Question

If (1 + 3 + 5+.....+p) + (1 + 3
+ 5+..+q) = (1 + 3 + 5 + .... +
r) where each set of
parentheses contains the sum
of consecutive odd integers as
shown, the smallest possible
value of p + q + r, (where p
> 6) is
(A) 12 (B) 21
(C) 45 (D) 54​

Answer 1

ANSWER

We have,

(1+3+5+...+p)+(1+3+5+...+q)=(1+3+5+...+r)⇒(1)

Sum of these AP's are:

1+3+5+...+p=4(p+1)2

1+3+5+...+q=4(q+1)2

1+3+5+...+r=4(r+1)2

Thus equation (1) becomes,

(p+1)2+(q+1)2=(r+1)2

i.e. p+1,q+1,r+1∈ Pythagoras triplet

For minimum p+q+r we must have q=5,p=7,r=9

Hence, (p+q+r)min=21

Correct option is B

Hope it helps

Mark it as BRAINLIEST if you are satisfied with the answer