Question

If (m+1)th , (n+1)th and (r+1)th terms of an A.P are in G.P and m , n , r are in H.P, then the ratio of the first term of the A.P to its common difference in terms of n is equal to ?

Answer 1

Answer:

= $-\frac{n}{2}$

Step-by-step explanation:

Let the first term of AP be a and common difference be d, then

    $t_{m+1}= a + md , t_{n+1} = a + nd\\t_{r+1} = a + rd$

Also, m , n , r in H.P

      => $n=\frac{2mr}{m + r}$

Given,

            $t_{m+1} , t_{n+1} , t_{r+1}$ are in G.P

=> ( a + nd )² = ( a + md )( a + rd )

=> a² = n²d² + 2and

    => a² + ad(m + r) + mrd²

=> n² ( d² + $\frac{2ad}{n}$) = ad × $\frac{2mr}{n} + mrd^2$

=> n² [ d² + $\frac{2ad}{n}$] = mr[$\frac{2ad}{n} + d^2$]

=> ( n² - mr )[d² + $\frac{2ad}{n}$] = 0

=> n² = mr

(OR)

=> d² = - $\frac{2ad}{n} => \frac{a}{d} = -\frac{n}{2}$

GOOD LUCK !!

Answer 2

Tm+1 = a + md

Tn+1 = a + nd

 Tr+1 = a + rd

Tm+1,Tn+1,Tr+1 are in gp therefore

(Tn+1)2  = (Tn+1)(Tr+1)

 (a+nd)2 = (a+md)(a+rd)

 (a/d +n)2 = (a/d + m)(a/d+ r)             ...............1

now   m,n,r are in hp therefore

 2/n =1/r + 1/m

 mr=n(m+r)/2.........2

let Z is the ratio of first term and common difference of AP then Z=a/d

putting Z in eq 1

 Z=mr - n2 /2n -(m+r) 

after putting value of mr from eq 2 we get

Z=a/d = -n/2