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11. Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54th term? Solution: 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+132=771 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 12. Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms? Solution: 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 a1−a2 = 100……………………………………………………………….. (i) Difference between 1000th terms of the two APs (a1+999d) − (a2+999d) = a1−a2 From equation (i), This difference, a1−a2 = 100 Hence, the difference between 1000th terms of the two A.P. will be 100. 13. How many three digit numbers are divisible by 7? Solution: 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. 14. How many multiples of 4 lie between 10 and 250? Solution: 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. 15. For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal? Solution: 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. 16. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12. Solutions: 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]−[a +(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, … 17. Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253. Solution: 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. 18. 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. Solution: 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. 19. 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? Solution: 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. 20. 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. Solution: 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, an = 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.
![Fallin For You [slowed + reverb] - Shrey Singhal | Endorphin |](/img/n.gif)
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11. Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54th term?
Solution:
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+132=771
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
12. Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?
Solution:
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
a1−a2 = 100……………………………………………………………….. (i)
Difference between 1000th terms of the two APs
(a1+999d) − (a2+999d) = a1−a2
From equation (i),
This difference, a1−a2 = 100
Hence, the difference between 1000th terms of the two A.P. will be 100.
13. How many three digit numbers are divisible by 7?
Solution:
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.
14. How many multiples of 4 lie between 10 and 250?
Solution:
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.
15. For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal?
Solution:
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.
16. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solutions:
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]−[a +(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, …
17. Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.
Solution:
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.
18. 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.
Solution:
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.
19. 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?
Solution:
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.
20. 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.
Solution:
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,
an = 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.
Is this math?❓
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