Sequence And Series Question 398
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, what is the smallest possible value of (p + q + r) where $ p>6 $ ?
Options:
A) 12
B) 21
C) 45
D) 54
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Since nth term of $ A.P=a+(n-1)d $
$ \therefore ,p=1+(n-1)2 $ ( $ \because $ First term $ =a=1 $ and common difference $ =d=2 $ )
$ \Rightarrow ,n=\frac{p+1}{2} $
$ \therefore (1+3+5+….+p)+(1+3+5+….+q) $ $ =(1+3+5+….+r) $
$ \Rightarrow \frac{\frac{p+1}{2}}{2}[ 2\times 1+( \frac{p+1}{2}-1 )2 ] $ $ +\frac{( \frac{q+1}{2} )}{2}[ 2\times 1+( \frac{q+1}{2}-1 )2 ] $ $ =\frac{r+1}{4}[ 2\times 1+( \frac{r+1}{2}-1 )2 ] $
$ \Rightarrow ,\frac{p+1}{4}[2+(p-1)]+\frac{q+1}{4}[2+(q-1)] $ $ =\frac{r+1}{4}[2+r-1] $
$ \Rightarrow {{(p+1)}^{2}}+{{(q+1)}^{2}}={{(r+1)}^{2}} $ This is the possible only when $ p=7,q=5,r=9 $
$ \therefore ,p+q+r=7+5+9=21 $
BETA
