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Fill in the blanks in the following table, given that a is the first term, d is a common difference and the nth term of A.P:
(i)a = 7, d = 3, n = 8
= 7 + (8 − 1) 3
= 28
Hence, an = 28
(ii) a = −18, n = 10, an = 0,
0 = − 18 + (10 − 1) d
Hence, common difference, d = 2
(iii)d = −3, n = 18, an = −5
−5 = a + (18 − 1) (−3)
−5 = a + (17) (−3)
a = 51 − 5 = 46
Hence, a = 46
(iv) a = −18.9, d = 2.5, an = 3.6, n = ?
3.6 = − 18.9 + (n − 1) 2.5
3.6 + 18.9 = (n − 1) 2.5
22.5 = (n − 1) 2.5
Hence, n = 10
(v) a = 3.5, d = 0, n = 105, an = ?
an = 3.5 + (105 − 1) 0
an = 3.5 + 104 × 0
an = 3.5
Hence, an = 3.5
Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
a = 3
d = a2 − a1 = 15 − 3 = 12
a54 = a + (54 − 1) d
= 3 + (53) (12) = 771
Let nth term be 771.
= 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.
Find the number of terms in each of the following APs :
( i) first term = a = 7 ; common difference =d= 13-7=19-13=6; n th term =tn=205 ; number of terms =n
Now, tn = a +(n-1).d
Or, 205 = 7 +(n-1).6,
Or, 198 = (n-1). 6
Or, 33 = n-1. So, n = 34 (Answer).
( ii) first term = a = 18, common difference =d= 15.5 -18= - 2.5; number of terms = n; n th term = tn= -47.
Now, tn = a +(n-1).d
Or, - 47 = 18 +(n-1).(-2.5)
Or, - 65 = (n-1).(-2.5)
Or, 26 = (n-1). So, n = 27 (Answer)
Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
First term = a = 3; common difference = d =8-3=13-8=5; number of terms = n; n th term = tn = 78.
We know, tn = a + (n-1).d
So, 78 = 3 +(n-1).5
Or, 75 = (n-1).5,
Or, 15 = n - 1, or, n =16 (Ans)
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Let, first term = a; common difference = d; number of terms = n; n th term = tn.
We know, tn = a +(n-1).d
Putting, n=11 and n =16, we get,
a + 10.d = 38 .... ( i); a + 15.d = 73 ..( ii)
By ( ii) - ( i), 5.d = 35, or, d = 7.
Putting the value of d in ( i) , a+70=38, or, a = - 32.
So, 31st term = a +(n-1).d= -32 + (31-1).7 = - 32 + 210 = 178.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Given
= 12, n = 50, a50= 106
= a+2d= 12——(1)
a50=a+49d= 106——(2)
(2)-(1)=> 47d= 94
d= 2
Substitute d=2 in (1)
a= 8
We have to find a29 =
=64
If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
First term = a; common difference =d, number of terms=n; n th term = tn = a +(n-1).d.
Putting n=3 we get, a+(3-1).d = 4 or, a + 2.d = 4 ...( i)
Putting n=9, a +(9-1).d = - 8 ,
a + 8d= - 8 ..( ii)
By ( ii) - ( i), 6.d = - 12,
d = -2;
Putting d = -2 in ( i), a - 4 = 4,
a=8.
Let nth term =0,
a +(n-1).d =0,
8 +(n-1).(-2)=0
n=5
Therefore, the term of the AP is zero.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
First term =a; common difference =d; number of terms =n; n th term tn= a +(n-1).d.
So, 17th term = a+(167-1).d = a+16d ....( i); 10th term= a+(10-1).d = a + 9d ..( ii)
By (i) - ( ii), (a+16d) -(a+9d) =7 (given), or, 7d = 7, or, d=1 (Ans)
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Let the first terms of the two APs be a and b and the common difference of each be d. If number of terms = n and n th term is tn, then, tn = a+(n-1).d and b+(n-1)d.
So, difference between the 100th terms = [ a+(100-1).d] -[b +(100-1)d] = (a+99d) -(b+99d) = 100 (given), or, a- b = 100 ... ( i)
So, the difference between the 1000th terms =[a+(1000-1)d] - [b+(1000-1)d] = a - b =100 (Ans) [from ( i)].
Ans : 100
How many three-digit numbers are divisible by 7?
The first 3 digit number divisible by 7 is 105
Last 3 digit number divisible by 7 is 994
formula- =a+(n-1)d
Number of terms n = (l-a/d) +1
=(994-105/7) +1
=889/7 +1
=127+1
=128
Hence there are 128 three digit numbers divisible by 7.
How many multiples of 4 lie between 10 and 250?
Multiples of 4 between 10 and 250 are
12,16,20,24,.....248
Here first term a=12,common difference d=4,
Last term an=248, no.of terms n=?
Wkt an=a+(n-1)d
248=12+(n-1)4
248=12+4n-4
248=8+4n
248-8=4n
240=4n
n=240/4
Therefore n=60
No.of multiples between 10 and 250 are 60.
Sana For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
a = 63
d = a2 − a1 = 65 − 63 = 2
nth term of this A.P.
= a + (n − 1) d
= 63 + (n − 1) 2
= 63 + 2n − 2
= 61 + 2n................... (1)
3, 10, 17, …
a = 3
d = a2 − a1 = 10 − 3 = 7
nth term of this A.P. = 3 + (n − 1) 7
= 3 + 7n − 7
= 7n − 4 ..............(2)
It is given that, 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.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
a3 = 16
a + (3 − 1) d = 16
a + 2d = 16 ..............(1)
a7 − a5 = 12
[a+ (7 − 1) d] − [a + (5 − 1) d]= 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
From equation (1), we obtain
a + 2 (6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be
4, 10, 16, 22, …
Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
This A.P. can be written in reverse order as
253, 248, 243, …, 13, 8, 3
For this A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a20 = a + (20 − 1) d
= 253 + (19) (−5)
a20= 158
Therefore, 20th term from the last term is 158.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
First term =a; common difference = d; number of terms = n; n th term = tn = a+(n-+).d.
So, sum of 4th and 8th terms = a+(4-1).d + a+(8-1).d = 2a+10d =24 (given) .... ( i)
Sum of 6th and 10th terms = a+(6-1)d + a+(10-1)d = 2a + 14d = 44 ....( ii)
By ( ii) - ( i), 4d = 20, or, d =5.
So, putting d in ( i), 2a+ 50 = 24,
Or, 2a = - 26, or, a = -13.
So, the first 3 terms of the AP are: a, a+d, a+2d = -13,-8,-3 ( Ans)
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?
Subba Rao's salary is in the form of AP:
5000, 5200, 5400, …
Here, a = 5000
d = 200
Let after nth year, his salary be Rs 7000.
Therefore, 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 savings by Rs. 1.75. If in the nth week, her weekly savings become Rs. 20.75, find n.
Weekly Savings are 5, 6.75, 8.5, ..... So, it forms am AP.
First term = a = 5 ; common difference =1.75= d ; n th term = tn = 20.75.
Now, tn = a +( n-1).d
So, 20.75 = 5 +( n-1).(1.75).,
Or, (n-1)= 15.75 ÷ 1.75 =9
So, n = 10 (Answer)
Hence, in the 10th week, Ramkali's savings will be Rs. 20.75
Choose the correct choice in the following and justify :
(i) 30th term of the AP: 10, 7, 4, . . . , is (A) 97 (B) 77 (C) –77 (D) – 87
(i) a = 10
d = a2 − a1 = 7 − 10 = −3
= 10 + (30 − 1) (−3)
= 10 + (29) (−3)
= 10 − 87 = −77
Hence, the correct answer is C.
(ii) a = −3, d=
Hence, the answer is B.
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