(i) Given,
First term, a = 7
Common difference, d = 3
Number of terms, n = 8,
We have to find the $N^{th}$ term, $a_n$ = ?
As we know, for an A.P.,
$a_n$ = a+(n−1)d |
Putting the values,
7+(8 −1) 3
7+(7) 3
7+21 = 28
Hence, $a_n$ = 28
(ii) Given,
First term, a = -18
Common difference, d = ?
Number of terms, n = 10
$N^{th}$ term, $a_n$ = 0
As we know, for an A.P.,
$a_n$ = a+(n−1)d |
Putting the values,
0 = − 18 +(10−1)d
18 = 9d
d = $\dfrac{18}{9}$ = 2
Hence, common difference, d = 2
(iii) Given,
First term, a = ?
Common difference, d = -3
Number of terms, n = 18
$N^{th}$ term, $a_n$ = -5
As we know, for an A.P.,
$a_n$ = 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
(iv) Given,
First term, a = -18.9
Common difference, d = 2.5
Number of terms, n = ?
$N^{th}$ term, $a_n$ = 3.6
As we know, for an A.P.
$a_n$ = 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) = $\dfrac{22.5}{2.5}$
n – 1 = 9
n = 10
Hence, n = 10
(v) Given,
First term, a = 3.5
Common difference, d = 0
Number of terms, n = 105
$N^{th}$ term, $a_n$ = ?
As we know, for an A.P.
$a_n$ = a+(n −1)d |
Putting the values,
$a_n$ = 3.5+(105−1) 0
$a_n$ = 3.5+104×0
$a_n$ = 3.5
Hence, $a_n$ = 3.5
Choose the correct choice in the following and justify:
(i) 30th term of the A.P: 10,7, 4, …, is
(A) 97
(B) 77
(C) −77
(D) −87
(ii) 11th term of the A.P. -3, -$\dfrac{1}{2}$, ,2 …. is
(A) 28
(B) 22
(C) – 38
(D) – 48 $\dfrac{1}{2}$
(i) Answer - Option C
Given here,
A.P. = 10, 7, 4, …
Therefore, we can find,
First term, a = 10
Common difference, $d = a_2 − a_1 = 7−10 = −3$
As we know, for an A.P.,
$a_n$ = a +(n−1)d |
Putting the values;
$a_{30}$ = 10+(30−1)(−3)
$a_{30}$ = 10+(29)(−3)
$a_{30}$ = 10−87 = −77
Hence, the correct Answer is option C.'
(ii) Answer : Option B
Given here,
A.P. = -3, $-\dfrac{1}{2}$, ,2 …
Therefore, we can find,
First term a = – 3
Common difference, $d = a_2 − a_1 = (-\dfrac{1}{2}) -(-3)$
$(-\dfrac{1}{2}) + 3 = \dfrac{5}{2}$
As we know, for an A.P.,
$a_n$ = a+(n−1)d |
Putting the values;
$a_{11} = -3+(11-1)(\dfrac{5}{2})$
$a_{11} = -3+(10)(\dfrac{5}{2})$
$a_{11} = -3+25$
$a_{11} = 22$
Hence, the Answer is option B.
(i) For the given A.P : 2,__, 26
The first and third term are;
a = 2
$a_3$ = 26
As we know, for an A.P.
$a_n = a+(n −1)d$ |
Therefore, putting the values here,
$a_3$ = 2+(3-1)d
26 = 2+2d
24 = 2d
d = 12
$a_2$ = 2+(2-1)12 = 14
Therefore, 14 is the missing term.
(ii) For the given A.P : __ , 13,__ ,3
$a_2$ = 13 and
$a_4$ = 3
As we know, for an A.P.,
$a_n = a+(n−1) d$ |
Therefore, putting the values here,
$a_2$ = a +(2-1)d
13 = a+d ………………. (i)
$a_4$ = a+(4-1)d
3 = a+3d ………….. (ii)
On subtracting equation (i) from (ii), we get,
– 10 = 2d
d = – 5
From equation (i), putting the value of d,we get
13 = a+(-5)
a = 18
$a_3$ = 18+(3-1)(-5)
= 18+2(-5) = 18-10 = 8
Therefore, the missing terms are 18 and 8 respectively.
(iii) For the given A.P : 5, __, __, 9$\dfrac{1}{2}$
a = 5 and
$a_4 = \dfrac{19}{2}$
As we know, for an A.P.,
$a_n = a+(n −1)d$ |
Therefore, putting the values here,
$a_4$ = a+(4-1)d
$\dfrac{19}{2}$ = 5+3d
$(\dfrac{19}{2})$ – 5 = 3d
3d = $\dfrac{9}{2}$
d = $\dfrac{3}{2}$
$a_2$ = a+(2-1)d
$a_2 = 5+\dfrac{3}{2}$
$a_2 = \dfrac{13}{2}$
$a_3 = a+(3-1)d$
$a_3 = 5+2×\dfrac{3}{2}$
$a_3$ = 8
Therefore, the missing terms are $\dfrac{13}{2}$ and 8 respectively.
(iv) For the given A.P : -4, __, __, __, __, 6
a = −4 and
$a_6$ = 6
As we know, for an A.P.,
$a_n = a+(n −1)d$ |
Therefore, putting the values here,
$a_6$ = a+(6−1)d
6 = − 4+5d
10 = 5d
d = 2
$a_2$ = a+d = − 4+2 = −2
$a_3$ = a+2d = − 4+2(2) = 0
$a_4$ = a+3d = − 4+ 3(2) = 2
$a_5$ = a+4d = − 4+4(2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
(v) For the given A.P : __,38,__,__,__22
$a_2$ = 38
$a_6$ = −22
As we know, for an A.P.,
$a_n = a+(n −1)d$ |
Therefore, putting the values here,
$a_2$ = a+(2−1)d
38 = a+d ……………………. (i)
$a_6$ = a+(6−1)d
−22 = a+5d …………………. (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d = −15
a = $a_2$ − d = 38 − (−15) = 53
$a_3$ = a + 2d = 53 + 2 (−15) = 23
$a_4$ = a + 3d = 53 + 3 (−15) = 8
$a_5$ = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.
Given the A.P. series as 3, 8, 13, 18, …
First term, a = 3
Common difference, $d = a_2 − a_1 = 8 − 3 = 5$
Let the nth term of given A.P. be 78. Now as we know,
$a_n = a+(n−1)d$ |
Therefore,
78 = 3+(n −1)5
75 = (n−1)5
(n−1) = 15
n = 16
Hence, $16^{th}$ term of this A.P. is 78.
(i) 7, 13, 19, . . . , 205
(ii) 18,15 $\dfrac{1}{2}$, 13, . . . , – 47
(i) Given, 7, 13, 19, …, 205 is the A.P
Therefore
First term, a = 7
Common difference, $d = a_2 − a_1$ = 13 − 7 = 6
Let there are n terms in this A.P.
$a_n = 205$
As we know, for an A.P.
$a_n = a + (n − 1) d$ |
Therefore, 205 = 7 + (n − 1) 6
198 = (n − 1) 6
33 = (n − 1)
n = 34
Therefore, this given series has 34 terms in it.
(ii) 18,15 $\dfrac{1}{2}$, 13, . . . , – 47
First term, a = 18
Common difference, $d = a_2-a_1 $
d = 5 $\dfrac{1}{2}$ - 18
d = $\dfrac{(31-36)}{2} = -\dfrac{5}{2}$
Let there are n terms in this A.P.
$a_n = -47$
As we know, for an A.P.
$a_n = a+(n−1)d$ |
-47 = 18+(n-1)(-$\dfrac{5}{2})$
-47-18 = (n-1)(-$\dfrac{5}{2})$
-65 = (n-1)(-$\dfrac{5}{2})$
(n-1) = -$\dfrac{130}{-5}$
(n-1) = 26
n = 27
Therefore, this given A.P. has 27 terms in it.
For the given series, A.P. 11, 8, 5, 2..
First term, a = 11
Common difference, $d = a_2−a_1$ = 8−11 = −3
Let −150 be the $n^{th}$ term of this A.P.
As we know, for an A.P.,
$a_n = a+(n−1)d$ |
-150 = 11+(n -1)(-3)
-150 = 11-3n +3
-164 = -3n
n = $\dfrac{164}{3}$
Clearly, n is not an integer but a fraction.
Therefore, – 150 is not a term of this A.P.
Given that,
$11^{th}$ term, $a_{11}$ = 38
and $16^{th}$ term, $a_{16}$ = 73
We know that,
$a_n$ = a+(n−1)d |
$a_{11}$ = a+(11−1)d
38 = a+10d ………………………………. (i)
In the same way,
$a_{16}$ = 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
$a_{31}$ = a +(31−1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, $31^{st}$ term is 178
Given that,
$3^{rd}$ term, $a_3$ = 12
$50^{th}$ term, $a_{50}$ = 106
We know that,
$a_n = a+(n−1)d$ |
$a_3$ = a+(3−1)d
12 = a+2d ……………………………. (i)
In the same way,
$a_{50}$ = 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
$a_{29}$ = a+(29−1) d
$a_{29}$ = 8+(28)2
$a_{29}$ = 8+56 = 64
Therefore, $29^{th}$ term is 64.
Given that,
$3^{rd}$ term, $a_3$ = 4
and $9^{th}$ term, $a_9$ = −8
We know that,
$a_n = a+(n−1)d$ |
Therefore,
$a_3$ = a+(3−1)d
4 = a+2d ……………………………………… (i)
$a_9$ = 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 $n^{th}$ term of this A.P. be zero.
$a_n = a+(n−1)d$ |
0 = 8+(n−1)(−2)
0 = 8−2n+2
2n = 10
n = 5
Hence, $5^{th}$ term of this A.P. is 0.
We know that, for an A.P series;
$a_n = a+(n−1)d$ |
$a_{17}$ = a+(17−1)d
$a_{17}$ = a +16d
In the same way,
$a_{10}$ = a+9d
As it is given in the question,
$a_{17} − a_{10}$ = 7
Therefore,
(a +16d)−(a+9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1.