Question

Let a1, a2 ,a3 ...... be terms of an A.P. If (a1+a2+a3+...+ap) / (a1+a2+a3+...+aq)=p2/q2 ,P not equal to q, then a6/a21 equals to

a) 41/11 b)7/2 c)2/7 d)11/41

Answer 1

Let a1, a2, a3....be terms of an AP. If (a1 +a2.....ap)/(a1 ... a2, a3....be terms of an AP.

If (a1 +a2.....ap)/(a1 +a2.....aq) =p^2/q^2.

p is not equal to q. Then a6/a21 must be. A) 2/7 B) 11/41 C) 7/2 ...

q(a1+(p-1)d)=p(a1+(q-1)d).

(p-q)a1=(q-p)d.

a1=-d. a6/a21=(a1+5d)/(a1+20d).

=-4a1/-19a1.

=4/19. It is equal to terms that has to identify with A and P.

Answer 2

The sum of n terms of an AP with first term aa and common difference d is
Sn=n/2[2a+(n-1)d]
So,
Sp=p/2[2a+(p-1)d]
Sq=q/2[2a+(q-1)d]
It is given that,
Sq/Sp=q^2/p^2
2q[2a+(q−1)d]}/2p[2a+(p−1)d]=q2/p2
(2a+(q−1)d)/(2a+(p−1)d)=q/p
[2a+(p−1)d]q=[2a+(q−1)d]p
2a(q−p)=d(q−p)
2a=d
Now,
a21/a6=a+20d/a+5d
a21/a6=a+20(2a)/a+5(2a)
a21/a6=41a/11a
a21/a6=41/11

a6/a21=11/41