For a G.P., $a=\dfrac43$ and $t_7=\dfrac{243}{1024}$, find the value of $r$.

For a G.P., a=dfrac43 and t₇=(243)/(1024), find the value of r.

MISCELLANEOUS EXERCISE - 2 (II) · Q106

a=\dfrac43

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Concept understanding — General Term of a G.P.

General Term of a Geometric Progression

The Intuition: What a G.P. feels like

Imagine you start with a single grain of rice on the first square of a chessboard. On the second square, you put 2 grains. On the third, 4 grains. On the fourth, 8 grains. Each time, you multiply the previous number by 2.

That is a geometric progression. The defining rule is simple: you get the next term by multiplying the current term by a fixed number (called the common ratio, usually denoted $r$ ). In the rice example, $r = 2$ .

So if the first term is $a$ , the sequence looks like:

a, \ ar, \ ar^2, \ ar^3, \ ar^4, \ \dots

Notice the pattern: the exponent on $r$ is always one less than the position of the term. The 1st term has $r^0$ , the 2nd has $r^1$ , the 3rd has $r^2$ , and so on.

Tip

A quick way to see this: the 5th term is $a$ multiplied by $r$ four times — that's $a \times r^4$ . The exponent is always (position − 1).

The Precise Statement

For a geometric progression with first term $a$ and common ratio $r$ ( $r \neq 0$ ), the $n$ th term (also called the general term) is given by:

T_n = a \, r^{\,n-1}

where $n$ is a positive integer ( $n = 1, 2, 3, \dots$ ).

T_n = a r^{n-1}

Why This Formula Works

Let's verify it for the first few terms:

$n = 1$ : $T_1 = a r^{0} = a$ (correct)
$n = 2$ : $T_2 = a r^{1} = ar$ (correct)
$n = 3$ : $T_3 = a r^{2} = ar^2$ (correct)

The formula is simply a compact way of saying: "start with $a$ , then multiply by $r$ exactly $(n-1)$ times."

A Common Mistake to Avoid

Watch out

Many students write $T_n = a r^n$ by mistake. That would give $T_1 = ar$ , which is wrong. The exponent is always $n-1$ , not $n$ .

What If $r$ Is Negative or a Fraction?

The formula works for any real $r$ (except $r=0$ , which makes the sequence trivial). If $r = -2$ , the terms alternate sign: $a, -2a, 4a, -8a, \dots$ . If $r = \frac{1}{3}$ , the terms shrink: $a, \frac{a}{3}, \frac{a}{9}, \frac{a}{27}, \dots$ . The same $T_n = a r^{n-1}$ handles both cases perfectly.

A Quick Example

Find the 10th term of the G.P.: $3, 6, 12, 24, \dots$

Here $a = 3$ and $r = 6/3 = 2$ . Using $T_n = a r^{n-1}$ :

T_{10} = 3 \times 2^{9} = 3 \times 512 = 1536

The 10th term is 1536.

The Big Picture

The general term is the DNA of the progression — it tells you the value at any position without having to list all the previous terms. Once you know $a$ and $r$ , you know the entire infinite sequence. This idea (describing a pattern with a single formula) is one of the most powerful in mathematics.

$t_7=ar^6=\dfrac{243}{1024} \Rightarrow r^6=\dfrac{243/1024}{4/3}=\dfrac{729}{4096}$ . Since $729=3^6$ and $4096=4^6$ , $r^6=\left(\dfrac34\right)^6 \Rightarrow r=\dfrac34$ .

✓Final answer

r=\dfrac34

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Q.For a G.P., a=43a=\dfrac43a=34​ and t7=2431024t_7=\dfrac{243}{1024}t7​=1024243​, find the value of rrr.

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Q.For a G.P., $a=\dfrac43$ and $t_7=\dfrac{243}{1024}$ , find the value of $r$ .

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