A) \[{{S}_{1}}+{{S}_{2}}\]
B) \[{{S}_{1}}{{S}_{2}}\]
C) \[{{S}_{1}}-{{S}_{2}}\]
D) \[{{S}_{1}}/{{S}_{2}}\]
Correct Answer: A
Solution :
From \[S=ut+\frac{1}{2}a\ {{t}^{2}}\] \[{{S}_{1}}=\frac{1}{2}a{{(P-1)}^{2}}\] and \[{{S}_{2}}=\frac{1}{2}a\ {{P}^{2}}\] \[[As\ u=0\]] From \[{{S}_{n}}=u+\frac{a}{2}(2n-1)\] \[{{S}_{{{({{P}^{2}}-P+1)}^{th}}}}=\frac{a}{2}\left[ 2({{P}^{2}}-P+1)-1 \right]\] \[=\frac{a}{2}\left[ 2{{P}^{2}}-2P+1 \right]\] It is clear that \[{{S}_{{{({{P}^{2}}-P+1)}^{th}}}}={{S}_{1}}+{{S}_{2}}\]
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