# GSEB Solutions for ધોરણ ૧૦ English

#### GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

Find the middle term of the AP 213,205,197,…, 37.21

જવાબ : Given Ap is 213, 205, 197, ... , 37. Here, first term, a=213; common difference, d=205-213=-8, nth term,  an =37 => a+(n-1)d=37

For what value of k, will the consecutive terms 2k + 1, 3k + 3 and 5k -1 form an A.P.?

જવાબ : Given that 2K+1, 3K+3 ad 5K+1 are in A.P. so, (3K+3) - (2k+1) = (5K-1) - (3k+3) => K+2 = 2K-4 => 2K-K  =  2 + 4 =>  K=6

For what value of k will k + 9,2k -1 and 2k + 7 are the consecutive terms of an A.P.?

જવાબ : Given that K+9, 2K-1 ad 2K+7 are in A.P. so, (2K-1) - (k+9) = (2K+7) - (2k-1) => K-10 = 8  =>  K=18

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

જવાબ : Reversing the given A.P., we get 185,181,174,...,9,5 Now, first term(a) =185 Common difference, (d) =181 -185 = -4 We know that nth term of an A.P. is given by a+(n-1)d Ninth term a9 = a+(9-1)d                           =185+8x(-4) = 185-32= 153

The 5th term of an AP exceeds its 12th term by 14. If its 7th term is 4, find the AP.

જવાબ : How many terms of the A.P. 18,16,14,… be taken so that their sum is zero?

જવાબ : How many terms of the A.P. 27,24,21,… should be taken so that their sum is zero?

જવાબ : How many terms of the A.P. 65,60, 55,… be taken so that their sum is zero?

જવાબ : The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

જવાબ : If the ratio of the sum of the first m and n terms of an A.P. is m2 : n2 , show that the ratio of its mth and nth terms is (2m -1): (2n -1).

જવાબ : If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P

જવાબ : The sum of first n terms of three A.Ps’ are S1, S2 and S3. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S1 + S3 = 2Sr

જવાબ : Divide 56 in four parts in A.R such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5 : 6.

જવાબ : The sums of first n terms of three arithmetic progressions are S1, S2 and S3 respectively. The first term of each A.P. is 1 and their common differences are 1,2 and 3 respectively. Prove that S2 + S3 = 2Sr

જવાબ : The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.

જવાબ : If the ratio of the sum of first n terms of two A.P.’s is (7n + 1): (4n + 27), find the ratio of their mth terms.

જવાબ : Find the middle term of the AP 6,13,20,…, 216.

જવાબ : Find the 25th term of the A.P. – 5, -5/2, 0, 5/2……………

જવાબ : Reshma wanted to save at least Rs 6,500 for sending her daughter to school next year (after 12 months). She saved Rs 450 in the first month and raised her savings by Rs 20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year?

જવાબ : The sum of three numbers in A.P. is 12 and sum of their cubes is 288, Find the numbers.

જવાબ : A thief, after committing a theft, runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in the first minute and increases his speed by 5 m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?

જવાબ : In an AP, if S5 + S7 = 167 and S10 = 235, then find the AP, where Sn denotes the sum of its first n terms.

જવાબ : The fourth term of an A.P. is 11. The sum of the fifth and seventh terms of the A.P. is 34. Find its common difference

જવાબ : The fifth term of an A.P. is 20 and the sum of its seventh and eleventh terms is 64. Find the common difference of the A.P.

જવાબ : The ninth term of an A.P is -32, and the sum of eleventh and thirteenth terms is -94.find the common difference of the A.P

જવાબ :  જવાબ :  જવાબ : If Sn, denotes the sum of first n-terms of an AP. Prove that: S12 = 3 (S8 – S4)

જવાબ : The 14th term of an AP is twice its 8th term. If its 6th term is -8, then find the sum of its first 20 terms.

જવાબ : The 16th term of an AP is five times its third term. If its 10th term is 41, then find the sum of its first fifteen terms.

જવાબ : The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.

જવાબ : Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.

જવાબ : The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.

જવાબ : The first and the last terms of an AP are 7 and 49 respectively. If the sum of all its terms is 420, find its common difference.

જવાબ : The first and the last terms of an AP are 8 and 65 respectively. If sum of all its terms is 730, find its common difference.

જવાબ : Find the middle term of the sequence formed by all three-digit numbers which leave a remainder of 3, when divided by 4. Also, find the sum of all numbers on both sides of the middle terms separately.

જવાબ : An arithmetic progression 5, 12, 19,… has 50 terms. Find its last term. Hence find the sum of its last 15 terms.

જવાબ : 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.

જવાબ : In an A.P., if the 12th term is -13 and the sum of its first four terms is 24, find the sum of its first ten terms.

જવાબ : The 13th term of an AP is four times its 3rd term. If its fifth term is 16, then find the sum of its first ten terms.

જવાબ : If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3(S20-S10)

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. Which value is shown in this question?

જવાબ : 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.

જવાબ : 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.

જવાબ : The sum of the 5th and the 9th terms of an AP is 30. If its 25th term is three times its 8th term, find the AP.

જવાબ : The sum of the first seven terms of an AP is 182. If its 4tji and the 17th terms are in the ratio 1: 5, find the AP.

જવાબ : The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is I less than twice its 8th term, find the AP.

જવાબ : If the seventh term of an AP is and its ninth term is , find its 63rd term.

જવાબ : The sum of the first n terms of an AP is 4n2 + 2n. Find the nth term of this AP.

જવાબ : The sum of the first n terms of an AP is 5n – n2. Find the nth term of this AP.

જવાબ : Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.

જવાબ : Given that,

11th term, a11 = 38

and 16th term, a16 = 73

We know that,

an = a+(n−1)d

a11 = a+(11−1)d

38 = a+10d ………………………………. (i)

In the same way,

a16 = 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 = −32

a31 = a +(31−1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

Hence, 31st term is 178.

An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

જવાબ : Given that,

3rd term, a3 = 12

50th term, a50 = 106

We know that,

an = a+(n−1)d

a3 = a+(3−1)d

12 = a+2d ……………………………. (i)

In the same way,

a50 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,

12 = a+2(2)

a = 12−4 = 8

a29 = a+(29−1) d

a29 = 8+(28)2

a29 = 8+56 = 64

Therefore, 29th term is 64.

Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.

જવાબ : Given,

Third term, a3 = 16

As we know,

a +(3−1)d = 16

a+2d = 16 ………………………………………. (i)

It is given that, 7th term exceeds the 5th term by 12.

a7 − a5 = 12

[a+(7−1)d]−[+(5−1)d]= 12

(a+6d)−(a+4d) = 12

2d = 12

d = 6

From equation (i), we get,

a+2(6) = 16

a+12 = 16

a = 4

Therefore, A.P. will be4, 10, 16, 22, …

For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal?

જવાબ : Given two APs as; 63, 65, 67,… and 3, 10, 17,….

Taking first AP,

63, 65, 67, …

First term, a = 63

Common difference, d = a2−a1 = 65−63 = 2

We know, nth term of this A.P. = an = a+(n−1)d

an= 63+(n−1)2 = 63+2n−2

an = 61+2n ………………………………………. (i)

Taking second AP,

3, 10, 17, …

First term, a = 3

Common difference, d = a2 − a1 = 10 − 3 = 7

We know that,

nth term of this A.P. = 3+(n−1)7

an = 3+7n−7

an = 7n−4 ……………………………………………………….. (ii)

Given, nth term of these A.P.s are equal to each other.

Equating both these equations, we get,

61+2n = 7n−4

61+4 = 5n

5n = 65

n = 13

Therefore, 13th terms of both these A.P.s are equal to each other.

How many multiples of 4 lie between 10 and 250?

જવાબ : The first multiple of 4 that is greater than 10 is 12.

Next multiple will be 16.

Therefore, the series formed as;

12, 16, 20, 24, …

All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.

When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.

The series is as follows, now;

12, 16, 20, 24, …, 248

Let 248 be the nth term of this A.P.

first term, a = 12

common difference, d = 4

an = 248

As we know,

an = a+(n−1)d

248 = 12+(n-1)×4

236/4 = n-1

59  = n-1

n = 60

Therefore, there are 60 multiples of 4 between 10 and 250.

How many three-digit numbers are divisible by 7?

જવાબ : First three-digit number that is divisible by 7 are;

First number = 105

Second number = 105+7 = 112

Third number = 112+7 =119

Therefore, 105, 112, 119, …

All are three digit numbers are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.

As we know, the largest possible three-digit number is 999.

When we divide 999 by 7, the remainder will be 5.

Therefore, 999-5 = 994 is the maximum possible three-digit number that is divisible by 7.

Now the series is as follows.

105, 112, 119, …, 994

Let 994 be the nth term of this A.P.

first term, a = 105

common difference, d = 7

an = 994

n = ?

As we know,

an = a+(n−1)d

994 = 105+(n−1)7

889 = (n−1)7

(n−1) = 127

n = 128

Therefore, 128 three-digit numbers are divisible by 7.

Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?

જવાબ : Let, the first term of two APs be a1 and a2 respectively

And the common difference of these APs be d.

For the first A.P.,we know,

an = a+(n−1)d

Therefore,

a100 = a1+(100−1)d

a1 + 99d

a1000 = a1+(1000−1)d

a1000 = a1+999d

For second A.P., we know,

an = a+(n−1)d

Therefore,

a100 = a2+(100−1)d

a2+99d

a1000 = a2+(1000−1)d

a2+999d

Given that, difference between 100th term of the two APs = 100

Therefore, (a1+99d) − (a2+99d) = 100

a1a2 = 100……………………………………………………………….. (i)

Difference between 1000th terms of the two APs

(a1+999d) − (a2+999d) = a1a2

From equation (i),

This difference, a1a= 100

Hence, the difference between 1000th terms of the two A.P. will be 100.

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

જવાબ : Given A.P. is 3, 15, 27, 39,

First term, a = 3

Common difference, d = a2 − a1 = 15 − 3 = 12

We know that,

an = a+(n−1)d

Therefore,

a54 = a+(54−1)d

3+(53)(12)

3+636 = 639

a54 = 639

We have to find the term of this A.P. which is 132 more than a54, i.e.771.

Let nth term be 771.

an = a+(n−1)d

771 = 3+(n −1)12

768 = (n−1)12

(n −1) = 64

n = 65

Therefore, 65th term was 132 more than 54th term.

Or another method is;

Let nth term be 132 more than 54th term.

n = 54 + 132/2

= 54 + 11 = 65th term

If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.

જવાબ : We know that, for an A.P series;

an = a+(n−1)d

a17 = a+(17−1)d

a17 = a +16d

In the same way,

a10 = a+9d

As it is given in the question,

a17 − a10 = 7

Therefore,

(a +16d)−(a+9d) = 7

7d = 7

d = 1

Therefore, the common difference is 1.

If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.

જવાબ : Given that,

3rd term, a3 = 4

and 9th term, a9 = −8

We know that,

an = a+(n−1)d

Therefore,

a3 = a+(3−1)d

4 = a+2d ……………………………………… (i)

a9 = a+(9−1)d

−8 = a+8d ………………………………………………… (ii)

On subtracting equation (i) from (ii), we will get here,

−12 = 6d

d = −2

From equation (i), we can write,

4 = a+2(−2)

4 = a−4

a = 8

Let nth term of this A.P. be zero.

aa+(n−1)d

0 = 8+(n−1)(−2)

0 = 8−2n+2

2= 10

n = 5

Hence, 5th term of this A.P. is 0.

Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.

જવાબ : Given A.P. is3, 8, 13, …, 253

Common difference, d= 5.

Therefore, we can write the given AP in reverse order as;

253, 248, 243, …, 13, 8, 5

Now for the new AP,

first term, a = 253

and common difference, d = 248 − 253 = −5

n = 20

Therefore, using nth term formula, we get,

a20 = a+(20−1)d

a20 = 253+(19)(−5)

a20 = 253−95

a = 158

Therefore, 20th term from the last term of the AP 3, 8, 13, …, 253.is 158.

The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.

જવાબ : We know that, the nth term of the AP is;

an = a+(n−1)d

a4 = a+(4−1)d

a4 = a+3d

In the same way, we can write,

a8 = a+7d

a6 = a+5d

a10 = a+9d

Given that,

a4+a8 = 24

a+3d+a+7d = 24

2a+10d = 24

a+5d = 12 …………………………………………………… (i)

a6+a10 = 44

a +5d+a+9d = 44

2a+14d = 44

a+7d = 22 …………………………………….. (ii)

On subtracting equation (i) from (ii), we get,

2d = 22 − 12

2d = 10

d = 5

From equation (i), we get,

a+5d = 12

a+5(5) = 12

a+25 = 12

a = −13

a2 = a+d = − 13+5 = −8

a3 = a2+d = − 8+5 = −3

Therefore, the first three terms of this A.P. are −13, −8, and −3.

Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

જવાબ : It can be seen from the given question, that the incomes of Subba Rao increases every year by Rs.200 and hence, forms an AP.

Therefore, after 1995, the salaries of each year are;

5000, 5200, 5400, …

Here, first term, a = 5000

and common difference, d = 200

Let after nth year, his salary be Rs 7000.

Therefore, by the nth term formula of AP,

an = a+(n−1) d

7000 = 5000+(n−1)200

200(n−1)= 2000

(n−1) = 10

n = 11

Therefore, in 11th year, his salary will be Rs 7000.

Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.

જવાબ : Given that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75.

Hence,

First term, a = 5

and common difference, d = 1.75

Also given,

a= 20.75

Find, n = ?

As we know, by the nth term formula,

an = a+(n−1)d

Therefore,

20.75 = 5+(n -1)×1.75

15.75 = (n -1)×1.75

(n -1) = 15.75/1.75 = 1575/175

= 63/7 = 9

n -1 = 9

n = 10

Hence, n is 10.

A thief runs with a uniform speed of 100 m/minute. After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief.

જવાબ : The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses proceeding the house numbered X is equal to sum of the numbers of houses following X.

જવાબ : Find the middle term of the sequence formed by all numbers between 9 and 95, which leave a remainder 1 when divided by 3. Also find the sum of the numbers on both sides of the middle term separately.

જવાબ : Find the middle term of the sequence formed by all three-digit numbers which leave a remainder 5 when divided by 7. Also find the sum of all numbers on both sides of the middle term separately.

જવાબ : The sum of the first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.

જવાબ : Sum of the first 20 terms of an A.P. is – 240, and its first term is 7. Find its 24th term.

જવાબ : Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term.

જવાબ : The sum of the third and seventh terms of an A.P. is 40 and the sum of its sixth and 14th terms is 70. Find the sum of the first ten terms of the A.P.

જવાબ : The sum of the first five terms of an A.P. is 25 and the sum of-its next five terms is – 75. Find the 10th term of the A.P.

જવાબ : The sum of 4th and 8th terms of an A.P. is 24 and the sum of its 6th and 10th terms is 44. Find the sum of first ten terms of the A.P.

જવાબ : A sum of Rs 1600 is to be used to give ten cash prizes to students of a school for their overall academic performance. If each prize is Rs20 less than its preceding prize, find the value of each of the prizes.

જવાબ : 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. [All India]

જવાબ : Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.

જવાબ : Is -150 a term of the AP 17,12, 7, 2,…?

જવાબ : Find the number of two-digit numbers which are divisible by 6.

જવાબ : Which term of the A.P. 3,14,25,36,… will be 99 more than its 25th term

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