A) 4 : 5
B) 7 : 9
C) 16 : 25
D) 1 : 1
Correct Answer: B
Solution :
\[{{S}_{n}}th=u+\frac{a}{2}\left( 2n-1 \right)\] \[\text{So, }\frac{{{S}_{4}}}{{{S}_{5}}}=\frac{0+\frac{g}{2}\left( 7 \right)}{0+\frac{g}{2}\left( 9 \right)}=\frac{7}{9}\]
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