A) 12
B) 21
C) 45
D) 54
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\[=d=2\]) |
| \[\Rightarrow \,\,\,n=\frac{p+1}{2}\] |
| \[\therefore \,\,(1+3+5+....+p)+(1+3+5+....+q)\] |
| \[=(1+3+5+....+r)\] |
| \[\Rightarrow \,\,\frac{\frac{p+1}{2}}{2}\left[ 2\times 1+\left( \frac{p+1}{2}-1 \right)2 \right]\] |
| \[+\frac{\left( \frac{q+1}{2} \right)}{2}\left[ 2\times 1+\left( \frac{q+1}{2}-1 \right)2 \right]\] |
| \[=\frac{r+1}{4}\left[ 2\times 1+\left( \frac{r+1}{2}-1 \right)2 \right]\] |
| \[\Rightarrow \,\,\,\frac{p+1}{4}[2+(p-1)]+\frac{q+1}{4}[2+(q-1)]\] |
| \[=\frac{r+1}{4}[2+r-1]\] |
| \[\Rightarrow \,\,{{(p+1)}^{2}}+{{(q+1)}^{2}}={{(r+1)}^{2}}\] |
| This is the possible only when \[p=7,q=5,r=9\] |
| \[\therefore \,\,\,p+q+r=7+5+9=21\] |
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