# CBSE Solutions for Class 11 English

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

જવાબ : √3(1/3)7

Find the 9th term of the G.P. 1/(a3 x3), ax, a5 x5 , ...

જવાબ : a29x29

Find the 10th term of the G.P. 0.3, 0.06, 0.012, ...

જવાબ : 0.3(0.2)9

Find the 11th term of the G.P. −3/4, 1/2, −1/3, 2/9, ...

જવાબ : ½ (2/3)9

Find the 6th term of the G.P. 1, 4, 16, 64, ...

જવાબ : 1024

જવાબ : √3(1/3)8

Find the 10th term of the G.P. 1/(a3 x3), ax, a5 x5 , ...

જવાબ : a33x33

Find the 6th term of the G.P. 0.3, 0.06, 0.012, ...

જવાબ : 0.3(0.2)5

Find the 8th term of the G.P. −3/4, 1/2, −1/3, 2/9, ...

જવાબ : ½ (2/3)6

Find the 7th term of the G.P. 1, 4, 16, 64, ...

જવાબ : 4096

જવાબ : √3(1/3)9

Find the 11th term of the G.P. 1/(a3 x3), ax, a5 x5 , ...

જવાબ : a37x37

Find the 7th term of the G.P. 0.3, 0.06, 0.012, ...

જવાબ : 0.3(0.2)6

Find the 9th term of the G.P. −3/4, 1/2, −1/3, 2/9, ...

જવાબ : ½ (2/3)7

Find the 8th term of the G.P. 1, 4, 16, 64, ...

જવાબ : 16384

Find the 9th term from the end of the G.P. 2/27, 2/9, 2/3, ..., 162

જવાબ : 486

Find the 8th term from the end of the G.P. 2/27, 2/9, 2/3, ..., 162

જવાબ : 162

Find the 7th term from the end of the G.P. 2/27, 2/9, 2/3, ..., 162

જવાબ : 54

Find the 7th term from the end of the G.P. 2/27, 2/9, 2/3, ..., 162

જવાબ : 54

Find the 6th term from the end of the G.P. 2/27, 2/9, 2/3, ..., 162

જવાબ : 18

Find the 5th term from the end of the G.P. 2/27, 2/9, 2/3, ..., 162

જવાબ : 6

Find 11th term of the A.P. 1, 4, 7, 10, ....

જવાબ : 31

Find 12th term of the A.P. 1, 4, 7, 10, ...

જવાબ : 34

Find 13th term of the A.P. 1, 4, 7, 10,...

જવાબ : 37

Find 14th term of the A.P. 13, 8, 3, −2, ...

જવાબ : -52

Find 13th term of the A.P. 13, 8, 3, −2,...

જવાબ : -47

Find 12th term of the A.P. 13, 8, 3, −2, ...

જવાબ : -42

Find 11th term of the A.P. 13, 8, 3, −2, ...

જવાબ : -37

Find 10th term of the A.P. 13, 8, 3, −2, ...

જવાબ : -32

Find 14th term of the A.P. 1, 4, 7, 10,...

જવાબ : 40

Find 15th term of the A.P. 1, 4, 7, 10,...

જવાબ : 43

Find 16th term of the A.P. 1, 4, 7, 10,...

જવાબ : 46

Find 17th term of the A.P. 1, 4, 7, 10,...

જવાબ : 49

Find 18th term of the A.P. 1, 4, 7, 10,...

જવાબ : 52

Find 19th term of the A.P. 1, 4, 7, 10,...

જવાબ : 55

Find 20th term of the A.P. 1, 4, 7, 10,...

જવાબ : 58

Find 17th term of the A.P. √2, 3√2, 5√2,...

જવાબ : 33√2

Find 16th term of the A.P. √2, 3√2, 5√2,...

જવાબ : 31√2

Find 15th term of the A.P. √2, 3√2, 5√2,...

જવાબ : 29√2

Find 14th term of the A.P. √2, 3√2, 5√2,...

જવાબ : 27√2

Find 13th term of the A.P. √2, 3√2, 5√2,...

જવાબ : 25√2

Find 19th term of the A.P. √2, 3√2, 5√2,...

જવાબ : 37√2

Find 20th term of the A.P. √2, 3√2, 5√2,...

જવાબ : 39√2

Find 21th term of the A.P. √2, 3√2, 5√2,...

જવાબ : 41√2

Find 22th term of the A.P. √2, 3√2, 5√2,...

જવાબ : 43√2

Find 23th term of the A.P. √2, 3√2, 5√2,...

જવાબ : 45√2

Find 5th term of the A.P. 13, 8, 3, −2,...

જવાબ : -7

Find 6th term of the A.P. 13, 8, 3, −2,...

જવાબ : -12

Find 7th term of the A.P. 13, 8, 3, −2,...

જવાબ : -17

Find 8th term of the A.P. 13, 8, 3, −2,...

જવાબ : -22

Find 9th term of the A.P. 13, 8, 3, −2,...

જવાબ : -27

Which term of the G.P. :

1/3, 1/9, 1/27 ...is 1/729 ?

જવાબ : Here, first term, a=1/3 and common ratio, r=1/3

Let the nth term be 1/729.

∴ an =1/729

⇒ arn-1 = 1/729

⇒ (1/3)(1/3)n-1 = 1/729

⇒ (1/3)n-1 = 3/(3)6 = (1/3)5

⇒ n-1 = 5

n=6

Thus, the 6th term of the given G.P. is 1/729.

Which term of the G.P. :

√3, 3, 3√3, ... is 27√3 ?

જવાબ : Here, first term, a=√3 and common ratio, r=√3

Let the nth term be 27√3.

∴ an=27√3

⇒ arn-1 = 27√3

⇒ (√3)(√3)n-1 = 27√3

⇒ (√3)n-1 = (√3)6

n-1 = 6

n=7

Thus, the 7th term of the given G.P. is 27√3.

Which term of the G.P. :

1/3, 1/9, 1/27 ...is 1/2187 ?

જવાબ : Here, first term, a=1/3 and common ratio, r=1/3

Let the nth term be 1/2187.

∴ an =1/2187

⇒ arn-1 = 1/2187

⇒ (1/3)(1/3)n-1 = 1/2187

⇒ (1/3)n-1 = 3/(3)7 = (1/3)6

⇒ n-1 = 6

n=7

Thus, the 7th term of the given G.P. is 1/2187.

Which term of the G.P. :

√3, 3, 3√3, ... is 81√3 ?

જવાબ : Here, first term, a=√3 and common ratio, r=√3

Let the nth term be 81√3.

∴ an=81√3

⇒ arn-1 = 81√3

⇒ (√3)(√3)n-1 = 81√3

⇒ (√3)n-1 = (√3)8

n-1 = 8

n=9

Thus, the 9th term of the given G.P. is 81√3.

Which term of the G.P. :

2, 2√2, 4, ... is 64√2 ?

જવાબ : Here, first term, a=2 and common ratio, r=√2

Let the nth term be 64√2.

∴ an = 64√2

⇒ arn-1 = 64√2

⇒ (2)(√2)n-1 = 64√2

⇒2 (√2)n-1 = 64√2

⇒ (√2)n-1 = 32√2

⇒ (√2)n-1 = (√2)11

n-1 = 11

n = 12

Thus, the 12th term of the given G.P. is 64√2.

Which term of the G.P. :

√2,1/√2, 1/2√2 ,1/4√2, ... is 1/128√2?

જવાબ : Here, first term, a= √2  and common ratio, r=1/2

Let the nth term be 1/128√2.

∴ an = 1/128√2

⇒ arn-1 = 1/128√2

⇒ (√2)(1/2)n-1 = 1/128√2

⇒ (1/2)n-1 = 1/256

⇒ (1/2)n-1 = (1/2)8

n-1 = 8

n=9

Thus, the 9th term of the given G.P. is 1/128√2.

Which term of the G.P. :

1/3, 1/9, 1/27 ...is 1/6561 ?

જવાબ : Here, first term, a=1/3 and common ratio, r=1/3

Let the nth term be 1/6561.

∴ an =1/6561

⇒ arn-1 = 1/6561

⇒ (1/3)(1/3)n-1 = 1/6561

⇒ (1/3)n-1 = 3/(3)8 = (1/3)7

⇒ n-1 = 7

n=8

Thus, the 8th term of the given G.P. is 6561.

Which term of the G.P. :

√3, 3, 3√3, ... is 729√3 ?

જવાબ : Here, first term, a=√3 and common ratio, r=√3

Let the nth term be 729√3.

∴ an=729√3

⇒ arn-1 = 729√3

⇒ (√3)(√3)n-1 = 729√3

⇒ (√3)n-1 = (√3)12

n-1 = 12

n=13

Thus, the 13th term of the given G.P. is 729√3.

Which term of the G.P. :

2, 2√2, 4, ... is 128√2 ?

જવાબ : Here, first term, a=2 and common ratio, r=√2

Let the nth term be 128√2.

∴ an = 128√2

⇒ arn-1 = 128√2

⇒ (2)(√2)n-1 = 128√2

⇒2 (√2)n-1 = 128√2

⇒ (√2)n-1 = 64√2

⇒ (√2)n-1 = (√2)13

n-1 = 13

n = 14

Thus, the 14th term of the given G.P. is 128√2.

Which term of the G.P. :

√2,1/√2, 1/2√2 ,1/4√2, ... is 1/256√2?

જવાબ : Here, first term, a= √2  and common ratio, r=1/2

Let the nth term be 1/256√2.

∴ an = 1/256√2

⇒ arn-1 = 1/256√2

⇒ (√2)(1/2)n-1 = 1/256√2

⇒ (1/2)n-1 = 1/512

⇒ (1/2)n-1 = (1/2)9

n-1 = 9

n=10

Thus, the 10th term of the given G.P. is 1/256√2.

Which term of the A.P. 3, 8, 13, ... is 243?

જવાબ : 3, 8, 13...
Here, we have:
a = 3
d = (8-3)  =5
Let an = 243

⇒ a+(n-1)d = 243

⇒3+(n-1)5 = 243

⇒(n-1)5 =240

n-1 = 48

⇒ n= 49

Hence, 243 is the 49th term of the given A.P.

Which term of the A.P. 80, 76, ... is 0?

જવાબ : 84, 80, 76...
Here, we have:
a = 80
d = (76-80) = -4
Let an =0

a+(n-1)d = 0

⇒80 + (n-1)(-4) = 0

⇒ (n-1)(-4) = -80

⇒(n-1) = 20⇒ n = 21

Hence, 0 is the 21nd term of the given A.P.

Which term of the A.P. 4, 9, 14, ... is 249?

જવાબ : 4, 9, 14...
Here, we have:
a = 4
d = (9-4) = 5
Let an = 249

a+(n-1) d = 249

⇒4+(n-1) 5 = 249

⇒(n-1) 5 =245

⇒(n-1) = 49⇒ n= 50

Hence, 249 is the 50th term of the given A.P.

Which term of the A.P. 3, 8, 13, ... is 253?

જવાબ : 3, 8, 13...
Here, we have:
a = 3
d = (8-3)  =5
Let an = 253

⇒ a+(n-1)d = 253

⇒3+(n-1)5 = 253

⇒(n-1)5 =250

n-1 = 40

⇒ n= 51

Hence, 253 is the 51th term of the given A.P.

Which term of the A.P. 80, 76, ... is 4?

જવાબ : 84, 80, 76...
Here, we have:
a = 80
d = (76-80) = -4
Let an =4

a+(n-1)d = 4

⇒80 + (n-1)(-4) = 4

⇒ (n-1)(-4) = -76

⇒(n-1) = 19

⇒ n = 20

Hence, 4 is the 20th term of the given A.P.

Which term of the A.P. 4, 9, 14, ... is 259?

જવાબ : 4, 9, 14...
Here, we have:
a = 4
d = (9-4) = 5
Let an = 259

a+(n-1) d = 259

⇒4+(n-1) 5 = 259

⇒(n-1) 5 =255

⇒(n-1) = 51⇒ n= 52

Hence, 255 is the 52th term of the given A.P.

Is 67 a term of the A.P. 7, 10, 13, ...?

જવાબ : 7, 10, 13...
Here, we have:
a = 7
d = (10-7) = 3

Let an = 67

⇒ a+(n-1) d = 67

⇒7+(n-1)(3) = 67

⇒(n-1)(3) = 60

⇒(n-1) = 60/3

n =  20 +1 = 21

Since n is a natural number. So, 67 is a term of the given A.P.

Is 301 a term of the A.P. 3, 8, 13, ...?

જવાબ : 3, 8, 13...
Here, we have:
a  = 3
d = (8-3)=5

Let an = 301

⇒ a+(n-1)d = 301

⇒3+(n-1)5 = 301

⇒(n-1)5 =298

⇒(n-1) = 298/5

n = 298/5 + 1 = 303/5
Since n is not a natural number.So, 301 is not a term of the given A.P.

Is 69 a term of the A.P. 7, 10, 13, ...?

જવાબ : 7, 10, 13...
Here, we have:
a = 7
d = (10-7) = 3

Let an = 69

⇒ a+(n-1) d = 69

⇒7+(n-1)(3) = 69

⇒(n-1)(3) = 62

⇒(n-1) = 62/3

n =  62/3 +1 = 65/3

Since n is not a natural number. So, 69 is not a term of the given A.P.

Is 303 a term of the A.P. 3, 8, 13, ...?

જવાબ : 3, 8, 13...
Here, we have:
a  = 3
d = (8-3)=5

Let an = 303

a+(n-1)d = 303

3+(n-1)5 = 303

(n-1)5 =300

(n-1) = 300/5

n = 60 + 1 = 61
Since n is a natural number. So, 303 is a term of the given A.P.