Find the common difference of the AP . (2013D)

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જવાબ :

Find the common difference of the A.P. . (2013D)

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જવાબ : d = a_{2} – a_{1} =

= = -3

Find the common difference of the A.P. . (2013D)

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જવાબ : d = a_{2} – a_{1} =

= = -2

Calculate the common difference of the A.P. . (2013D)

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જવાબ : d = a_{2} – a_{1}=

=

Calculate the common difference of the A.P. (2013OD)

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જવાબ : d = a_{2} – a_{1} =

= = -b

What is the common difference of an A.P. in which a_{21} – a_{7} = 84? (2017OD )

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જવાબ : a_{21} – a_{7} = 84 …[Given in the question

∴ (a + 20d) – (a + 6d) = 84 …[a_{n} = a + (n – 1)d

20d – 6d = 84

14d = 84 ⇒ d = 6

Find the 9th term from the end (towards the first term) of the A.P. 5,9,13, …, 185. (2016D)

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જવાબ : Here, a = 5, d = 9 – 5 = 4, l = 185

9^{th} term from the end = 185 – (9 – 1)4

= 185 – 8 × 4 = 185 – 32 = 153

The angles of a triangle are in A.P., the least being half the greatest. Find the angles. (2011D)

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જવાબ : 40°, 60°, 80°

Is -150 a term of the A.P. 17, 12, 7, 2, … ? (2011D)

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જવાબ : no

Which term of the progression 4, 9, 14, 19, … is 109? (2011D)

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જવાબ : 22nd term

Which term of the progression 20, 192, 183, 17 … is the first negative term? (2017OD)

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જવાબ : 28^{th} term

The 4^{th} term of an A.P. is zero. Will the 25^{th} term of the A.P. is three times its 11^{th }term. (2016OD)

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જવાબ : Yes
Let 1^{st} term = a, Common difference = d

a4 = 0 a + 3d = 0 ⇒ a = -3d … (A)

To prove: a_{25} = 3 × a11

a + 24d = 3(a + 10d) …[From (A)

⇒ -3d + 24d = 3(-3d + 10d)

⇒ 21d = 21d

From above, a_{25} = 3(a_{11}) (Hence proved)

The 7^{th} term of an A.P. is 20 and its 13^{th} term is 32. Find the A.P. (2012 OD)

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જવાબ : 8, 10, 12,14
Solution:

Let ‘a’ be the 1st term and ‘d’ be the common difference.

Is the 10^{th} term from end of the A.P. 4,9, 14, …, 254., 209?

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જવાબ : Yes

Common difference d = 9 – 4

= 14 – 9 = 5

Given that the last term, l = 254, n = 10

nth term from the end = l – (n – 1) d

∴ 10th term from the end = 254 – (10 – 1) × 5

= 254 – 45 = 209

Find how many two-digit numbers are divisible by 6? (2011OD)

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જવાબ : 6

Solution:

12, 18, 24, …,96

Here,First term a = 12, common diff. d = 18 – 12 = 6, an = 96

a + (n – 1)d = an

∴ 12 + (n – 1)6 = 96

⇒ (n − 1)6 = 96 – 12 = 84

⇒ n – 1 = = 14

⇒ n = 14 + 1 = 15

∴ There are 15 two-digit numbers divisible by 6.

How many natural numbers are there between 200 and 500, which are divisible by 7? (2011OD)

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જવાબ : 43

Solution:

203, 210, 217, …, 497

Here, First term a = 203, common diff. d = 210 – 203 = 7, an = 497

∴ a + (n – 1) d = an

203 + (n – 1) 7 = 497

(n – 1) 7 = 497 – 203 = 294

n – 1 = = 42 ∴ n = 42 + 1 = 43

∴ There are 43 natural nos. between 200 and 500 which are divisible by 7.

How many two-digit numbers are divisible by 3? (2012OD)

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જવાબ : 30

Solution:

Two-digit numbers divisible by 3 are:

12, 15, 18, …, 99

Here, First term a = 12, common diff. d = 15 – 12 = 3, an = 99

∴ a + (n – 1)d = an

12 + (n – 1) (3) = 99

(n – 1) (3) = 99 – 12 = 87

n – 1 = = 29

∴ n = 29 + 1 = 30

∴ There are 30 such numbers.

Are there total 128 three-digit natural numbers are divisible by 7?

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જવાબ : Yes

Solution:

“3 digits nos.” are 100, 101, 102, …, 999

3 digits nos. “divisible by 7” are:

105, 112, 119, 126, …, 994

first term a = 105, common diff. d = 7, an = 994, n = ?

As a + (n – 1)d = 994 = an

∴ 105 + (n – 1)7 = 994

(n − 1)7 = 994 – 105

(n – 1) = = 127

∴ n = 127 + 1 = 128

Are there total 100 three-digit natural numbers which are divisible by 9?

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જવાબ : Yes

Solution:

To find: Number of terms of A.P., i.e., n.

A.P. = 108 + 117 + 126 + … + 999

a = 108

d = 117 – 108 = 9

a_{n} = 999

a + (n – 1)d = an

∴ 108 + (n – 1) 9 = 999

⇒ (n − 1) 9 = 999 – 108 = 891

⇒ (n − 1) = = 99

∴ n = 99 + 1 = 100

Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5. (2014OD)

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જવાબ : 89

Solution:

Numbers divisible by both 2 and 5 are 110, 120, 130, …, 990.

Here first term a = 110, common diff. d = 120 – 110 = 10, an = 990

As a + (n – 1)d = an = 990

110 + (n – 1)(10) = 990

(n – 1)(10) = 990 – 110 = 880

(n − 1) = = 88

∴ n = 88 + 1 = 89

Do in an AP with first term *a* and common difference *d*, the *nth* term is given by *a*_{n}=*a*+(*n*−1)*d* ?

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જવાબ : Yes

State whether the statement True or False : 10th term of A.P. 10, 20, 30, 40 ......is 100

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જવાબ : True

State whether the statement True or False : 20th term of A.P. whose first term = 10, common difference = 3 is 67

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જવાબ : True

State whether the statement True or False : The number of bacteria in a certain food item after each second forms an AP. when they double in every second

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જવાબ : False

State whether the statement True or False : The amount of money in the account of Varun at the end of every year forms an AP when Rs 1000 is deposited at simple interest of 10% per annum

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જવાબ : True

If the nth term of the A. P. 9, 7, 5.... is same as the nth term of the A. P. 15, 12, 9...... find n

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જવાબ : n=7

Which term of the A. P. 3, 15, 27, 39...... will be 132 more than its 54th term?

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જવાબ : 65^{th}

Which term of the arithmetic progression 5, 15, 25, ----- will be 130 more than its 31st term?

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જવાબ : 44^{th}

In an A. P, the sum of first ten terms is -150 and the sum of its next ten terms are -550. Find the A. P.

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જવાબ : 6, 2, -2, -6 …….

In an A. P., the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference of the A. P.

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જવાબ : d=3

The first and the last term of an A. P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

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જવાબ : n=38 & S_{n }=6973

In an A. P. the first term is 8, nth term is 33 and the sum to first n terms is 123. Find n and d, the common differences.

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જવાબ : n=6 & d=5

In an A. P., the sum of first n terms is 3n^{2}/2+13n/2. Find its 25th term.

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જવાબ : T_{25}=S_{25}−S_{24}T_{25}=S_{25}−S_{24} = [3(25)^{2 }/2] +[13×25]/2−3(24)^{2}/2 + [13×24]/2= 80

The 4th term of an A.P is equal to 3 times the first term and the 7th term exceeds twice the 3rd term by 1. Find the A.P.

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જવાબ : The AP formed is 3, 5, 7, 9, .........

Find the value of the middle most term (s) of the AP

-11, -7, -3,..., 49

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-11, -7, -3,..., 49

જવાબ : middle most terms will 8 and 9

T8=−11+(8−1)4=17T8=−11+(8−1)4= 17

T9=−11+(9−1)4=21T9=−11+(9−1)4= 21

Find the sum

x−yx+y+3x−2yx+y+5x−3yx+y+x−yx+y+3x−2yx+y+5x−3yx+y+..upto 11 terms

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x−yx+y+3x−2yx+y+5x−3yx+y+x−yx+y+3x−2yx+y+5x−3yx+y+..upto 11 terms

જવાબ : 11(11x−6y)/x+y

If the numbers n - 2, 4n - 1 and 5n + 2 are in AP, find the value of n.

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જવાબ : 1

State whether the statement True or False : The fee charged every month by a school from Classes I to XII forms an AP, when the monthly fee for Class I is Rs 250, and it increases by Rs 50 for the next higher class.

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જવાબ : True

State whether the statement True or False : The fee charged from a student every month by a school for the whole session forms an AP, when the monthly fee is Rs 400.

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જવાબ : False

State whether the statement True or False : Is the sequence 5, 10, 15, 20,25...... an A.P.? If yes, find the Common Difference.

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જવાબ : Yes, d=5

State whether the statement True or False : Is the sequence 2, 2, 2, 2,2...... an A.P.? If yes, find the Common Difference.

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જવાબ : Yes, d=0

State whether the statement True or False : Is the sequence 2, 4, 8, 16,32...... an A.P.? If yes, find the Common Difference.

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જવાબ : No

State whether the statement True or False : Is the sequence 3, 14, 25, 36...... an A.P.? If yes, find the Common Difference.

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જવાબ : Yes, d=11

State whether the statement True or False : Is the sequence 11, 8, 5, 2, ..... an A.P.? If yes, find the Common Difference.

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જવાબ : Yes, d=-3

Write next two terms of given A.P. -3, -5, -7,-9......

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જવાબ : Next 2 terms are -11, -13

For what value of a, are (2a-1), 7 and 3a three consecutive terms of an A.P?

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જવાબ : 3

How many multiples of 4 lie between 10 and 250?

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જવાબ : 60

If S_{n}, the sum of first n terms of an A.P is given by

Sn=3n2−4nSn=3n2−4n

then find its nth term

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Sn=3n2−4nSn=3n2−4n

then find its nth term

જવાબ : 6n -7

Find the sum of the following A.P 1 + 3 + 5 + ..... + 199.

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જવાબ : 10000

Solve the equation:

1+4+7+10+...+x=2871+4+7+10+...+x=287

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1+4+7+10+...+x=2871+4+7+10+...+x=287

જવાબ : 40

If p^{th}, q^{th} and r^{th} terms of an A.P. are a, b, c respectively, then show that (a – b)r + (b – c)p+ (c – a)q = 0. (2011D)

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જવાબ : Let A be the first term and D be the common difference of the given A.P.

p^{th} term = A + (p – 1)D = a …(A)

q^{th} term = A + (q – 1)D = b …(B)

r^{th} term = A + (r – 1)D = c … (C)

L.H.S. = (a – b)r + (b – c)p + (c – a)q

= [A + (p – 1)D – (A + (q – 1)D)]r + [A + (q – 1)D – (A + (r – 1)D)]p + [A + (r – 1)D – (A + (p – 1)D)]q

= [(p – 1 – q + 1)D]r + [(q – 1 – r + 1)D]p + [(r – 1 – p + 1)D]q

= D[(p – q)r + (q – r)p + (r – p)q]

= D[pr – qr + qp – rp + rq – pq]

= D[0] = 0 = R.H.S.

The 17^{th} term of an AP is 5 more than twice its 8^{th} term. If the 11^{th} term of the AP is 43, then find its n^{th} term. (2012D)

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જવાબ :

From (i),

a = 2(4) – 5 = 8 – 5 = 3

As a_{n} = a + (n – 1) d

∴ a_{n} = 3 + (n – 1) 4 = 3 + 4n – 4

a_{n} = (4n – 1)

The 15^{th} term of an AP is 3 more than twice its 7^{th} term. If the 10^{th} term of the AP is 41, then find its n^{th} term. (2012D)

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જવાબ :

From (i),

a = 2(4) – 3

= 8 – 3 = 5

n^{th} term = a + (n – 1) d

∴ n^{th} term = 5 + (n – 1) 4

= 5 + 4n – 4 = (4n + 1)

The 16^{th} term of an AP is 1 more than twice its 8^{th} term. If the 12^{th} term of the AP is 47, then find its n^{th} term. (2012D)

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જવાબ :

From (A) and (B), a = 4 – 1 = 3

As n^{th} term = a + (n – 1) d

∴ n^{th} term = 3 + (n – 1) 4

= 3 + 4n -4 = 4n -1

Find the 60th term of the AP 8, 10, 12, …, if it has a total of 60 terms and hence find the sum of its last 10 terms. (2015OD)

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જવાબ : a = 8, d = a_{2} – a_{1} = 10 – 8 = 2, n = 60

a_{60} = a + 59d = 8 + 59(2) = 126

∴ Sum of its last 10 terms = S_{60} – S_{50}

= (a + a_{n}) – (2a + (n – 1)d)

= (8 + a_{60}) – (2 × 8 + (50 – 1)2)

= 30 (8 + 126) – 25 (16 + 98)

= 4020 – 25 × 114

= 4020 – 2850 = 1170

An Arithmetic Progression 5, 12, 19, … has 50 terms. Find its last term. Hence find the sum of its last 15 terms. (2015OD)

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જવાબ : Let, ‘a’ be the first term and ‘d’ be the common difference of A.P. respectively,

a = 5, d = 12 – 5 = 7, n = 50

∴ a_{n} = a + (n – 1)d

a_{50} = 5 + 49(7) = 5 + 343 = 348

∴ Sum of its last 15 terms = S_{50} – S_{35}

= (a + a_{n}) – (2a + (n − 1)d)

= (5 + 348) – [2(5) + (35 – 1)7]

= 25(353) – (10 + 238)

= 8825 – 35 × 124

= 8825 – 4340 = 4485

If the sum of first 4 terms of an A.P. is 40 and that of first 14 terms is 280, find the sum of its first n terms. (2011D)

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જવાબ : S_{n} = [2a + (n − 1) d] …(A)

Putting the value of d = 2 in (B), we get a = 7

∴ S_{n} = [2(7) + (n – 1). 2]

= .2 [7 + n – 1]

= n (n + 6) or n^{2} + 6n

A sum of ₹1,600 is to be used to give ten cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes. (2012OD)

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જવાબ : Here, Sum of 10 prizes = 1600

SS

∴ [2a + (10 – 1)(-20)] = 1600

2a – 180 = 320

2a = 320 + 180 = 500

a = 250

Hence, 1

2

3

4

5

6

7

8

9

10

= ₹ 1,600

The first and the last terms of an A.P. are 8 and 350 respectively. If its common difference is 9, how many terms are there and what is their sum? (2011D)

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જવાબ : Here a = 8, l = 350, d = 9

As we know, a + (n − 1) d = a_{2}

∴ 8 + (n – 1) 9 = 350

(n − 1) 9 = 350 – 8 = 342

n – 1 = = 38

n = 38 + 1 = 39

∴ There are 39 terms.

∴ S_{n} = (a + l)

∴ S_{39} = (8 + 350) = × 358

= 39 × 179 = 6981

Sum of the first 20 terms of an AP is -240, and its first term is 7. Find its 24^{th} term. (2012D)

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જવાબ : Given: a = 7, S_{20} = -240

Here, S_{n} = [2a + (n – 1)d]

∴ S_{20} = [2(7) + (20 – 1)d]

-240 = 10(14 + 19d)

– = 14 + 19d = -24

19d = -24 – 14 = -38

⇒ d = = -2 …(A)

Again, a_{n} = a + (n – 1)d

∴ a_{24} = 7 + (24 – 1) (-2) … [From (A)

= 7 – 46 = -39

Find the common difference of an A.P. whose first term is 5 and the sum of its first four terms is half the sum of the next four terms. (2012OD)

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જવાબ : Here a = 5 … (A)

Here, a_{1} + a_{2} + a_{3} + a_{4} = (a_{5} + a_{6} + a_{7} + a_{8})

∴ a + (a + d) + (a + 2d) + (a + 3d)

= [(a + 4d) + (a + 5d) + (a + 6d) + (a + 780] …[a_{n} = a(n – 1)d

∴ 4a + 6d = (4a + 22d)

8a + 12d = 4a + 22d

8a – 4a = 22d – 12d

4a = 100 ⇒ 4(5) = 10d

d = = 2 …[From (A)

∴ Common difference, d = 2

If the sum of the first 7 terms of an A.P. is 119 and that of the first 17 terms is 714, find the sum of its first n terms. (2012OD)

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જવાબ : As S_{n} = [2a + (n − 1)d]

S_{2} = 119

From (A) and (B), a = 17 – 3(5) = 17 – 15 = 2

∴S_{n} = [2(2) + (n – 1)5]

= [4 + 5n – 5]

= (5n – 1)

The 24^{th} term of an AP is twice its 10th term. Show that its 72^{nd} term is 4 times its 15th term. (2013D)

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જવાબ : Solution:

a_{24} = 2 (10) … [Given

a + 23d = 2 (a +9d) [∵ a_{n} = a + (n – 1)d)

23d = 2a + 18d – a

23d – 18d = a ⇒ a = 5d …(A)

To prove: a_{72} = 4 (a_{15})

From (ii) and (iii), L.H.S = R.H.S …Hence Proved

Find the number of terms of the A.P. -12, -9, – 6, …, 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained. (2013D)

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જવાબ : Given that, a = -12, d=-9 + 12 = 3

Here, a_{n} = a + (n – 1)d = 21

∴ -12 + (n – 1).3 = 21

(n – 1).3 = 21 + 12 = 33

∴ n – 1 = 11

Total number of terms,

n = 11 + 1 = 12

New A.P. is

Here 1st term, a = -11

Common difference, d = -8 + 11 = 3;

Last term, a_{n} = 22; Number of terms, n= 12

In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the AP. (2014D)

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જવાબ : Here, n = 50,

Here, S_{10} = 210

= (2a + 9d) = 210 ….[S_{n} = [2a+(n – 1)2]

5(2a + 9d) = 210

2a + 9d = = 42

⇒ 2a = 42 – 9d ⇒ a = …(A)

Now, 50 = (1 + 2 + 3 + …) + (36 + 37 + … + 50) Sum = 2565

Sum of its last 15 terms = 2565 …[Given

S_{50} – S_{35} = 2565

(2a + 49d) – (2a + 34d) = 2565

100a + 2450d – 70a – 1190d = 2565 × 2

30a + 1260d = 5130

3a + 1260 = 513 …[Dividing both sides by 10

If the ratio of the sum of the first n terms of two A.Ps is (7n + 1): (4n + 27), then find the ratio of their 9^{th} terms. (2017OD)

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જવાબ :

In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two Sections, find how many trees were planted by the students. (2014OD)

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જવાબ : Classes: 1 + I + II + … + XII

Sections: 2(I) + 2(II) + 2(III) + … + 2(XII)

Total no. of trees

= 2 + 4 + 6 … + 24

= (2 × 2) + (2 × 4) + (2 × 6) + … + (2 × 24)

= 4 + 8 + 12 + … + 48

:: S_{12} = (4 + 48) = 6(52) = 312 trees

Ramkali required ₹500 after 12 weeks to send her daughter to school. She saved ₹100 in the first week and increased her weekly saving by ₹20 every week. Find whether she will be able to send her daughter to school after 12 weeks. (2015D)

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જવાબ : Money required by Ramkali for admission of her daughter = ₹2500

A.P. formed by saving

100, 120, 140, … upto 12 terms ….(A)

Let, a, d and n be the first term, common difference and number of terms respectively. Here, a = 100, d = 20, n = 12

S_{n} = (2a + (n − 1)d)

⇒ S_{12} = (2(100) + (12 – 1)20)

S_{12} = [2(100) + 11(20)] = 6[420] = ₹2520

The sum of first n terms of an A.P. is 5n^{2} + 3n. If its m^{th} term is 168, find the value of m. Also find the 20^{th} term of this A.P. (2013D)

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જવાબ :

The sum of first m terms of an AP is 4m^{2} – m. If its n^{th} term is 107, find the value of n. Also find the 21^{st} term of this A.P. (2013D)

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જવાબ : We have, S_{m} = 4m^{2} – m

Put m = 1,

S_{1} = 4(1)^{2} – (1)

= 4 – 1 = 3

Put m = 2,

S_{2} = 4(2)^{2} – 2

= 16 – 2 = 14

∴ a_{1} = S_{1} = 3 = a

a_{2} = S_{2} – S_{1} = 14 – 3 = 11

d = a_{2} – a_{1}

d = 11 – 3 = 8

a_{n} = a + (n – 1)d = 107 …[Given in the ques.

∴ 3+ (n – 1)8 = 107

(n – 1)8 = 107 – 3 = 104

(n – 1) = = 13

n = 13 + 1 = 14

∴ a_{21} = a + 20d

= 3 + (20)8

= 3 + 160 = 163

The sum of first g terms of an A.P. is 63q – 3q^{2}. If its p^{th} term is -60, find the value of p. Also find the 11^{th} term of this A.P. (2013D)

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જવાબ : We have, S_{q} = 63q – 3q^{2}

Put q = 1 in the equation

S_{7} = 63(1) – 3(1)^{2}

= 63 – 3 = 60

Put q = 2 in the equation

S2 = 63(2) – 3(2)^{2}

= 126 – 12 = 114

∴ a = a_{1} = S_{1} = 60

a_{2} = S_{2} – S7_{1} = 114 – 60 = 54

d = a_{2} – a_{1} = 54 – 60 = -6

p^{th} term = -60

a + (p – 1)d = -60

60 + (p – 1)(-6) = -60

(p – 1)(-6) = -60 – 60 = -120

(p – 1) = = 20

p= 20 + 1 = 21

∴ a_{11} = a + 10d

= 60 + 10(-6)

= 60 – 60 = 0

Find the 10^{th }term

1 |
1,3,5,7…. |
A |
57 |

2 |
2,4,6,8…. |
B |
51 |

3 |
2,7,12,17…. |
C |
19 |

4 |
21,24,27,30…. |
D |
20 |

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જવાબ :

1-C, 2-D, 3-A, 4-B

Find the n^{th }term

1 |
2,6,10,14…. |
A |
8+10n |

2 |
13,9,5,1…. |
B |
4n-2 |

3 |
18,28,38,48…. |
C |
26-n |

4 |
25,24,23,22,…. |
D |
17-4n |

જવાબ :

1-B, 2-D, 3-A, 4-C

Find the 2^{nd }term from last

1 |
-2,5,…….,68 |
A |
900 |

2 |
-100,0,….., 1000 |
B |
98 |

3 |
28,30,…… , 100 |
C |
15 |

4 |
75,45,….-15 |
D |
61 |

Hide | Show

જવાબ :

1-D, 2-A, 3-B, 4-C

જવાબ :

1-B, 2-A, 3-D, 4-C

Find the common differences

1 |
89,85,81,77 |
A |
11 |

2 |
100, 89, 78, 67 |
B |
-4 |

3 |
100, 111,122,133 |
C |
4 |

4 |
89, 93, 97, 101 |
D |
-11 |

Hide | Show

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-B, 2-A, 3-D, 4-C

જવાબ :

1-D, 2-A, 3-B, 4-C

Find the n^{th }term

1 |
a=2, d=4 |
A |
12-2n |

2 |
a=4, d=2 |
B |
14-4n |

3 |
a=10, d=-2 |
C |
4n-2 |

4 |
a=10, d=-4 |
D |
2+2n |

Hide | Show

જવાબ :

1-C, 2-D, 3-A, 4-B

Find the S_{8}

1 |
1,3,5,7…. |
A |
252 |

2 |
4,6,8,10…. |
B |
156 |

3 |
2,7,12,17…. |
C |
88 |

4 |
21,24,27,30…. |
D |
72 |

Hide | Show

જવાબ :

1-D, 2-C, 3-B, 4-A

જવાબ :

1-B, 2-A, 3-D, 4-C

Math

Chapter 05 : Arithmetic Progressions

- Math Book for CBSE Class 10
- History Book for CBSE Class 10
- Geography Book for CBSE Class 10
- Political Science Book for CBSE Class 10
- Chemistry Book for CBSE Class 10
- Economics Book for CBSE Class 10
- Biology Book for CBSE Class 10
- Physics Book for CBSE Class 10

The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.