Step-by-Step Solution
Key Concept: Case study on arithmetic progressions.
(a) How much does Reeta deposit in the 12th month? [1 Mark]
$a=500,d=100$. $a_{12}=500+11(100)=500+1100=1600$. She deposits Rs $1600$ in the $12$th month. [1.0 Mark]
(b) What is her total savings at the end of 12 months? [1 Mark]
$S_{12}=\dfrac{12}{2}[2(500)+11(100)]=6[1000+1100]=6\times2100=12600$. Her total savings are Rs $12{,}600$. [1.0 Mark]
(c) In which month will her deposit first exceed Rs 2000? [1 Mark]
$500+(n-1)100>2000\Rightarrow(n-1)100>1500\Rightarrow n-1>15\Rightarrow n>16$. So the smallest such month is the $17$th month. [1.0 Mark]
(d) If Reeta wants her total savings to reach at least Rs 50,000, is 20 months of saving under this scheme enough? Justify with a calculation. [1 Mark]
$S_{20}=\dfrac{20}{2}[2(500)+19(100)]=10[1000+1900]=10\times2900=29000$. Since Rs $29{,}000<$ Rs $50{,}000$, $20$ months is NOT enough. [1.0 Mark]
Correct Answer: $a=500,d=100$. $a_{12}=500+11(100)=500+1100=1600$. She deposits Rs $1600$ in the $12$th month. [1.0 Mark] | $S_{12}=\dfrac{12}{2}[2(500)+11(100)]=6[1000+1100]=6\times2100=12600$. Her total savings are Rs $12{,}600$. [1.0 Mark] | $500+(n-1)100>2000\Rightarrow(n-1)100>1500\Rightarrow n-1>15\Rightarrow n>16$. So the smallest such month is the $17$th month. [1.0 Mark] | $S_{20}=\dfrac{20}{2}[2(500)+19(100)]=10[1000+1900]=10\times2900=29000$. Since Rs $29{,}000<$ Rs $50{,}000$, $20$ months is NOT enough. [1.0 Mark]