Question

Bcoz I don’t know about these questions???

Answer 1

Question - 8 :

Given A.P. : 3 , 5 , 7 , 9 , 11

• first term, a = 3
• common difference, d = 5 - 3 = 2

(i) nth term of an AP is given by,

$\underline{\boxed{\bf a_n=a+(n-1)d}}$

Substitute the values,

aₙ = 3 + (n - 1)(2)

aₙ = 3 + 2n - 2

aₙ = 2n + 1

∴ nth term = 2n + 1

(ii) 16th term,

Substitute n = 16,

a₁₆ = 2(16) + 1

a₁₆ = 32 + 1

a₁₆ = 33

∴ 16th term = 33

Question - 9 :

Given,

• 7th term, a₇ = -4
• 13th term, a₁₃ = -16

nth term of an AP is given by,

$\underline{\boxed{\bf a_n=a+(n-1)d}}$

7th term :

a₇ = a + (7 - 1)d

-4 = a + 6d ---[1]

13th term :

a₁₃ = a + (13 - 1)d

-16 = a + 12d ---[2]

equation [2] - equation [1],

a + 12d - (a + 6d) = -16 - (-4)

a + 12d - a - 6d = -16 + 4

6d = -12

 d = -12/6

 d = -2

∴ Common difference = -2

Substitute d = -2, in equation [1]

-4 = a + 6(-2)

-4 = a - 12

a = 12 - 4

a = 8

∴ first term = 8

Second term = a + d = 8 + (-2) = 8 - 2 = 6

Third term = a + 2d = 8 + 2(-2) = 8 - 4 = 4

Fourth term = a + 3d = 8 + 3(-2) = 8 - 6 = 2

The required A.P. is 8 , 6 , 4 , 2 , ...

Question - 10 :

The first two digit number divisible by 3 = 12

The second two digit number divisible by 3 = 15

The last two digit number divisible by 3 = 99

Thus we get the sequence 12 , 15 , ..... , 99 which is in A.P.

To find how many two digit numbers are divisible by 3, we have to find which term is 99.

nth term of an AP is given by,

$\underline{\boxed{\bf a_n=a+(n-1)d}}$

Put

• a = 12
• d = 15 - 12 = 3
• aₙ = 99

and solve for n

99 = 12 + (n - 1)(3)

99 - 12 = (n - 1)(3)

87 = 3n - 3

87 + 3 = 3n

3n = 90

n = 90/3

n = 30

30th term is 99

∴ The two digit numbers divisible by 3 are 30.