Question

1.sum of first n term of an AP if first term is a and common difference is d=__

2.the nth term of G.P if first term is a and commom ratio is r =__

3.the geometric mean of 4 and 16 =__

4.value of 0!=__

5.sum of first term of G.P if first term is a and common ratio r>1=__​

Answer 1

1. Sum of first n terms of an AP , if first term is a and common difference is d =

$\pink{\bigstar}\ \; \sf S_n=\dfrac{n}{2}\ \big\{\ 2a+(n-1)d\ \big\}$

2. The nth term of GP , if first term is a and common ratio is r =

$\green{\bigstar}\ \; \sf a_n=ar^{n-1}$

3. The Geometric mean of 4 and 16 = 8

Geometric mean (G) b/w two numbers a and b is given by :

$\orange{\bigstar}\ \; \sf G= \sqrt{ab}$

$\implies \sf G_{ 4 \& 16}=\sqrt{4 \times 16}=\sqrt{64}=8$

4. Value of 0! = 1  

Small proof :

$\sf \red{\bigstar}\ \; n!=n \times (n-1)!$

Sub. n = 1 ,

⇒ 1! = 1 × (1 - 1)!

⇒ 1! = 1 × 0!

⇒ 1! = 0!

⇒ 0! = 1  $\red{\bigstar}$

5. Sum of first term of G.P , if first term is a and common ratio , r > 1 =

nth term of GP , $\sf a_n=ar^{n-1}$

We know that ,

S₁ = a₁

⇒ S₁ = ar¹⁻¹

⇒ S₁ = ar⁰

⇒ S₁ = a $\purple{\bigstar}$    [ ∵ n⁰ = 1 , where n ≠ 0 ]

Answer 2

Answer:

Required Answers :-

It is given that

1) First term = a

Common Difference = d

Term = n

Now,

We know that

Nth term of any AP = n/2 [2a + (n-1)d]

2) First term = a

Common Ratio = r

Term = n

$\sf \: a_n \: = ar {}^{n - 1}$

3)

$\begin{gathered} \sf \: G \: = \sqrt{ab} \\ \sf \: G = \sqrt{16 \times 4} \sf \: \\ \sf \: G = \sqrt{64} = 8 \end{gathered}$

4) Value of 0! = 1

n!=n×(n−1)!

1! = 1 × (1 - 1)!

1! = 1 × 0!

Therefore,

0! = 1

5) As we know that

$\sf \: S_1 = a_1$

$\sf \: S_1 = ar {}^{1 - 1}$

$\sf \: S_1 = ar{}^0$

$\sf \: S_1 = a$

∵ n⁰ = 1