A) 9 cm
B) 35 cm
C) 18 cm
D) 28 cm
Correct Answer: A
Solution :
Let AB be\[{{t}_{(m+n)}}=a+(m+n-1)d\]cm. Then\[=n+m-1-m-n+1=0\]\[pS_{2}^{2}=9{{\left[ \frac{n(n+1)(2n+1)}{6} \right]}^{2}}\] \[=9\frac{{{n}^{2}}{{(n+1)}^{2}}{{(2n+1)}^{2}}}{36}\]
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