of `ya n dz`, then prove that the A.M. of `A_1a n dA_2i sy`. |
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Answer» Since x,y,z and in AP, we have x + z= 2y Since, ` A_(1)` is the AM between x and y, we have ` A_(1) =(x +y)/ 2` Since ` A_(2) ` is the AM between y and z, we have `A_(2) = (y +z)/ 2` AM between ` A_(1) and A_(2)= 1/2 (A_(1) +A_(2)) = 1/2{(x+y)/2 + (y+z)/2}` ` = ( x+2y +z)/4 = (4y)/4 =y` Hence, the AM between ` A_(1) and A_(2)` is y . |
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| 147. |
If `a_1,a_2,a_3, ,a_n`are an A.P. ofnon-zero terms, prove that`1/(a_1+a_2)+1/(a_1+a_3)++1/(a_(n-1)+a_n)=(n-1)/(a_1+a_n)dot` |
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Answer» Let a be the common difference of the given AP. Then `(a_(2) a_(1)) = (a_(3) -a_(2)) = (a_(4) - a_(3)) = …..= ( a_(n) -a_(n-1)) =d` ` 1/ (a_(1)a_(2)) + 1/(a_(2) a_(3)) + 1/(a_(3) a_(4)) +….+ 1/(a_(n-1) a_(n)) ` ` = 1/d . { d/(a_(1)a_(2)) + d/(a_(2)a_(3)) + d/ (a_(3) a_(4)) +.....+ d/(a_(n-1) a_(n)) }` ` =1/d .{ ((a_(2) -a_(1))/(a_(1)a_(2)) + ((a_(3) -a_(2))/(a_(2) a_(3)) + ((a_(4) -a_(3)))/(a_(3)a_(4)) +....+((a_(n) -a_(n-1)))/(a_(n-1)a_(n))}` `[ because(a_(2)-a_(1))=(a_(3)-a_(2))=....=(a_(n)-a_(n-1))=d]` `=1/d.{(1/a_(1)-1/a_(2))+(1/a_(2)-1/a_(3))+(1/a_(3)-1/a_(4))+....+(1/a_(n-1) -1/a_(n))}` `= 1/d .[ ({a_(1) + (n-1) d} -a_(1))/(a_(1)a_(n))]` ` 1/d. ((n-1)d)/(a_(1)a_(n))= ((n-1))/(a_(1) a_(n))` Hence, the result follows. |
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| 148. |
find the nth term of the AP 8,3,-2,-7,,-12,…., |
| Answer» Correct Answer - `T_(n)= (13-5n)` | |
| 149. |
Find the nth term of the AP `1, 5/6,3/2,1/2` ,…., |
| Answer» Correct Answer - `T_(n)=1/6(7-n)` | |
| 150. |
Is -150 a term of the AP 11, 8, 5, 2,…? |
| Answer» Correct Answer - No | |