\nIn the given AP,a = 10 and d = 7-10
\nThus, the 30th term is t30<\/sub>= 10 + (30 -1) (-3) = -77<\/p>\n
![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions.png?resize=159%2C13&ssl=1\")
<\/p>\n
Question 2.<\/p>\n
11th term of the A.P.,-3,-i, 2,… is:<\/h2>\n
(A) 28
\n(B) 22
\n(C) -38
\n(D) -48
\nAnswer:
\n(B) 22<\/p>\n
Explanation: Question 3.<\/p>\n (A) 6 Explanation: Question 4.<\/p>\n (A) 0 Explanation: Explanation: Question 5.<\/p>\n (A) an A.P., with d = – 16 Explanation: Question 6.<\/p>\n (A) -20 Explanation: Question 7.<\/p>\n (A) -2,0,2,4 Explanation: Question 8.<\/p>\n (A) 17 Explanation: Question 9.<\/p>\n (A) Pythagoras Explanation: Question 10.<\/p>\n (A) 0 Explanation: Question 11.<\/p>\n (A) -320 Explanation: Question 12.<\/p>\n (A) 19 Explanation: an<\/sub> = a+ (n – 1)d Question 13.<\/p>\n (A) 45 Explanation: Question 14.<\/p>\n (A) 180 Explanation: Assertion and Reason Based MCQs<\/span><\/p>\n Direction: In the. following question, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. Question 1.<\/p>\n Answer: Explanation: Question 2.<\/p>\n Answer: Explanation: Question 3.<\/p>\n Answer: Explanation: Question 4.<\/p>\n Answer: Explanation: Question 5.<\/p>\n Answer: Explanation: Case – Based MCQs<\/span><\/p>\n Attempt any four sub-parts from each question. Each sub-part carries 1 mark. Question 1.<\/p>\n (A) 51,53,55…. Explanation: Question 2.<\/p>\n (A) 10 Explanation: Question 3.<\/p>\n (A) 41 Question 4.<\/p>\n (A) 2 Question 5.<\/p>\n (A) 6 Explanation: II. Read the following text and answer the below questions: Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of \u20b9 1,18,000 by paying every month starting with the first instalment of \u20b9 1000. If he increases the instalment by \u20b9 100 every month. Question 1.<\/p>\n The amount paid by him in 30th<\/sup> installment is<\/p>\n (A) 3900 Explanation: Question 2.<\/p>\n (A) 37000 Explanation: Question 3.<\/p>\n (A) 45500 Question 4.<\/p>\n (A) 4900 Explanation: Question 5.<\/p>\n (A) 1: 49 III. Read the following text and answer the below questions: Jaspal Singh takes a loan from a bank for his car. Jaspal Singh repays his total loan of \u20b9 118000 by paying every month starting with the first installment of \u20b9 1000. If he increases the installment by \u20b9 100 every month. Question 1.<\/p>\n (A) 1000,100 Explanation: Question 2.<\/p>\n (A) \u20b9 3300 Explanation: Question 3.<\/p>\n (A) \u20b9 3900 Explanation: Question 4.<\/p>\n (A) \u20b9 7500 Explanation: Question 5.<\/p>\n (A) \u20b9400 Explanation: IV. Read the following text and answer the below questions: Question 1.<\/p>\n (A) 200 cm Explanation: Question 2.<\/p>\n (A) 20 Exploration: Question 3.<\/p>\n (A) 25 Explanation: Question 4.<\/p>\n (A) 25 Explanation: Question 5.<\/p>\n (A) 385 Explanation: Arithmetic Progressions Class 10 MCQ Questions with Answers Question 1. 30th term of the A. P., 10, 7,4,…………, is: (A) 97 (B) 77 (C) -77 (D) -87 Answer: (B) 77 Explanation: In the given AP,a = 10 and d = 7-10 Thus, the 30th term is t30= 10 + (30 -1) (-3) = -77 Question …<\/p>\n
\nIn the given A.P., a = -3 and
\nd = – \\(\\frac{1}{2}\\) + 3 = \\(\\frac{5}{2}\\)
\nThus, the 11th term is t11<\/sub> =-3 + (11-1)\\(\\left(\\frac{5}{2}\\right)\\) = 22<\/p>\nIn an A.P., if d = -4, n = 7, an = 4, then a is;<\/h2>\n
\n(B) 7
\n(C) 20
\n(D) 28
\nAnswer:
\n(D) 28<\/p>\n
\nIn the given A.P., d = -4, n = 7,
\nan<\/sub> = 4
\nan<\/sub> =a + (n – 1)d \u21d2 4 = a + (7 – 1)(-4) \u21d2 a = 28<\/p>\nIn an A.P., if a = 3.5, d = 0, n = 101, then an will be<\/h2>\n
\n(B) 3.5
\n(C) 103.5
\n(D) 104.5
\nAnswer:
\n(B) 3.5<\/p>\n
\nIn the given A.P., a = 3.5, d = 0, n = 101 than an<\/sub> will be
\n(A) 0
\n(B) 3.5
\n(C) 103. 5
\n(D) 104.5<\/p>\n
\nIn the given A.p., = 3.5, d = 0,
\nn = 101
\nan<\/sub> = a + (n – 1) d \u21d2an<\/sub> = 3.5 + (101 -1) 0 \u21d2an<\/sub> = 3.5<\/p>\nThe list of numbers – 10, – 6, – 2, 2,…………is:<\/h2>\n
\n(B) an A.P., with d = 4
\n(C) an A.P., with d = – 4
\n(D) not an A.P., 0
\nAnswer:
\n(B) an A.P., with d = 4<\/p>\n
\nIn the given numbers
\n-10,-6,-2, 2,…
\n(-6)-(-10) = 4
\n(-2)-(-6) = 4
\n2 – (-2) = 4
\nSince, (-6) – (-10) = (-2) – (-6) = 2 – (-2) = 4, thus, the given numbers are in AP with d = 4.<\/p>\n<\/p>\nThe 11th<\/sup> term of the A.P., -5, \\(-\\frac{5}{2}\\), 0, \\(\\frac{5}{2}\\),… is:<\/h2>\n
\n(B) 20
\n(C) -30
\n(D) 30
\nAnswer:
\n(B) 20<\/p>\n
\nIn the given A.P.,
\na = -5, d = \\(-\\frac{5}{2}\\)-(-5) = \\(\\frac{5}{2}\\) , n = 11
\ntn<\/sub> = a + (n – 1)d \u21d2 t11<\/sub> = -5 + -(11 1) \\(\\left(\\frac{5}{2}\\right)\\) \u21d2t11<\/sub> = 20<\/p>\nThe first four terms of an A.P., whose first term is – 2 and the common difference is -2, are:<\/h2>\n
\n(B)-2,4,-8,16
\n(C) – 2, – 4, – 6, – 8
\n(D)-2,-4,-8,-16 [1
\nAnswer:
\n(C) – 2, – 4, – 6, – 8<\/p>\n
\nIn the given AP, a = -2, d = -2,
\ntn<\/sub> =a + (n -1)d
\nt1<\/sub> = (-2) + (1 – 1) (-2) = -2
\nt2<\/sub> = (-2) + (2 – 1)(-2) = -4
\nt3<\/sub> = (-2) + (3 -1)(-2) = -6
\nt4<\/sub>= (-2) + (4 -1)(-2) = -8<\/p>\nThe 21st<\/sup> term of the A.P., whose first two terms are -3 and 4 is :<\/h2>\n
\n(B) 137
\n(C) 143
\n(D) -143
\nAnswer:
\n(B) 137<\/p>\n
\nIn the given A.P., t1<\/sub> = \u20143 and t2<\/sub> = 4
\n\u21d2 d =t2<\/sub> – t1<\/sub> = 4 – (-3) = 7
\ntn = a +(n – 1)d
\n\u21d2 t21<\/sub> = (-3) + (21 – 1) (7) = 137<\/p>\nThe famous mathematician associated with finding the sum of the first 100 natural numbers is :<\/h2>\n
\n(B) Newton
\n(C) Gauss
\n(D) Euclid
\nAnswer:
\n(C) Gauss<\/p>\n
\nThe famous mathematician associated with finding the sum of the first 100 natural numbers is Gauss.<\/p>\nIf the first term of an A.P. is -5 and the common difference is 2, then the sum of the first 6 terms is :<\/h2>\n
\n(B) 5
\n(C) 6
\n(D) 15
\nAnswer:
\n(A) 0<\/p>\n
\nIn the given A.P., a = -5 and d = 2 Thus,
\nS<\/span>n<\/sub> = \\(\\frac{n}{2}\\) [2a+(n – 1)d]
\nSn<\/sub> = \\(\\frac{5}{2}\\)[2x(-5)+(6-1)x2] = 0<\/p>\nThe sum of first 16 terms of the A.P., 10,6,2,… is :<\/h2>\n
\n(B) 320
\n(C) -352
\n(D) -400
\nAnswer:
\n(A) -320<\/p>\n
\nIn the given A.P.,a = 10,d = 6- 10 = -4
\nThus,
\nSn<\/sub> = \\(\\frac{n}{2}\\) [2a + (n-1)d]
\nS16<\/sub> = \\(\\frac{n}{2}\\) [2 \u00d7 10 + (16 – 1) (-4)]
\n= -320<\/p>\nIn an A.P., if a = 1, an<\/sub> = 20 and Sn<\/sub> = 399, then n is :<\/h2>\n
\n(B) 21
\n(C) 38
\n(D) 42
\nAnswer:
\n(C) 38<\/p>\n
\nIn the given A.P., a = 1, an<\/sub> = 20 and Sn<\/sub> = 399<\/p>\n
\n\u21d220 – 1 + (n -1)d
\n(n – 1)d = 19
\nSn<\/sub> = \\(\\frac{n}{2}\\) [2a + (n -1)d]
\n\u21d2399 = \\(\\frac{n}{2}\\) [2 +19]
\n\u21d2n = 38<\/p>\n<\/p>\nThe sum of tirst live multiples of 3 is :<\/h2>\n
\n(B) 55
\n(C) 65
\n(D) 75
\nAnswer:
\n(A) 45<\/p>\n
\nIn the given AP, a = 3, d = 3 and n = 5
\nThus,
\nSn<\/sub> = \\(\\frac{n}{2}\\) [2a + (n -1)d]
\n5<\/sub> = \\(\\frac{5}{2}\\) [2 3 + (5 -1) 3] =45<\/p>\nThe sum of first five positive integers divisible by 6 is:<\/h2>\n
\n(B) 90
\n(C) 45
\n(D) 30
\nAnswer:
\n(B) 90<\/p>\n
\nPositive integers divisible by 6 are 0,12,18,24,30 Since difference is same, its an AP We need to find sum of first 5 integers We can use formula
\nSn = \\(\\frac{5}{2}\\)[(2a + (n – 1)d)
\nn = 5, d = 6, a =6
\nS<\/span>5<\/sub> = \\(\\frac{n}{2}\\)(2 x 6 + (5 – 1) x 6)
\n<\/span>S<\/span>5<\/sub> = \\(\\frac{5}{2}\\) (12 + 24)
\n<\/span>S<\/span>5<\/sub> =\\(\\frac{5}{2}\\) \u00d7 36 = 90.<\/span><\/p>\n
\n(A) BothAandR are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\nAssertion (A): If the second term of an A.P., is 13 and the fifth term is 25, then its 7th term is 33.
\nReason (R): If the common difference of an A.P. is 5, then a18<\/sub> – a13<\/sub> is 25.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion:
\nIn the given A.P.,t2<\/sub> = 13 and t5<\/sub> = 25
\na + d = 13
\na + 4d – 25
\nSolving these equations, we get a = 9 and d = 4 Thus,
\ntn = a +(n -1)dayt7 = 9 +(7 – 1)4 = 33
\n\u2234 Assertion is correct:
\nIn case o freason
\nIn the given A.P.,d = 5 thus,
\na18 -a13 = a + 17d – a 12d = 5d = 25
\n\u2234Reason is correct.
\nHence, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\nAssertion (A): If a18<\/sub> – a14<\/sub> = 32, then the common difference of an A.P., is – 8.
\nReason (R): If 7 times the 7th<\/sup> term of an A.P. is equal to 11 times its 11th<\/sup> term, then its 18th<\/sup> term will be 0.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of assertion:
\nIn the given A.P., a18 <\/sub>– a14<\/sub> = 32
\nThus,
\na18<\/sub> – a14<\/sub> = 32
\n\u21d2 a + 17d – a – 13d = 32
\n\u21d2 4d = 32
\n\u21d2 d = 8
\n\u2234 Assertion is incorrect.
\nIn case of reason:
\nAccording to question,
\n7t7<\/sub> = 11t11<\/sub>
\n\u21d2 7(a +6d) = 11(a + 10d)
\n\u21d2 4a + 68d = 0
\n\u21d2 4(a + 17d) = 0
\n\u21d2 (a + 17d) = 0
\nt18<\/sub> = 0
\n\u2234 Reason is correct.
\nHence, assertion is incorrect but reason is correct.<\/p>\n<\/p>\nAssertion (A): The fourth term from the end of the A.P., -11, -8, -5,…, 49 is 40.
\nReason (R): If the nth term of an A.P., – 1,4,9,14,… is 129, then the value of n is 100.<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nIn case of assertion:
\nIn the given A.P., the last term l= 49 and common difference d = -8 + 11 = 3
\n4th<\/sup> term from last is t4<\/sub> = 49 – (4 – 1) x 3 = 40
\n\u2234 Assertion is correct.
\nIn case of reason:
\nGiven, a = – 1 and d = 4 – (- 1) = 5
\nan = -1 + (n- 1) x 5 – 129
\nor, (n – 1)5 = 130
\n(n – 1) = 26
\nn = 27
\nHence, 27th <\/sup>term = 129.
\n\u2234 Reason is incorrect.
\nHence, assertion is correct but reason is incorrect.<\/p>\nAssertion (A): If nlh temrof an A.P. is (2n + 1), then the sum of its first three terms is 15.
\nReason (R): The sum of first 16 terms of the A.P. 10, 6,2,… is – 320.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion:
\n\u2234 an<\/sub> = (2n + 1)
\n\u2234 a1<\/sub> = 2 \u00d7 1 + 1 = 3
\nl = a3<\/sub> = 2 \u00d73 + 1 = 7
\nSince, Sn<\/sub> = \\(\\frac{n}{2}\\)[a + l]
\nHence, S3<\/sub> = \\(\\frac{3}{2}\\)[3 + 7]
\nS3<\/sub> = 15
\n\u2234 Assertion is correct.
\nIn case of reason:
\nHwere, a = 10, d = 6 – 10 = -4 and n = 16
\nSn<\/sub> = \\(\\frac{n}{2}\\) [2a + (n – 1)d]
\nS16<\/sub> = \\(\\frac{16}{2}\\) [2 \u00d7 10 + (16 – 1)(-4)]
\n= 8[20 + 15 \u00d7(-4)]
\n= 8[20 – 60]
\n= 8 \u00d7 (-40)
\n= – 320
\n\u2234 Reason is correct.
\nHence, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\nAssertion (A): If the sum of first k terms of an A.P. is 3k2<\/sup>– k and its common difference is 6, then the first term is 5.
\nReason (R): If the \u00abth term of an A.P. is 7 – 3n, then the sum of 25 terms is – 800.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of assertion:
\nLet the sum of k terms of A.P. is Sk<\/sub> = 3k2<\/sup> – k Now kih term of A.P.
\nak<\/sub> = Sk <\/sub>– Sk-1<\/sub>
\nak<\/sub> = (3k2<\/sup>-k)-[3(k – 1)2<\/sup> – (k – 1)]
\n= 3k2<\/sup> – k – [3k2<\/sup> – 6k + 3 – k + 1]
\n= 3k2<\/sup> – k – 3k2<\/sup> + 7k – 4
\n= 6k – 4
\nHence, first term a = 6 x 1- 4 = 2
\n\u2234 Assertion is incorrect.
\nIn case of reason:
\nHere n = 25 and an<\/sub> = 7 – 3n<\/sub>
\nTaking n = 1,2,3,…………, we get
\na1<\/sub> = 7 – 3 x 1 = 4
\na2<\/sub> = 7 – 3 x 2 = 1
\na3<\/sub> = 7 – 3 x 3 = – 2
\n\u2234 GivenA.P. is 4, 1, -2,……….. .
\nHere, a = 4 and d = 1 – 4 = – 3
\nSince, Sn<\/sub> = \\(\\frac{n}{2}\\) [2a + (n – 1)d]
\nNow, S25<\/sub> = \\(\\frac{25}{2}\\) [2 x 4 +(25 – 1)(- 3)]
\n= \\(\\frac{25}{2}\\)[ 8 + 24 (-3)]
\n= \\(\\frac{25}{2}\\) (8 – 72)
\n= – 800
\n\u2234 Reason is correct.
\nHence, assertion is incorrect but reason is correct :<\/p>\n
\nI. Read the following text and answer the below questions:
\nYour friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do in 31 seconds.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-1.png?resize=259%2C221&ssl=1\")
<\/p>\nWhich of the following terms are in AP for the given situation<\/h2>\n
\n(B) 51,49,47….
\n(C) -51,-53,-55….
\n(D) 51,55,59…
\nAnswer:
\n(B) 51,49,47….<\/p>\n
\na = 51
\nd = – 2
\nAP = 51,49,47 ……<\/p>\n<\/p>\nWhat is the minimum number of days he needs to practice till his goal is achieved ?<\/h2>\n
\n(B) 12
\n(C) 11
\n(D) 9
\nAnswer:
\n(C) 11<\/p>\n
\nGoal = 31 second
\nn = number of days
\nan<\/sub> = 31
\na +(n – 1) d = 31
\n51 +(n – 1)(- 2) = 31
\n51 – 2n + 2 = 31
\n-2n = 31 – 53
\n-2n = -22
\nn = 11<\/p>\nWhich of the following term is not in the AP of the above given situation<\/h2>\n
\n(B) 30
\n(C) 37
\n(D) 39
\nAnswer:
\n(B) 30<\/p>\nIf nth<\/sup> term of an AP is given by an<\/sub> = 2n + 3 then common difference of an AP is<\/h2>\n
\n(B) 3
\n(C) 5
\n(D) 1
\nAnswer:
\n(A) 2<\/p>\nThe value of x, for which 2x, x + 10,3x + 2 are three consecutive terms of an AP<\/h2>\n
\n(B) -6
\n(C) 18
\n(D) -18
\nAnswer:
\n(A) 6<\/p>\n
\nSince, 2x,x + 10,3x + 2 are in AP, this common difference will remain same.
\nx + 10 – 2x = (3x + 2) – (x + 10)
\n10 – x = 2x – 8
\n2x = 18
\nx = 6<\/p>\n
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-2.png?resize=356%2C191&ssl=1\")
<\/p>\n
\n(B) 3500
\n(C) 3700
\n(D) 3600
\nAnswer:
\n(A) 3900<\/p>\n
\na = 1000
\nd = 100
\na80<\/sub>= a + (n -1 )d
\n= 1000 – (30 – 1)100
\n= 1000 + 2900<\/p>\nThe amount paid by him in the 30 installments is<\/h2>\n
\n(B) 73500
\n(C) 75300
\n(D) 75000
\nAnswer:
\n(B) 73500<\/p>\n
\nSum of 30 installments
\n= \\(\\frac{n}{2}\\)[2a + (n – 1 ) d ]
\n= \\(\\frac{30}{2}\\) [2 x 1000 + (30-1)100]
\n= 15[2000 + 2900]
\n= 15 x 4900
\n= 73500
\nTotal Amount paid in 30 installments = \u20b9 73500<\/p>\n<\/p>\nWhat amount does he still have to pay after 30th<\/sup> installment ?<\/h2>\n
\n(B) 49000
\n(C) 44500
\n(D) 54000
\nAnswer:
\n(C) 44500<\/p>\nIf total installments are 40 then amount paid in the last installment ?<\/h2>\n
\n(B) 3900
\n(C) 5900
\n(D) 9400
\nAnswer:
\n(A) 4900<\/p>\n
\nAmount paid in 40th<\/sup> installment, a40<\/sub>
\nFormula a + (n -1 )d
\na40<\/sub> = 1000 + (40 – 1)100
\n= 1000 + 3900
\n= 4900<\/p>\nThe ratio of the 1st<\/sup> installment to the last installment is<\/h2>\n
\n(B) 10: 49
\n(C) 10: 39
\n(D) 39:10
\nAnswer:
\n(B) 10: 49<\/p>\n
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-3.png?resize=329%2C184&ssl=1\")
<\/p>\nIf the given problem is based on A.P., then what is the first term and common difference ?<\/h2>\n
\n(B) 100,1000
\n(C) 100,100
\n(D) 1000,1000
\nAnswer:
\n(A) 1000,100<\/p>\n
\nThe number involved in this case form an A.P. in which first term (a) = 1000 and common difference (d) = 100.<\/p>\nThe amount paid by him in 25th<\/sup> installment is:<\/h2>\n
\n(B) \u20b9 3200
\n(C) \u20b9 3400
\n(D) \u20b9 3500
\nAnswer:
\n(C) \u20b9 3400<\/p>\n
\nThe amount paid by him in 25th<\/sup> installment is:
\nT25<\/sub> = a + 24d
\n= 1000 + 24 \u00d7 100
\n= 1000 + 2400
\n= \u20b9 3400.<\/p>\nThe amount paid by him in 30th<\/sup> installment is<\/h2>\n
\n(B) \u20b9 3500
\n(C) \u20b9 3000
\n(D) \u20b9 3600
\nAnswer:
\n(A) \u20b9 3900<\/p>\n
\nThe amount paid by him in 30th<\/sup> installment,
\nT30<\/sub> = a + 29d
\n= 1000 + 29 \u00d7 100
\n= 1000 + 2900 = \u20b9 3900.<\/p>\n<\/p>\nThe total amount paid by him in 25th<\/sup> and 30th<\/sup> installment is:<\/h2>\n
\n(B) \u20b9 7300
\n(C) \u20b9 7800
\n(D) \u20b9 7600
\nAnswer:
\n(B) \u20b9 7300<\/p>\n
\nTotal amount paid by him in 25th<\/sup> and 30th<\/sup> installment = \u20b9 (3400 + 3900) = \u20b9 7300.<\/p>\nThe difference amount paid by him in 26th<\/sup> and 28th<\/sup> installment is:<\/h2>\n
\n(B) \u20b9100
\n(C) \u20b9 500
\n(D) \u20b9 200
\nAnswer:
\n(D) \u20b9 200<\/p>\n
\nThe amount paid by him in 26th<\/sup> installment,
\nT26<\/sub> = a + 25 d
\n= 1000 + 25 x 100
\n= 1000 + 2500 = \u20b9 3500
\nThe amount paid by him in 28th<\/sup> installment,
\nT28 = a + 27d
\n= 1000 + 27 \u00d7 100
\n= 1000 + 2700
\n= \u20b9 3700
\n\u2234 The difference amount paid by him in 26th<\/sup> and 28th<\/sup> installment is:
\n= \u20b9 (3700-3500)
\n= \u20b9 200.<\/p>\n
\nA ladder has rungs 25 cm apart, (see the below).
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-4.png?resize=192%2C360&ssl=1\")
\nThe rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. The top and the bottom rungs are \\(2 \\frac{1}{2}\\) m apart.<\/p>\nThe top and bottom rungs are apart at a distance:<\/h2>\n
\n(B) 250 cm
\n(C) 300 cm
\n(D) 150 cm
\nAnswer:
\n(B) 250 cm<\/p>\n
\nSince the top and the bottom rungs are apart by \\(2 \\frac{1}{2}\\) m = \\(\\frac{5}{2}\\)m
\n= \\(\\frac{5}{2}\\) x 100m
\n= 250 cm<\/p>\nTotal number of the rungs is:<\/h2>\n
\n(B) 25
\n(C) 11
\n(D) 15
\nAnswer:
\n(C) 11<\/p>\n
\nThe distance between the two rungs is 25 cm.
\nHence, the total number of rungs = \\(\\frac{250}{25}\\) +1 = 11.<\/p>\nThe given problem is based on A.P. find its first term.<\/h2>\n
\n(B) 45
\n(C) 11
\n(D) 13
\nAnswer:
\n(A) 25<\/p>\n
\nThe length of the rungs increases from 25 to 45 and total number of rungs is 11. Thus, this is in the form of an A.P., whose first term is 25.<\/p>\nWhat is the last term of A.P. ?<\/h2>\n
\n(B) 45
\n(C) 11
\n(D) 13
\nAnswer:
\n(B) 45<\/p>\n
\nTotal number of terms, n = 11 and the last term, T11<\/sub> = 45.<\/p>\n<\/p>\nWhat is the length of the wood required for the rungs ?<\/h2>\n
\n(B) 538
\n(C) 532
\n(D) 382
\nAnswer:
\n(A) 385<\/p>\n
\nThe required length of the wood,
\nS11<\/sub> = \\(\\frac{11}{2}\\)[25 + 45]
\n= \\(\\frac{11}{2}\\)\u00d7 70
\n= 385 cm.<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"