A) 10 cm
B) 32 cm
C) 6 cm
D) 2 cm
Correct Answer: B
Solution :
In\[111=n\]and\[{{t}_{1}}=a;{{t}_{n+2}}=b\], \[{{t}_{n+2}}=a+(n+2-1)d\]and \[\Rightarrow \] \[b=a+(n+2-1)d\] (Vertically opposite angles) \[\Rightarrow \] \[\frac{b-a}{n+1}=d\]. (S.A.S. similarity) \[{{1}^{st}}\] Corresponding sides are proportional \[=a+\frac{b-a}{n+1}=\frac{an+b}{n+1}\]\[=b-\frac{b-a}{n+1}=\frac{bn+a}{n+1}\] \[=\frac{n}{2}\left[ \frac{na+b}{n+1}+\frac{bn+a}{n+1} \right]\] \[=\frac{n}{2}\left[ \frac{(n+1)(a+b)}{n+1} \right]\] \[=\frac{n}{2}(a+b)\]
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