Question

if n terms are in between a and b in AP then find the common difference
if n terms are in between x and y in GP then find the common ratio
answer only if u know​

Answer 1

Answer

common difference for AP = $\sf\frac{b-a}{n+1}$

common ratio for GP = $\sf (\frac{b}{a})^{\frac{1}{n+1}}$

Explanation

Given, there are n terms between a and b in A.P.

⇒ Including a and b, there are a total of (n + 2) terms

This means that b is the (n + 2)th term of the A.P.

Now, we know that nth term = a + (n - 1)d

where, a = first term

n = number of term

d = common difference

Hence,

⇒ b = a + (n + 2 - 1)d

⇒ d = $\sf\frac{b-a}{n+1}$

$\rule{100}1$

Now, again for GP, b would be the (n + 2)th term

We know that, $\sf T_n = ar^{n-1}$

So, b = $\sf ar^{n +2-1}$

⇒ b = $\sf ar^{n +1}$

⇒ r = $\sf (\frac{b}{a})^{\frac{1}{n+1}}$

Answer 2

||✪✪ QUESTION ✪✪||

if n terms are in between a and b in AP then find the common difference

if n terms are in between x and y in GP then find the common ratio ?

|| ✰✰ ANSWER ✰✰ ||

AP series :- a , x1 , x2, x3, x4 , ___________ xn , b

So, Total Number of Terms in The AP series are = (n + 2) .

Now, we can see That, The last Term of AP series is b .

So,

→ (n+2)th Term = b

→ First Term = a

→ Number of terms = (n+2)

→ Let Common Difference = d

Than :-

→ Tn = a + (n - 1)d

→ b = a + [(n+2) - 1]d

→ b = a + ( n +1)d

→ (n+1)d = (b - a)

→ d = [ (b - a)/(n+1) ] (Ans).

______________________________

Similarly we can say That now,

→ GP series :- a , x1 , x2, x3, x4 , ___________ xn , b

Total Terms = (n +2) .

So,

→ (n+2)th Term = b

→ First Term = a

→ Number of terms = (n+2)

→ Let Common Ratio = r

Than :-

→ Tn = a*r^(n-1)

→ b = a * r^[(n+2) - 1)]

→ b = a * r^(n+1)

→ r^(n+1) = (b/a)

→ r = (b/a)^{1/(n+1)} (Ans).

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