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Shub doc (maths) - CBSE 10th lesson Arithmetic progression notes

CBSE 10th lesson Arithmetic progression notes

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Zoolagy (CBSE)

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Arithmetic progression

INTRODUCTION:

In which of the following situations, does the list of numbers involved make as arithmetic progression and why? (i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km. Solution: We can write the given condition as; Taxi fare for 1 km = 15 Taxi fare for first 2 kms = 15+8 = 23 Taxi fare for first 3 kms = 23+8 = 31 Taxi fare for first 4 kms = 31+8 = 39 And so on...... Thus, 15, 23, 31, 39 ... forms an A. because every next term is 8 more than the preceding term. (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time. Solution: Let the volume of air in a cylinder, initially, be V litres. In each stroke, the vacuum pump removes 1/4th of air remaining in the cylinder at a time. Or we can say, after every stroke, 1-1/4 = 3/4th part of air will remain. Therefore, volumes will be V, 3V/4 , (3V/4) 2 , (3V/4) 3 .. so on Clearly, we can see here, the adjacent terms of this series do not have the common difference between them. Therefore, this series is not an A. (iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre. Solution: We can write the given condition as; Cost of digging a well for first metre = Rs. Cost of digging a well for first 2 metres = Rs+50 = Rs.

Cost of digging a well for first 3 metres = Rs+50 = Rs. Cost of digging a well for first 4 metres =Rs+50 = Rs. And so on.. Clearly, 150, 200, 250, 300 ... forms an A. with a common difference of 50 between each term. (iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum. Solution: We know that if Rs. P is deposited at r% compound interest per annum for n years, the amount of money will be: P(1+r/100)n Therefore, after each year, the amount of money will be; 10000(1+8/100), 10000(1+8/100) 2 , 10000(1+8/100) 3 ...... Clearly, the terms of this series do not have the common difference between them. Therefore, this is not an A. 2. Write first four terms of the A. when the first term a and the common difference are given as follows: (i) a = 10, d = 10 (ii) a = -2, d = 0 (iii) a = 4, d = – 3 (iv) a = -1 d = 1/ (v) a = – 1, d = – 0. Solutions: (i) a = 10, d = 10 Let us consider, the Arithmetic Progression series be a 1 , a 2 , a 3 , a 4 , a 5 ... a 1 = a = 10 a 2 = a 1 +d = 10+10 = 20 a 3 = a 2 +d = 20+10 = 30 a 4 = a 3 +d = 30+10 = 40 a 5 = a 4 +d = 40+10 = 50 And so on... Therefore, the A. series will be 10, 20, 30, 40, 50 ...

a 4 = a 3 + d = – 1.75-0 = – 2. Therefore, the A series will be 1, – 1, – 1, – 2. ........ And first four terms of this A. will be – 1, – 1, – 1 and – 2. 3. For the following A.P, write the first term and the common difference. (i) 3, 1, – 1, – 3 ... (ii) -5, – 1, 3, 7 ... (iii) 1/3, 5/3, 9/3, 13/ .... (iv) 0, 1, 2, 3. ... Solutions (i) Given series, 3, 1, – 1, – 3 ... First term, a = 3 Common difference, d = Second term – First term ⇒ 1 – 3 = - ⇒ d = - (ii) Given series, – 5, – 1, 3, 7 ... First term, a = - Common difference, d = Second term – First term ⇒ ( – 1)-( – 5) = – 1+5 = 4 (iii) Given series, 1/3, 5/3, 9/3, 13/ .... First term, a = 1/ Common difference, d = Second term – First term ⇒ 5/3 – 1/3 = 4/ (iv) Given series, 0, 1, 2, 3. ... First term, a = 0. Common difference, d = Second term – First term ⇒ 1 – 0. ⇒ 1. 4. Which of the following are APs? If they form an A. find the common difference d and write three more terms.

(i) 2, 4, 8, 16 ... (ii) 2, 5/2, 3, 7/ .... (iii) -1, -3, -5, -7. ... (iv) -10, – 6, – 2, 2 ... (v) 3, 3 + √2, 3 + 2√2, 3 + 3√ (vi) 0, 0, 0, 0. .... (vii) 0, – 4, – 8, – 12 ... (viii) -1/2, -1/2, -1/2, -1/ .... (ix) 1, 3, 9, 27 ... (x) a, 2a, 3a, 4a ... (xi) a, a 2 , a 3 , a 4 ... (xii) √2, √8, √18, √ ... (xiii) √3, √6, √9, √ ... (xiv) 1 2 , 3 2 , 5 2 , 7 2 ... (xv) 1 2 , 5 2 , 7 2 , 7 3 ... Solution (i) Given to us, 2, 4, 8, 16 ... Here, the common difference is; a 2 – a 1 = 4 – 2 = 2 a 3 – a 2 = 8 – 4 = 4 a 4 – a 3 = 16 – 8 = 8 Since, an+1 – an or the common difference is not the same every time. Therefore, the given series are not forming an A. (ii) Given, 2, 5/2, 3, 7/ .... Here, a 2 – a 1 = 5/2-2 = 1/ a 3 – a 2 = 3-5/2 = 1/ a 4 – a 3 = 7/2-3 = 1/ Since, an+1 – an or the common difference is same every time. Therefore, d = 1/2 and the given series are in A. The next three terms are; a 5 = 7/2+1/2 = 4 a 6 = 4 +1/2 = 9/

Therefore, d = √2 and the given series forms a A. Hence, next three terms are; a 5 = (3+√2) +√2 = 3+4√ a 6 = (3+4√2)+√2 = 3+5√ a 7 = (3+5√2)+√2 = 3+6√ (vi) 0, 0, 0, 0. ....

Here, a 2 – a 1 = 0.22-0 = 0. a 3 – a 2 = 0.222-0 = 0. a 4 – a 3 = 0.2222-0 = 0. Since, an+1 – an or the common difference is not same every time. Therefore, and the given series doesn’t forms a A. (vii) 0, -4, -8, - ...

Here, a 2 – a 1 = (-4)-0 = - a 3 – a 2 = (-8)-(-4) = - a 4 – a 3 = (-12)-(-8) = - Since, an+1 – an or the common difference is same every time. Therefore, d = -4 and the given series forms a A. Hence, next three terms are; a 5 = -12-4 = - a 6 = -16-4 = - a 7 = -20-4 = - (viii) -1/2, -1/2, -1/2, -1/ ....

Here, a 2 – a 1 = (-1/2) – (-1/2) = 0 a 3 – a 2 = (-1/2) – (-1/2) = 0 a 4 – a 3 = (-1/2) – (-1/2) = 0

Since, an+1 – an or the common difference is same every time. Therefore, d = 0 and the given series forms a A. Hence, next three terms are; a 5 = (-1/2)-0 = -1/ a 6 = (-1/2)-0 = -1/ a 7 = (-1/2)-0 = -1/ (ix) 1, 3, 9, 27 ...

Here, a 2 – a 1 = 3-1 = 2 a 3 – a 2 = 9-3 = 6 a 4 – a 3 = 27-9 = 18 Since, an+1 – an or the common difference is not same every time. Therefore, and the given series doesn’t form a A. (x) a, 2a, 3a, 4a ... Here, a 2 – a 1 = 2a–a = a a 3 – a 2 = 3a-2a = a a 4 – a 3 = 4a-3a = a Since, an+1 – an or the common difference is same every time. Therefore, d = a and the given series forms a A. Hence, next three terms are; a 5 = 4a+a = 5a a 6 = 5a+a = 6a a 7 = 6a+a = 7a (xi) a, a 2 , a 3 , a 4 ... Here, a 2 – a 1 = a 2 – a = a(a-1) a 3 – a 2 = a 3 – a 2 = a 2 (a-1) a 4 – a 3 = a 4 – a 3 = a 3 (a-1)

Or 1, 25, 49, 73 ... Here, a 2 − a 1 = 25−1 = 24 a 3 − a 2 = 49−25 = 24 a 4 − a 3 = 73−49 = 24 Since, an+1 – an or the common difference is same every time. Therefore, d = 24 and the given series forms a A. Hence, next three terms are; a 5 = 73+24 = 97 a 6 = 97+24 = 121 a 7 = 121+24 = 145

Exercise 5 Page: 105

Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the A.

Solutions: (i) Given, First term, a = 7 Common difference, d = 3 Number of terms, n = 8,

We have to find the nth term, an =? As we know, for an A., an = a+(n−1)d Putting the values, => 7+(8 −1) 3 => 7+(7) 3 => 7+21 = 28 Hence, an = 28 (ii) Given, First term, a = - Common difference, d =? Number of terms, n = 10 Nth term, an = 0 As we know, for an A., an = a+(n−1)d Putting the values, 0 = − 18 +(10−1)d 18 = 9d d = 18/9 = 2 Hence, common difference, d = 2 (iii) Given, First term, a =? Common difference, d = - Number of terms, n = 18 Nth term, an = - As we know, for an A., an = a+(n−1)d Putting the values, −5 = a+(18−1) (−3)

an = 3. Hence, an = 3. 2. Choose the correct choice in the following and justify: (i) 30th term of the A: 10,7, 4, ..., is (A) 97 (B) 77 (C) −77 (D) − (ii) 11th term of the A. -3, -1/2, ,2 .... is (A) 28 (B) 22 (C) – 38 (D)

Solutions: (i) Given here, A. = 10, 7, 4, ... Therefore, we can find, First term, a = 10 Common difference, d = a 2 − a 1 = 7−10 = − As we know, for an A., an = a +(n−1)d Putting the values; a 30 = 10+(30−1)(−3) a 30 = 10+(29)(−3) a 30 = 10−87 = − Hence, the correct answer is option C. (ii) Given here, A. = -3, -1/2, , ... Therefore, we can find, First term a = – 3 Common difference, d = a 2 − a 1 = (-1/2) -(-3) ⇒(-1/2) + 3 = 5/ As we know, for an A., an = a+(n−1)d

Putting the values; a 11 = -3+(11-1)(5/2) a 11 = -3+(10)(5/2) a 11 = -3+ a 11 = 22 Hence, the answer is option B. 3. In the following APs find the missing term in the boxes.

Solutions: (i) For the given A., 2,2 , 26 The first and third term are; a = 2 a 3 = 26 As we know, for an A., an = a+(n −1)d Therefore, putting the values here, a 3 = 2+(3-1)d 26 = 2+2d 24 = 2d d = 12 a 2 = 2+(2-1)

(19/2) – 5 = 3d 3d = 9/ d = 3/ a 2 = a+(2-1)d a 2 = 5 +3/ a 2 = 13/ a 3 = a+(3-1)d a 3 = 5 +2×3/ a 3 = 8 Therefore, the missing terms are 13/2 and 8 respectively. (iv) For the given A., a = −4 and a 6 = 6 As we know, for an A., an = a +(n−1) d Therefore, putting the values here, a 6 = a+(6−1)d 6 = − 4+5d 10 = 5d d = 2 a 2 = a+d = − 4+2 = − a 3 = a+2d = − 4+2(2) = 0 a 4 = a+3d = − 4+ 3(2) = 2 a 5 = a+4d = − 4+4(2) = 4 Therefore, the missing terms are −2, 0, 2, and 4 respectively. (v) For the given A., a 2 = 38 a 6 = − As we know, for an A.,

an = a+(n −1)d Therefore, putting the values here, a 2 = a+(2−1)d 38 = a+d ......................... (i) a 6 = a+(6−1)d −22 = a+5d ...................... (ii) On subtracting equation (i) from (ii), we get − 22 − 38 = 4d −60 = 4d d = − a = a 2 − d = 38 − (−15) = 53 a 3 = a + 2d = 53 + 2 (−15) = 23 a 4 = a + 3d = 53 + 3 (−15) = 8 a 5 = a + 4d = 53 + 4 (−15) = − Therefore, the missing terms are 53, 23, 8, and −7 respectively. 4. Which term of the A. 3, 8, 13, 18, ... is 78? Solutions: Given the A. series as3, 8, 13, 18, ... First term, a = 3 Common difference, d = a 2 − a 1 = 8 − 3 = 5 Let the nth term of given A. be 78. Now as we know, an = a+(n−1)d Therefore, 78 = 3+(n −1) 75 = (n−1) (n−1) = 15 n = 16 Hence, 16th term of this A. is 78. 5. Find the number of terms in each of the following A.

-65 = (n-1)(-5/2) (n-1) = -130/- (n-1) = 26 n = 27 Therefore, this given A. has 27 terms in it. 6. Check whether -150 is a term of the A. 11, 8, 5, 2, ... Solution: For the given series, A. 11, 8, 5, 2.. First term, a = 11 Common difference, d = a 2 −a 1 = 8−11 = − Let −150 be the nth term of this A. As we know, for an A., an = a+(n−1)d -150 = 11+(n -1)(-3) -150 = 11-3n + -164 = -3n n = 164/ Clearly, n is not an integer but a fraction. Therefore, – 150 is not a term of this A. 7. Find the 31st term of an A. whose 11th term is 38 and the 16th term is 73. Solution: Given that, 11 th term, a 11 = 38 and 16th term, a 16 = 73 We know that, an = a+(n−1)d a 11 = a+(11−1)d 38 = a+10d ..................................... (i) In the same way,

a 16 = a +(16−1)d 73 = a+15d ................................................ (ii) On subtracting equation (i) from (ii), we get 35 = 5d d = 7 From equation (i), we can write, 38 = a+10×(7) 38 − 70 = a a = − a 31 = a +(31−1) d = − 32 + 30 (7) = − 32 + 210 = 178 Hence, 31st term is 178. 8. An A. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29 th term. Solution: Given that, 3 rd term, a 3 = 12 50 th term, a 50 = 106 We know that, an = a+(n−1)d a 3 = a+(3−1)d 12 = a+2d .................................. (i) In the same way, a 50 = a+(50−1)d 106 = a+49d ............................... (ii) On subtracting equation (i) from (ii), we get 94 = 47d d = 2 = common difference From equation (i), we can write now,

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Shub doc (maths) - CBSE 10th lesson Arithmetic progression notes

Course: Zoolagy (CBSE)

10 Documents

Students shared 10 documents in this course

University: Mangalore University

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Arithmetic progression

INTRODUCTION:

1. In which of the following situations, does the list of numbers involved make as

arithmetic progression and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each

additional km.

Solution:

We can write the given condition as;

Taxi fare for 1 km = 15

Taxi fare for first 2 kms = 15+8 = 23

Taxi fare for first 3 kms = 23+8 = 31

Taxi fare for first 4 kms = 31+8 = 39

And so on……

Thus, 15, 23, 31, 39 … forms an A.P. because every next term is 8 more than the preceding

term.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air

remaining in the cylinder at a time.

Solution:

Let the volume of air in a cylinder, initially, be V litres.

In each stroke, the vacuum pump removes 1/4th of air remaining in the cylinder at a time.

Or we can say, after every stroke, 1-1/4 = 3/4th part of air will remain.

Therefore, volumes will be V, 3V/4 , (3V/4)2 , (3V/4)3…and so on

Clearly, we can see here, the adjacent terms of this series do not have the common

difference between them. Therefore, this series is not an A.P.

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first

metre and rises by Rs 50 for each subsequent metre.

Solution:

We can write the given condition as;

Cost of digging a well for first metre = Rs.150

Cost of digging a well for first 2 metres = Rs.150+50 = Rs.200

Shub doc (maths) - CBSE 10th lesson Arithmetic progression notes

Zoolagy (CBSE)

Mangalore University

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Arithmetic progression

INTRODUCTION:

Shub doc (maths) - CBSE 10th lesson Arithmetic progression notes

Course: Zoolagy (CBSE)

University: Mangalore University

Students also viewed

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