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(w h e r ep >6)` is `12` b. `21` c. `45` d. `54`

Options

A. 12
B. 21
C. 45
D. 54

Answer

B

Explanation

Since `n^(th)` term of `A.P=a+(n-1)d`
`:.p=1+(n-1)2`
`(because" First term = a = 1 and common difference = d = 2")`
`impliesn=(p+1)/(2)`
`:.(1+3+5+……+p)+(1+3+5+…..+q)`
`=(1+3+5+……+r)`
`implies((p+1)/(2))/(2)[2xx1+((p+1)/(2)-1)2]+(((q+1)/(2)))/(2)[2xx1+((q+1)/(2)-1)2]`
`=(r+1)/(4)[2xx1+((r+1)/(2)-1)2]`
`implies(p+1)/(4)[2+(p-1)]+(q+1)/(4)[2+(q-1)]`
`=(r+1)/(4)[2+r-1]`
`implies(p+1)^(2)+(q+1)^(2)=(r+1)^(2)`
This is the possible only when `p=7,q=5,r=9`
`:.p+q+r=7+5+9=21`