a) A parent places in the saving bank Rs. 500 on his child's first birth, Rs. 1000 on his second, Rs. 1500 on his third and so on increasing the amount by Rs. 500 on each birthday. How much will he…

a) Savings Calculation Using Arithmetic Progression (AP)

Given:

• The parent deposits money in an arithmetic progression (AP) on each birthday.
• First deposit (a₁) = Rs. 500 (on the 1st birthday)
• Second deposit (a₂) = Rs. 1000 (on the 2nd birthday)
• Third deposit (a₃) = Rs. 1500 (on the 3rd birthday)
• Common difference (d) = Rs. 500 (since each deposit increases by Rs. 500)
• Number of deposits (n) = 16 (from 1st to 16th birthday, inclusive)

Objective: Find the total savings after 16 deposits.

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Step 1: Identify the AP Parameters

The deposits form an arithmetic sequence (AP) where:

• First term, $a_1=500$
• Common difference, $d=500$
• Number of terms, $n=16$

The general form of an AP is: $$a_n=a_1+(n-1)d$$

Step 2: Find the Last Term (a₁₆)

Using the formula for the $n$-th term of an AP: $$a_{16}=a_1+(16-1)d=500+15\times 500=500+7500=7500$$ So, the deposit on the 16th birthday is Rs. 7500.

Step 3: Calculate the Total Savings (Sum of AP)

The sum of the first $n$ terms of an AP is given by: $$S_n=\frac{n}{2}(a_1+a_n)$$ Substituting the known values: $$S_{16}=\frac{16}{2}(500+7500)=8\times 8000=64000$$

Final Answer: The parent will have saved a total of Rs. 64,000 by the child’s 16th birthday.

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b) Inserting 4 Geometric Means Between $\frac{2}{3}$ and $5\frac{1}{16}$

Given:

• First term ($a$) = $\frac{2}{3}$
• Last term ($l$) = $5\frac{1}{16}=\frac{81}{16}$ (converted to improper fraction)
• Number of geometric means to insert = 4
• Total number of terms in the sequence = $1+4+1=6$

Objective: Find the 4 geometric means between $\frac{2}{3}$ and $\frac{81}{16}$.

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Step 1: Understand the Geometric Sequence (GP)

A geometric sequence has the form: $$a,ar,a{r}^2,a{r}^3,\dots ,a{r}^{n-1}$$ where:

• $a$ = first term
• $r$ = common ratio
• $n$ = total number of terms

Here, the last term ($l$) is the 6th term: $$l=a{r}^{n-1}$$ $$\frac{81}{16}=\frac{2}{3}\times r^5$$

Step 2: Solve for the Common Ratio ($r$)

$$\frac{81}{16}=\frac{2}{3}\times r^5$$ $$r^5=\frac{81}{16}\times \frac{3}{2}=\frac{243}{32}$$ $$r={\left(\frac{243}{32}\right)}^{\frac{1}{5}}=\frac{3}{2}$$ (Since $3^5=243$ and $2^5=32$, $r=\frac{3}{2}$)

Step 3: Write the Complete Geometric Sequence

The sequence with 4 geometric means inserted will have 6 terms: $$a_1=\frac{2}{3},a_2=\frac{2}{3}\times \frac{3}{2}=1,a_3=1\times \frac{3}{2}=\frac{3}{2},a_4=\frac{3}{2}\times \frac{3}{2}=\frac{9}{4},a_5=\frac{9}{4}\times \frac{3}{2}=\frac{27}{8},a_6=\frac{27}{8}\times \frac{3}{2}=\frac{81}{16}$$

Final Answer: The four geometric means between $\frac{2}{3}$ and $5\frac{1}{16}$ are: 1, $\frac{3}{2}$, $\frac{9}{4}$, $\frac{27}{8}$ (or 1, 1.5, 2.25, 3.375 in decimal form).