The 1^st and 6^th terms of a G.P are 153 and 17/27. Find the sum of the first four terms.

The first term is $T_1 = a$. We are given $T_1 = 153$. So, $a = 153$.
The sixth term is $T_6 = a \cdot r^{6-1} = a \cdot r^5$. We are given $T_6 = \frac{17}{27}$.
So, we have the equation:

$$a \cdot r^5 = \frac{17}{27}$$

Substitute the value of $a=153$ into the equation from Step 1:

$$153 \cdot r^5 = \frac{17}{27}$$

To isolate $r^5$, divide both sides by 153:

$$r^5 = \frac{17}{27 \times 153}$$

We can simplify the fraction. Notice that $153 = 9 \times 17$.

$$r^5 = \frac{17}{27 \times (9 \times 17)}$$

Cancel out the 17 from the numerator and denominator:

$$r^5 = \frac{1}{27 \times 9}$$

Express the denominators as powers of 3: $27 = 3^3$ and $9 = 3^2$.

$$r^5 = \frac{1}{3^3 \times 3^2}$$

Combine the powers of 3 in the denominator:

$$r^5 = \frac{1}{3^{3+2}}$$ $$r^5 = \frac{1}{3^5}$$

Since $r^5 = \left(\frac{1}{3}\right)^5$, we can conclude that:

$$r = \frac{1}{3}$$

The formula for the sum of the first $n$ terms of a GP is $S_n = \frac{a(1 - r^n)}{1 - r}$. We use this form because $|r| = \left|\frac{1}{3}\right| < 1$.
We have $a=153$, $r=\frac{1}{3}$, and we want to find $S_4$, so $n=4$.
Substitute these values into the formula:

$$S_4 = \frac{153 \left(1 - \left(\frac{1}{3}\right)^4\right)}{1 - \frac{1}{3}}$$

First, calculate $\left(\frac{1}{3}\right)^4$:

$$\left(\frac{1}{3}\right)^4 = \frac{1^4}{3^4} = \frac{1}{81}$$

Next, calculate the denominator $1 - \frac{1}{3}$:

$$1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}$$

Now substitute these back into the $S_4$ formula:

$$S_4 = \frac{153 \left(1 - \frac{1}{81}\right)}{\frac{2}{3}}$$

Calculate the term in the parenthesis:

$$1 - \frac{1}{81} = \frac{81}{81} - \frac{1}{81} = \frac{80}{81}$$

Substitute this back:

$$S_4 = \frac{153 \left(\frac{80}{81}\right)}{\frac{2}{3}}$$

To simplify, we can multiply by the reciprocal of the denominator:

$$S_4 = 153 \times \frac{80}{81} \times \frac{3}{2}$$

Now, perform the multiplication and simplification:

$$S_4 = \frac{153 \times 80 \times 3}{81 \times 2}$$

We can simplify by dividing common factors. For example, $81 = 3 \times 27$, and $153 = 3 \times 51 = 9 \times 17$.

$$S_4 = \frac{(9 \times 17) \times (2 \times 40) \times 3}{(9 \times 9) \times 2}$$

Cancel out common factors (9 and 2):

$$S_4 = \frac{17 \times 40 \times 3}{9}$$

Further simplify by dividing 3 from the numerator and 9 from the denominator:

$$S_4 = \frac{17 \times 40 \times 1}{3}$$ $$S_4 = \frac{680}{3}$$

The sum of the first four terms is $\boxed{\frac{680}{3}}$.

7. If $T_2=35$ and $T_4=875$ respectively. Find the first and fifth terms.

• Understanding the problem: We are given the second term ($T_2$) and the fourth term ($T_4$) of a Geometric Progression (GP). We need to find the first term ($a$) and the fifth term ($T_5$). We will use the general term formula $T_n = a \cdot r^{n-1}$ to set up equations and solve for $a$ and $r$.

For the second term ($n=2$):

$$T_2 = a \cdot r^{2-1} = a \cdot r = 35 \quad \text{(Equation 1)}$$

For the fourth term ($n=4$):

$$T_4 = a \cdot r^{4-1} = a \cdot r^3 = 875 \quad \text{(Equation 2)}$$

To find $r$, we can divide Equation 2 by Equation 1. This eliminates $a$:

$$\frac{a \cdot r^3}{a \cdot r} = \frac{875}{35}$$

Simplify both sides:

$$r^{3-1} = \frac{875}{35}$$ $$r^2 = 25$$

To solve for $r$, take the square root of both sides. Remember that a square root can be positive or negative:

$$r = \pm \sqrt{25}$$ $$r = \pm 5$$

This means there are two possible values for the common ratio, $r=5$ and $r=-5$. We need to find the first and fifth terms for each case.

• Case 1: When $r=5$ Substitute $r=5$ into Equation 1 ($a \cdot r = 35$):

  $$a \cdot 5 = 35$$

  Divide by 5 to find $a$:

  $$a = \frac{35}{5}$$ $$a = 7$$

  So, the first term is $T_1 = a = 7$. Now, find the fifth term ($T_5$) using $T_n = a \cdot r^{n-1}$ with $a=7$, $r=5$, and $n=5$:

  $$T_5 = a \cdot r^{5-1} = a \cdot r^4$$ $$T_5 = 7 \cdot (5)^4$$

  Calculate $5^4$: $5 \times 5 \times 5 \times 5 = 25 \times 25 = 625$.

  $$T_5 = 7 \cdot 625$$ $$T_5 = 4375$$

• Case 2: When $r=-5$ Substitute $r=-5$ into Equation 1 ($a \cdot r = 35$):

  $$a \cdot (-5) = 35$$

  Divide by -5 to find $a$:

  $$a = \frac{35}{-5}$$ $$a = -7$$

  So, the first term is $T_1 = a = -7$. Now, find the fifth term ($T_5$) using $T_n = a \cdot r^{n-1}$ with $a=-7$, $r=-5$, and $n=5$:

  $$T_5 = a \cdot r^{5-1} = a \cdot r^4$$ $$T_5 = -7 \cdot (-5)^4$$

  Calculate $(-5)^4$: $(-5) \times (-5) \times (-5) \times (-5) = 25 \times 25 = 625$. (An even power of a negative number is positive).

  $$T_5 = -7 \cdot 625$$ $$T_5 = -4375$$

Therefore, there are two possible solutions:

• If $r=5$, the first term is $\boxed{7}$ and the fifth term is $\boxed{4375}$.
• If $r=-5$, the first term is $\boxed{-7}$ and the fifth term is $\boxed{-4375}$.