# Geometric Progression IIT JEE Study Material

A geometric progression or a geometric sequence is a sequence of numbers in which the ratio of consecutive terms is always constant. This constant term is called the common ratio. Also, the first term of a Geometric Progression is always a non-zero number. Let, a be the 1st term and r be the common ratio of a GP, then, the general form of a geometric progression can be written as:

a, a.r, a.r$_{2}$, a.r$_{3}$, a.r$_{4}$, a.r$_{5}$, . . . . . . . . , a.r $_{n-1}$
[ Where, n = total number of terms in a GP ]

Therefore, the nth term of a GP is given by $t_{n} = a r^{n}$ – 1 where, common ratio r = $\mathbf{\frac{a_{n}}{a_{n-1}}}$

The Sum of first n terms of a Geometric Progression is given by:

Case 1: When r = 1, Sn = na [a = First term]

Case 2: When r > 1,

Sn = $\mathbf{\frac{a\;\left ( r^{n}\;-\;1 \right )}{r\;-\;1}}$

## Sum of an infinite GP:

When $\mathbf{n\; \rightarrow \;\infty }$, $\boldsymbol{r^{n}\; \rightarrow \;0}$ [ If |r| < 1 ]

Therefore, if |r| < 1, then, $\mathbf{S_{\infty }\;=\;\frac{a}{1\;-\;r}}$

## Properties of Common Ratio (r):

The value of the common ratio (r) determines whether the GP is decreasing, increasing, positive or negative.

Case 1: If r > 1, then, the GP will show an exponential growth towards positive infinity.

Case 2: If r < – 1, the GP will show an unsigned (alternating sign) exponential growth towards infinity.

Case 3: If -1 < Common Ratio (r) < 1 and r $\neq$ 0, then the GP will decay exponentially toward zero.

## Properties of a Geometric Progression:

1. If m, n, and o are any three consecutive terms of a geometric sequence, then, n$^{2}$= mo.

2. If $t_{1}, t_{2}, t_{3}, t_{4}, t_{5}, . . . . . . . t_{n – 3}, t_{n – 2}, t_{n – 1}, t_{n}$ are in Geometric Progression then, the product of the terms equidistant from the beginning and the end is constant. Also, it is equal to the product of the 1st and the last term.

$t_{1}. t_{n} = t_{2}. t_{n – 1} = t_{3}. t_{n – 2} = t_{4}. t_{n – 3}$. . . . . . and so on.

3. If $t_{1}, t_{2}, t_{3}, . . . . . . . , t_{n}$ are in GP and $t_{1}, t_{2}, t_{3}, . . . . . . ., t_{n} >0$, then $log t_{1}, log t_{2}, . . . . . , log t_{n}$are in A.P. and vice versa.

4. If each term of a Geometric Progression is either divided, multiplied or raised to power by the same non-zero number then the resulting sequence will also form a GP.

5. If $t_{1}, t_{2}, t_{3}, . . . . . . . , t_{n}$ and $p_{1}, p_{2}, p_{3}, . . . . . . . , p_{n}$ are two Geometric progressions with the common ratio as a and b respectively, then, the resulting sequence $t_{1} p_{1}, t_{2} p_{2}, t_{3} p_{3}, . . . . . . . , t_{n} p_{n}$ will also form a geometric sequence with $a \times b$ as the common ratio (r).

Similarly, $\mathbf{\frac{t_{1}}{p_{1}}, \;\frac{t_{1}}{p_{1}}, \;\frac{t_{1}}{p_{1}}\;.\;.\;.\;. }$ will also form a GP with common ratio (r) = $\mathbf{\frac{a}{b} }$.

6. Three terms of a geometric progression can be assumed as $\mathbf{\frac{a}{r}}$, a, and ar. Similarly, four terms of a geometric sequence can be taken as $\mathbf{\frac{a}{r^{2}}}$, $\mathbf{\frac{a}{r}}, ar, and, ar^{2}$.

## Geometric Progression IIT JEE Problems

Example 1: Find the mth term of a Geometric Progression if its (m + n)th term is p and (m – n)th term is q.

Solution:

Given: $T_{m} + n = a r^{(m + n – 1)} = p$ . . . . . . (1)

And, $T_{m} – n = a r ^{(m – n – 1)} = q$ . . . . . . . . . (2)

On dividing Equation (1) and (2) we will get,

i.e. $\mathbf{\frac{p}{q}\;=\;\frac{a\;r^{m\;+\;n\;-\;1}}{a\;r^{m\;-\;n\;\;1}}\;=\;r^{m\;+\;n\;-\;1\;-\;(m\;-\;n\;-\;1)}}$

Therefore, $\mathbf{\frac{p}{q}\;=\;r^{\;2n}}$

Hence, $\mathbf{r \;= \;\left ( \;\frac{p}{q}\; \right )\;^{\frac{1}{2\;n}}}$ . . . . . . (3)

Now, using equation (1) and (3) we will get,

$\mathbf{a\;\left ( \;\frac{p}{q}\; \right )\;^{\frac{m\;+\;n\;-\;1}{2\;n}}\;=\;p}$

Therefore, $\mathbf{a\;=\;\frac{p}{\left ( \;\frac{p}{q}\; \right )\;^{\frac{m\;+\;n\;-\;1}{2\;n}}}}$

Now, the mth term = a r $^{(m-1)}$

= $\mathbf{\frac{p}{\left ( \;\frac{p}{q}\; \right )\;^{\frac{m\;+\;n\;-\;1}{2\;n}}}\; \times \;\left ( \;\frac{p}{q}\; \right )\;^{\frac{1}{2\;n}\; \times \;(m\;-\;1)}}$

= $\mathbf{p\left ( \frac{p}{q} \right )^{\frac{m\;-\;1}{2n}- \frac{m+n-1}{2}}=p\left ( \frac{p}{q} \right )^{\;-\;\frac{1}{2}}\;=\; p\;^{\frac{1}{2}}\;q\;^{\frac{1}{2}}}$

Therefore, the mth term of a Geometric Progression is $\mathbf{\sqrt{p\;q}}$

Example 2: Find the common ratio (r) of a geometric progression if all its terms are positive and each term is equal to the sum of its consecutive next two terms.

Solution:

According to the given condition: $T_{n} = T_{n+1} + T_{n+2}$

i.e. $a r ^{(n – 1)} = a r ^{[(n + 1) – 1]} + a r ^{[(n + 2) – 1]}$

Or, $r ^{(n – 1)} = r ^ {n} + r ^{n + 1} = r^{n(1+r)}$

Or, $r^{n} . r ^{- 1} = r^ {n (1 + r)}$

Or, $r^{2} . r ^{- 1} = 0$

r = $\mathbf{\frac{-\;1\;\pm \;\sqrt{1\;+\;4}}{2}\;=\;\frac{-\;1\;\pm \;\sqrt{5}}{2}}$

Therefore, the common ratio (r) = $\mathbf{\;\frac{-\;1\;+ \;\sqrt{5}}{2}}$ [Since, each term of GP is positive]

Example 3: Find the product of the first five terms of a GP, if its third term is 4.

Solution:

Given: $T_{3}= a r ^{(3 – 1)} = a r^{2} = 4$

Now, the product of first five terms, $a . a r . a r^{2} . a r^{3} . a r^{4} = a^{5} r_{10} = (a r^{2})5 = 45$.

Therefore, the product of the first five terms of a GP is 1024.

Example 4: Find the sum of the series 5 + 55 + 555 + . . . . . n terms.

Solution:

5 + 55 + 555 + . . . . . . + n = $\mathbf{\frac{5}{9}}$ ( 9 + 99 + 999 + 9999 + . . . . . n terms)

= $\mathbf{\frac{5}{9}}$ (10 + 102 + 103 + 104 + . . . . n terms – (1 + 1 + 1 + . . . + n terms ) )

= $\mathbf{\frac{5}{9}\;\left [\frac{10\;(10^{n}\;-\;1)}{10\;-\;1}\;-\;n \right ]\;=\;\frac{5}{81}\;\left [ \;10^{n\;+\;1}\; – \;10\; – \;9n\; \right ]}$

Therefore, 5 + 55 + 555 + . . . . . . + n = $\mathbf{\frac{5}{81}\;\left [ \;10^{n\;+\;1}\; -\; 10\; -\; 9n\; \right ]}$

Example 5: Find the value of m if, (10)9 + 2 (11)1 (10)8 + 3 (11)2 (10)7 + 4 (11)3 (10)6 + . . . . + (11)9 = m (10)9.

Solution:

Given: (10)9 + 2 (11)1 (10)8 + 3 (11)2 (10)7 + 4 (11)3 (10)6 + . . . + (11)9 = m (10)9 . . . . . (1)

On dividing equation (1) by 109 we will get,

$\mathbf{1\; +\; 2\;\left ( \frac{11}{10} \right )}\;+\;\mathbf{3\;\left ( \frac{11}{10} \right ) ^{2}}\;+\; \mathbf{4\;\left ( \frac{11}{10} \right ) ^{3}}\; + \;. \;. \;. \; +\; \mathbf{10\;\left ( \frac{11}{10} \right ) ^{9}\;=\;m}$ . . . . (2)

Now, multiplying the above equation by $\mathbf{\frac{11}{10}}$ we get,

$\mathbf{\frac{11}{10}\; +\; 2\;\left ( \frac{11}{10} \right )^{2}}\;+\;\mathbf{3\;\left ( \frac{11}{10} \right ) ^{3}}\; + \;. \;. \;. \; +\; \mathbf{10\;\left ( \frac{11}{10} \right ) ^{10}\;=\;\left ( \frac{11}{10} \right )\;m}$ . . . . (3)

Now, Equation (3) – Equation (2) we will get,

$\mathbf{1\; +\;\left ( \frac{11}{10} \right )}\;+\;\mathbf{\left ( \frac{11}{10} \right ) ^{2}\; + \;. \;. \;. \; +\;\left ( \frac{11}{10} \right ) ^{9}\; -\;10\;\left ( \frac{11}{10} \right ) ^{10}\;=\;m\;\left (1\;-\;\frac{11}{10} \right )}$

Since, a = 1, n = 10 and r = $\mathbf{\frac{11}{10}}$

$\mathbf{\frac{\left [ \;\left ( \frac{11}{10} \right )^{10} \;- \;1\; \right ]}{\frac{11}{10}\;-\;1}\;-\;10\;\left ( \frac{11}{10} \right )^{10}\;=\;\;m\;\left (1\;-\;\frac{11}{10} \right )}$

Or, $\mathbf{10\;\left [ \;\left ( \frac{11}{10} \right )^{10} \;- \;1\; \right ]\;-\;10\;\left ( \frac{11}{10} \right )^{10}\;=\;\;m\;\left (1\;-\;\frac{11}{10} \right )}$

Or, $\mathbf{-\;10=\;\;m\;\left (-\;\frac{1}{10} \right )}$

Therefore, the value of m = 100.