CBSE Class 11 Maths Sequence and Series NCERT Solutions

Let the first term of the AP be $a$ and the common difference be $d$. \nAccording to the problem statement, the sum of the first $p$ terms ($S\_p$) is 0.

Using the formula for the sum of an AP, $S\_n\; =\; \backslash frac\{n\}\{2\}[2a\; +\; (n-1)d]$, we have:

$$S\_p\; =\; \backslash frac\{p\}\{2\}[2a\; +\; (p-1)d]\; =\; 0$$

Since $p\; \backslash neq\; 0$ (as it represents the number of terms), we can infer that:

$$2a\; +\; (p-1)d\; =\; 0$$

From this equation, we can express the common difference $d$ in terms of $a$ and $p$:

$$d\; =\; \backslash frac\{-2a\}\{p-1\}$$

Now, we need to find the sum of the next $q$ terms. This sum is equal to the sum of the first $(p+q)$ terms minus the sum of the first $p$ terms. \nSum of next $q$ terms = $S\_\{p+q\}\; -\; S\_p$.

Since $S\_p\; =\; 0$, the sum of the next $q$ terms is simply $S\_\{p+q\}$.

Using the sum formula for $(p+q)$ terms:

$$S\_\{p+q\}\; =\; \backslash frac\{p+q\}\{2\}[2a\; +\; (p+q-1)d]$$

We can rewrite $(p+q-1)d$ as $(p-1)d\; +\; qd$. Substituting this into the equation:

$$S\_\{p+q\}\; =\; \backslash frac\{p+q\}\{2\}[2a\; +\; (p-1)d\; +\; qd]$$

We know from the earlier step that $2a\; +\; (p-1)d\; =\; 0$. Substitute this:

$$S\_\{p+q\}\; =\; \backslash frac\{p+q\}\{2\}[0\; +\; qd]$$

$$S\_\{p+q\}\; =\; \backslash frac\{p+q\}\{2\}[qd]$$

Now, substitute the expression for $d\; =\; \backslash frac\{-2a\}\{p-1\}$ into this equation:

$$S\_\{p+q\}\; =\; \backslash frac\{p+q\}\{2\}\; \backslash left[\; q\; \backslash left(\; \backslash frac\{-2a\}\{p-1\}\; \backslash right)\; \backslash right]$$

$$S\_\{p+q\}\; =\; \backslash frac\{p+q\}\{2\}\; \backslash left[\; \backslash frac\{-2aq\}\{p-1\}\; \backslash right]$$

Simplifying the expression:

$$S\_\{p+q\}\; =\; \backslash frac\{-a(p+q)q\}\{p-1\}$$

Thus, the sum of the next $q$ terms is $\backslash frac\{-a(p+q)q\}\{p-1\}$.

This problem describes a situation where the annual savings form an Arithmetic Progression (AP) because there is a constant increase each year. \nLet the amount saved in the first year be $a$ (in ₹). \nThe increase in savings each year is ₹ 200, so the common difference $d\; =\; 200$. \nThe total number of years is $n\; =\; 20$. \nThe total amount saved over 20 years is $S\_\{20\}\; =\; 66000$.

We use the formula for the sum of an AP: $S\_n\; =\; \backslash frac\{n\}\{2\}[2a\; +\; (n-1)d]$.

Substituting the given values:

$$S\_\{20\}\; =\; \backslash frac\{20\}\{2\}[2a\; +\; (20-1)d]$$

$$66000\; =\; 10[2a\; +\; 19\; \backslash times\; 200]$$

Now, simplify and solve for $a$:

$$66000\; =\; 10[2a\; +\; 3800]$$

Divide both sides by 10:

$$6600\; =\; 2a\; +\; 3800$$

Subtract 3800 from both sides:

$$2a\; =\; 6600\; -\; 3800$$

$$2a\; =\; 2800$$

Divide by 2:

$$a\; =\; \backslash frac\{2800\}\{2\}\; =\; 1400$$

Therefore, the man saved ₹ 1400 in the first year.

The monthly salary increases by a fixed amount each month, which means the salaries form an Arithmetic Progression (AP). \nHere, the first term (initial salary) is a = 5200. \nThe common difference (monthly increment) is d = 320.

(i) Salary for the tenth month:

To find the salary for the tenth month, we need to calculate the 10th term of the AP (a_{10}). \nWe use the formula for the n-th term of an AP: a_n = a + (n-1)d.

For the tenth month, n = 10:

a_{10} = 5200 + (10-1) \times 320

a_{10} = 5200 + 9 \times 320

a_{10} = 5200 + 2880

a_{10} = 8080

So, his salary for the tenth month is ₹ 8080.

(ii) Total earnings during the first year:

The first year consists of 12 months, so n = 12. We need to find the sum of the salaries for these 12 months (S_{12}).

We use the formula for the sum of an AP: S_n = \frac{n}{2}[2a + (n-1)d].

Substituting the values for the first year:

S_{12} = \frac{12}{2}[2 \times 5200 + (12-1) \times 320]

S_{12} = 6[10400 + 11 \times 320]

S_{12} = 6[10400 + 3520]

S_{12} = 6[13920]

S_{12} = 83520

Therefore, his total earnings during the first year are ₹ 83520.

Let the first term of the Geometric Progression (GP) be a and the common ratio be r.

The formula for the n-th term of a GP is a_n = a \cdot r^{n-1}.

According to the problem statement:

The p-th term is q: a \cdot r^{p-1} = q ... (i)

The q-th term is p: a \cdot r^{q-1} = p ... (ii)

To find the common ratio r, we divide equation (i) by equation (ii):

\frac{a \cdot r^{p-1}}{a \cdot r^{q-1}} = \frac{q}{p}

r^{(p-1) - (q-1)} = \frac{q}{p}

r^{p-q} = \frac{q}{p}

From this, we can express r as:

r = \left(\frac{q}{p}\right)^{\frac{1}{p-q}}

Now, we need to find the (p+q)-th term, which is a_{p+q} = a \cdot r^{p+q-1}.

To find a, we can use equation (i) or (ii). Let's use (i): a = \frac{q}{r^{p-1}}.

Substitute the expression for r into the expression for a:

a = \frac{q}{\left[\left(\frac{q}{p}\right)^{\frac{1}{p-q}}\right]^{p-1}} = \frac{q}{\left(\frac{q}{p}\right)^{\frac{p-1}{p-q}}}

Now, let's find the (p+q)-th term a_{p+q} = a \cdot r^{p+q-1}:

a_{p+q} = \left(\frac{q}{r^{p-1}}\right) \cdot r^{p+q-1}

a_{p+q} = q \cdot r^{(p+q-1) - (p-1)}

a_{p+q} = q \cdot r^{p+q-1-p+1}

a_{p+q} = q \cdot r^{q}

Substitute the expression for r = \left(\frac{q}{p}\right)^{\frac{1}{p-q}} into this equation:

a_{p+q} = q \cdot \left[\left(\frac{q}{p}\right)^{\frac{1}{p-q}}\right]^{q}

a_{p+q} = q \cdot \left(\frac{q}{p}\right)^{\frac{q}{p-q}}

This expression needs to be manipulated to match the required form \left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}. Let's re-evaluate the approach using powers of q/p.

We have r^{p-q} = \frac{q}{p}.

We want to show a_{p+q} = \left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}.

Let's express a and r in terms of p and q more directly.

From (i), a = q \cdot r^{-(p-1)}.

From (ii), a = p \cdot r^{-(q-1)}.

Consider a_{p+q} = a \cdot r^{p+q-1}.

Let's raise equation (i) to the power q and equation (ii) to the power p:

(i)^q: (a \cdot r^{p-1})^q = q^q \Rightarrow a^q \cdot r^{q(p-1)} = q^q

(ii)^p: (a \cdot r^{q-1})^p = p^p \Rightarrow a^p \cdot r^{p(q-1)} = p^p

This approach seems complicated. Let's use the derived r value.

We found r = \left(\frac{q}{p}\right)^{\frac{1}{p-q}}.

Let's find a using a = q \cdot r^{-(p-1)}:

a = q \cdot \left[\left(\frac{q}{p}\right)^{\frac{1}{p-q}}\right]^{-(p-1)} = q \cdot \left(\frac{q}{p}\right)^{-\frac{p-1}{p-q}}

Now, a_{p+q} = a \cdot r^{p+q-1}:

a_{p+q} = q \cdot \left(\frac{q}{p}\right)^{-\frac{p-1}{p-q}} \cdot \left(\frac{q}{p}\right)^{\frac{1}{p-q} \cdot (p+q-1)}

a_{p+q} = q \cdot \left(\frac{q}{p}\right)^{\frac{-(p-1) + (p+q-1)}{p-q}}

a_{p+q} = q \cdot \left(\frac{q}{p}\right)^{\frac{-p+1+p+q-1}{p-q}} = q \cdot \left(\frac{q}{p}\right)^{\frac{q}{p-q}}

This is the same expression we got before. Let's try to manipulate it.

a_{p+q} = q \cdot \frac{q^{\frac{q}{p-q}}}{p^{\frac{q}{p-q}}} = \frac{q^{1 + \frac{q}{p-q}}}{p^{\frac{q}{p-q}}} = \frac{q^{\frac{p-q+q}{p-q}}}{p^{\frac{q}{p-q}}} = \frac{q^{\frac{p}{p-q}}}{p^{\frac{q}{p-q}}}

Now, let's look at the target expression: \left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}

\left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}} = \frac{(q^p)^{\frac{1}{p-q}}}{(p^q)^{\frac{1}{p-q}}} = \frac{q^{\frac{p}{p-q}}}{p^{\frac{q}{p-q}}}

The expressions match. Thus, the (p+q)-th term is \left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}.