Arithmetic Progressions Ex 5.2 Solutions
Question - 1 : - Fill in the blanksin the following table, given that a is the first term, d thecommon difference and an the nth termof the A.P.
Answer - 1 : -
Solution
(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.P.,
an = a+(n−1)d
Putting the values,
=> 7+(8 −1) 3
=> 7+(7) 3
=> 7+21 = 28
Hence, an = 28
Solution
(ii) Given,
First term, a = -18
Common difference, d = ?
Number of terms, n = 10
Nth term, an = 0
As we know, for an A.P.,
an = a+(n−1)d
Putting the values,
0 = − 18 +(10−1)d
18 = 9d
d = 18/9 = 2
Hence, common difference, d =2
Solution
(iii) Given,
First term, a = ?
Common difference, d = -3
Number of terms, n = 18
Nth term, an = -5
As we know, for an A.P.,
an = a+(n−1)d
Putting the values,
−5 = a+(18−1) (−3)
−5 = a+(17) (−3)
−5 = a−51
a = 51−5 = 46
Hence, a = 46
Solution
(iv) Given,
First term, a = -18.9
Common difference, d = 2.5
Number of terms, n = ?
Nth term, an = 3.6
As we know, for an A.P.,
an = a +(n −1)d
Putting the values,
3.6 = − 18.9+(n −1)2.5
3.6 + 18.9 = (n−1)2.5
22.5 = (n−1)2.5
(n – 1) = 22.5/2.5
n – 1 = 9
n = 10
Hence, n = 10
Solution
(v) Given,
First term, a = 3.5
Common difference, d = 0
Number of terms, n = 105
Nth term, an = ?
As we know, for an A.P.,
an = a+(n −1)d
Putting the values,
an = 3.5+(105−1) 0
an = 3.5+104×0
an = 3.5
Hence, an = 3.5
Question - 2 : - Choose the correct choice in the following and justify:
Answer - 2 : -
(i) 30th term of the AP: 10, 7, 4, …, is
(a) 97 (b) 77 (c) -77 (d) -87
(ii) 11th term of the AP: -3, −12 , 2, …, is
(a) 28 (b) 22 (c) -38 (d) -48
Solution
Question - 3 : - In the following APs, find the missing terms in the boxes
Answer - 3 : - 
Solution:
Solution:
Solution:
Solution:
Solution:
Answer - 4 : -
Given the A.P. series as3, 8, 13, 18, …
First term, a = 3
Common difference, d = a2 − a1 =8 − 3 = 5
Let the nth termof given A.P. be 78. Now as we know,
an = a+(n−1)d
Therefore,
78 = 3+(n −1)5
75 = (n−1)5
(n−1) = 15
n = 16
Hence, 16th term of this A.P.is 78.
Question - 5 : - Find the number of terms in each of the following A.P.
Answer - 5 : -
(i) 7, 13, 19, …,205
Solutions:
(i) Given, 7, 13,19, …, 205 is the A.P
Therefore
First term, a = 7
Common difference, d = a2 − a1 =13 − 7 = 6
Let there are n terms in thisA.P.
an = 205
As we know, for an A.P.,
an = a + (n − 1) d
Therefore, 205 = 7 + (n − 1) 6
198 = (n − 1) 6
33 = (n − 1)
n = 34
Therefore, this given series has 34 terms init.
First term, a = 18
Common difference, d = a2-a1 =
d = (31-36)/2 = -5/2
Let there are n terms in this A.P.
an = 205
As we know, for an A.P.,
an = a+(n−1)d
-47 = 18+(n-1)(-5/2)
-47-18 = (n-1)(-5/2)
-65 = (n-1)(-5/2)
(n-1) = -130/-5
(n-1) = 26
n = 27
Therefore, this given A.P. has 27 terms in it.
Question - 6 : - Check whether -150 is a term of the A.P. 11, 8, 5, 2, …
Answer - 6 : -
For the given series, A.P. 11, 8, 5, 2..
First term, a = 11
Common difference, d = a2−a1 =8−11 = −3
Let −150 be the nth termof this A.P.
As we know, for an A.P.,
an = a+(n−1)d
-150 = 11+(n -1)(-3)
-150 = 11-3n +3
-164 = -3n
n = 164/3
Clearly, n is not an integerbut a fraction.
Therefore, – 150 is not a term of this A.P.
Question - 7 : - Find the 31_st term of an A.P. whose 11_th term is 38 and the 16_th term is 73.
Answer - 7 : -
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.
Question - 8 : - An A.P. consists of 50 terms of which 3_rd term is 12 and the last term is 106. Find the 29_th term.
Answer - 8 : -
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 = commondifference
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.
Question - 9 : - If the 3_rd and the 9_th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.
Answer - 9 : -
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 ofthis A.P. be zero.
an = a+(n−1)d
0 = 8+(n−1)(−2)
0 = 8−2n+2
2n = 10
n = 5
Hence, 5th termof this A.P. is 0.
Question - 10 : - The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Answer - 10 : - 