Q: The sum of first three terms of a Geometric Progression (G.P.) is $\frac{21}{2}$ and their product is 27. Given $r < 1$, find its: common ratio. first three terms.

Step 1 Let the three terms of G.P. be $\frac{a}{r}, a, ar$. Product $= \left(\frac{a}{r}\right)(a)(ar) = a^3 = 27 \implies \mathbf{a = 3}$. [1 Mark] Step 2 Sum of three terms $= \frac{21}{2}$: $\frac{3}{r} + 3 + 3r = \frac{21}{2} \implies 3\left(\frac{1}{r} + 1 + r\right) = \frac{21}{2} \implies \frac{r^2 + r + 1}{r} = \frac{7}{2}$ $2r^2 + 2r + 2 = 7r \implies 2r^2 - 5r + 2 = 0 \implies (2r - 1)(r - 2) = 0$. $\implies r = \frac{1}{2}$ or $r = 2$. Since given $r < 1$, choose $\mathbf{r = \frac{1}{2}}$. [1 Mark] Step 3 Determine the three terms: $T_1 = \frac{a}{r} = \frac{3}{1/2} = \mathbf{6}$ $T_2 = a = \mathbf{3}$ $T_3 = ar = 3 \times \frac{1}{2} = \mathbf{\frac{3}{2}}$ (or $1.5$). Answer: Common ratio $r = \frac{1}{2}$, terms are $6, 3, \frac{3}{2}$ . [1 Mark]

Q: The 4th term of a G.P. is 16 and the 7th term is 128. Find the first term and common ratio of the series.

Step 1 General term $T_n = a r^{n-1}$: $$T_4 = a r^3 = 16 \quad \text{--- (1)}$$ [1 Mark] $$T_7 = a r^6 = 128 \quad \text{--- (2)}$$ [1 Mark] Step 2 Divide (2) by (1): $$\frac{a r^6}{a r^3} = \frac{128}{16} \implies r^3 = 8 \implies \mathbf{r = 2}$$ Step 3 Substitute $r = 2$ in (1): $$a(2^3) = 16 \implies 8a = 16 \implies \mathbf{a = 2}$$ [1 Mark] Answer: First term $a = 2$, Common ratio $r = 2$.

Question 1

The 5th term of the Geometric Progression (G.P.) 2, 6, 18, 54, ... is:

Accepted Answer

Correct Option: (a) 162 GP Term a = 2, common ratio r = 6/2 = 3. T₅ = a × r⁴ = 2 × (3⁴) = 2 × 81 = 162 .

Question 2

The fourth term of a Geometric Progression (G.P.) is 16 and its seventh term is 128. Find its: (a) common ratio. (b) first term.

Accepted Answer

Setup T₄ = ar³ = 16 ... (1)
T₇ = ar⁶ = 128 ... (2) [1 Mark]
Ratio Dividing (2) by (1): ar⁶ / ar³ = 128 / 16 ⇒ r³ = 8 ⇒ r = 2. [1 Mark]
First Term From (1): a(2)³ = 16 ⇒ 8a = 16 ⇒ a = 2. [1 Mark]

Question 3

If 1701 is the nth term of the Geometric Progression (G.P.) 7, 21, 63 ..., find: (a) the value of 'n', (b) hence find the sum of the 'n' terms of the G.P.

Accepted Answer

a = 7, r = 21 / 7 = 3.
(a) Tn = a r^(n−1) ⇒ 1701 = 7 × 3^(n−1) ⇒ 3^(n−1) = 243 = 3&sup5 ⇒ n − 1 = 5 ⇒ n = 6. [1.5 Marks]
(b) S6 = a(r&sup6 − 1)/(r − 1) = 7(729 − 1)/2 = 7(728)/2 = 7 × 364 = 2548. [1.5 Marks]

Question 4

15, 30, 60, 120… are in G.P. (Geometric Progression). (a) Find the n th term of this G.P. in terms of n . (b) How many terms of the above G.P. will give the sum 945?

Accepted Answer

a = 15, r = 30/15 = 2. (a) T n = a × r n −1 = 15 × 2 n −1 . [1.5 Marks] (b) S n = 15(2 n − 1) / (2 − 1) = 945 ⇒ 2 n − 1 = 63 ⇒ 2 n = 64 ⇒ n = 6 terms . [2.5 Marks]

Question 5

The 4th, 6th and the last term of a geometric progression are 10, 40 and 640 respectively. If the common ratio is positive, find the first term, common ratio and the number of terms of the series.

Accepted Answer

Step 1 Let the first term be $a$ and common ratio be $r$ ($r > 0$).
$T_4 = a r^3 = 10$
$T_6 = a r^5 = 40$
Dividing: $\frac{T_6}{T_4} = \frac{a r^5}{a r^3} = r^2 = \frac{40}{10} = 4 \implies \mathbf{r = 2}$ (since $r > 0$). [1.5 Marks]
Step 2 Find first term $a$:
$a r^3 = 10 \implies a (2)^3 = 10 \implies 8a = 10 \implies \mathbf{a = \frac{5}{4} = 1.25}$. [1 Mark]
Step 3 Find number of terms $n$:
Last term $T_n = a r^{n-1} = 640 \implies \frac{5}{4} \times 2^{n-1} = 640$
$$2^{n-1} = \frac{640 \times 4}{5} = 128 \times 4 = 512 = 2^9 \implies n - 1 = 9 \implies \mathbf{n = 10}$$
Answer: First term $a = \frac{5}{4}$, common ratio $r = 2$, number of terms $n = 10$. [1.5 Marks]

Question 6

The first and last term of a Geometrical Progression (G.P.) are 3 and 96 respectively. If the common ratio is 2, find: '$n$' the number of terms of the G.P. Sum of the $n$ terms.

Accepted Answer

Step 1 State parameters: $a = 3$, $l = T_n = 96$, $r = 2$.
Formula for $n\text{th}$ term: $T_n = a r^{n-1}$.
$96 = 3(2^{n-1}) \implies 2^{n-1} = \frac{96}{3} = 32$.
Since $32 = 2^5$: $n - 1 = 5 \implies \mathbf{n = 6}$.
$\therefore \text{Number of terms } n = \mathbf{6}$. [1.5 Marks]
Step 2 Find sum of the $n$ terms ($S_6$):
Formula: $S_n = \frac{a(r^n - 1)}{r - 1}$ for $r > 1$.
$S_6 = \frac{3(2^6 - 1)}{2 - 1} = \frac{3(64 - 1)}{1} = 3 \times 63 = \mathbf{189}$.
$\therefore \text{Sum of } 6 \text{ terms} = \mathbf{189}$. [1.5 Marks]

Question 7

The sum of first three terms of a Geometric Progression (G.P.) is $\frac{21}{2}$ and their product is 27. Given $r < 1$, find its: common ratio. first three terms.

Accepted Answer

Step 1 Let the three terms of G.P. be $\frac{a}{r}, a, ar$.
Product $= \left(\frac{a}{r}\right)(a)(ar) = a^3 = 27 \implies \mathbf{a = 3}$. [1 Mark]
Step 2 Sum of three terms $= \frac{21}{2}$:
$\frac{3}{r} + 3 + 3r = \frac{21}{2} \implies 3\left(\frac{1}{r} + 1 + r\right) = \frac{21}{2} \implies \frac{r^2 + r + 1}{r} = \frac{7}{2}$
$2r^2 + 2r + 2 = 7r \implies 2r^2 - 5r + 2 = 0 \implies (2r - 1)(r - 2) = 0$.
$\implies r = \frac{1}{2}$ or $r = 2$.
Since given $r < 1$, choose $\mathbf{r = \frac{1}{2}}$. [1 Mark]
Step 3 Determine the three terms:
$T_1 = \frac{a}{r} = \frac{3}{1/2} = \mathbf{6}$
$T_2 = a = \mathbf{3}$
$T_3 = ar = 3 \times \frac{1}{2} = \mathbf{\frac{3}{2}}$ (or $1.5$).
Answer: Common ratio $r = \frac{1}{2}$, terms are $6, 3, \frac{3}{2}$. [1 Mark]

Question 8

The 4th term of a G.P. is 16 and the 7th term is 128. Find the first term and common ratio of the series.

Accepted Answer

Step 1 General term $T_n = a r^{n-1}$:
$$T_4 = a r^3 = 16 \quad \text{--- (1)}$$ [1 Mark]
$$T_7 = a r^6 = 128 \quad \text{--- (2)}$$ [1 Mark]
Step 2 Divide (2) by (1):
$$\frac{a r^6}{a r^3} = \frac{128}{16} \implies r^3 = 8 \implies \mathbf{r = 2}$$ Step 3 Substitute $r = 2$ in (1):
$$a(2^3) = 16 \implies 8a = 16 \implies \mathbf{a = 2}$$ [1 Mark]
Answer: First term $a = 2$, Common ratio $r = 2$.

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