Sequences and Series

The sum of three consecutive numbers of a G.P is 56. If we subtract 1, 7 and 21 from these numbers in order, the resulting numbers form an A.P. Find the numbers.

Let the three consecutive numbers in G.P be $\dfrac{a}{r}$, a, $ar$
where first term $= a$; common ratio $= r$
$\therefore$ $\dfrac{a}{r}+a+ar = 56$ $\implies$ $\dfrac{a}{r} + ar = 56 - a$ $\cdots$ (1)
On subtracting 1, 7 and 21 from these numbers in order gives $\dfrac{a}{r}-1$, $a-7$ and $ar-21$
Then, $\dfrac{a}{r}-1$, $a-7$, $ar-21$ are in A.P
Common difference $= a-7-\left(\dfrac{a}{r} - 1\right) = ar - 21 - \left(a-7\right)$
i.e. $a-\dfrac{a}{r}-6=ar-a-14$
i.e. $\dfrac{a}{r}+ar-2a= 8$
i.e. $\dfrac{a}{r} + ar = 8 + 2a$ $\cdots$ (2)
$\therefore$ From equations (1) and (2) we have
$56-a = 8 + 2a$ i.e. $3a = 48$ $\implies$ $a=16$
Therefore, from equation (2) we have,
$\dfrac{16}{r} + 16r = 8 + 32 = 40$
i.e. $16r^2 -40r + 16 = 0$
i.e. $2r^2 -5r+2 = 0$
i.e. $\left(2r-1\right) \left(r-2\right) = 0$
i.e. $r = 2$ or $r = \dfrac{1}{2}$
Therefore, when $a=16$, $r=2$, the numbers are $\dfrac{16}{2}$, 16, $16\times2$ i.e. 8, 16, 32
When $a=16$, $r = \dfrac{1}{2}$, the numbers are $\dfrac{16}{1/2}$, 16, $16 \times \dfrac{1}{2}$ i.e. 32, 16, 8

Sequences and Series

Two arithmetic progressions have the same common difference. The first term of one A.P is $-5$ and that of the other is $-9$. Find the difference between their $7^{th}$ terms.

Let common difference $= d$
Case 1: First term $= a = -5$
$\therefore$ $7^{th}$ term $= t_7 = a + 6d = -5 + 6d$
Case 2: First term $= A = -9$
$\therefore$ $7^{th}$ term $= T_7 = A + 6d = -9 + 6d$
$\therefore$ Difference between the two $7^{th}$ terms $= t_7 - T_7 = -5 + 6d - \left(-9+6d\right) = -5+6d+9-6d = 4$

Sequences and Series

If $\left(p+q\right)^{th}$ term and $\left(p-q\right)^{th}$ terms of a G.P are a and b respectively, prove that the $p^{th}$ term is $\sqrt{ab}$.

Let the first term of G.P $= A$ and common ratio $= R$
$n^{th}$ term of G.P $= t_n = AR^{n-1}$
$\therefore$ $\left(p+q\right)^{th}$ term $= t_{p+q} = AR^{p+q-1} = a$ (given) $\cdots$ (1)
and $\left(p-q\right)^{th}$ term $= t_{p-q} = AR^{p-q-1} = b$ (given) $\cdots$ (2)
$\therefore$ $\dfrac{t_{p+q}}{t_{p-q}} = \dfrac{AR^{p+q-1}}{AR^{p-q-1}} = \dfrac{a}{b}$
i.e. $R^{p+q-1-p+q+1} = \dfrac{a}{b}$
i.e. $R^{2q} = \dfrac{a}{b}$ $\implies$ $R = \sqrt[2q]{\dfrac{a}{b}} = \left(\dfrac{a}{b}\right)^{\frac{1}{2q}}$ $\cdots$ (3)
$\therefore$ From equations (1) and (3) we have
$A\left[\left(\dfrac{a}{b}\right)^{\frac{1}{2q}}\right]^{p+q-1} = a$
i.e. $A \left(\dfrac{a}{b}\right)^{\frac{p+q-1}{2q}} = a$
$\implies$ $A = a \times \left(\dfrac{b}{a}\right)^{\frac{p+q-1}{2q}}$
i.e. $A = a^{1-\frac{\left(p+q-1\right)}{2q}} b^{\frac{p+q-1}{2q}} = a^{\frac{1-p+q}{2q}} b^{\frac{p+q-1}{2q}}$
$ \begin{aligned} \therefore p^{th} \text{term} = t_p & = AR^{p-1} \\ & = a^{\frac{1-p+q}{2q}} b^{\frac{p+q-1}{2q}} \times \left(\dfrac{a}{b}\right)^{\frac{p-1}{2q}} \\ & = a^{\frac{1-p+q+p-1}{2q}} b^{\frac{p+q-1}{2q} - \frac{p-1}{2q}} \\ & = a^{\frac{1}{2}} b^{\frac{1}{2}} = \sqrt{ab} \end{aligned} $

Sequences and Series

Determine the $12^{th}$ term of a G.P whose $8^{th}$ term is 192 and common ratio is 2. Also find $t_8 : t_{12}$

$8^{th}$ term of G.P $= t_8 = ar^7 = 192$
Common ratio $= r = 2$
$\therefore$ $a \times 2^7 = 192$
i.e. $a = \dfrac{192}{128} = \dfrac{3}{2}$
$\therefore$ $12^{th}$ term $= t_{12} = ar^{11} = \dfrac{3}{2} \times 2^{11} = 3072$
$\dfrac{t_8}{t_{12}} = \dfrac{192}{3072} = \dfrac{1}{16}$

Sequences and Series

Find the $n^{th}$ term and deduce the sum to n terms of the series $4+11+22+37+56+\cdots$

$ \begin{aligned} \text{Let } S_n & = & 4 + & 11 + 22 + 37 + 56 + \cdots + t_{n-1} + t_n \\ & & & \text{Shifting every term one place to the right} \\ S_n & = & & 4 + 11 + 22 + 37 + \cdots \cdots \cdots + t_{n-1} + t_n \end{aligned} $
Subtracting, we get $ \begin{aligned} 0 & = \left[4 + 7 + 11 + 15 + 19 + \cdots \text{to n terms}\right] - t_n \\ \text{i.e. } t_n & = 4 + \left[7 + 11 + 15 + 19 + \cdots \text{to } \left(n-1\right) \text{ terms}\right] \\ & = 4 + \left(\dfrac{n-1}{2}\right)\left[2 \times 7 + \left(n-2\right) \times 4\right] \\ & \left[\text{Note: Sum to n terms of an A.P } = \dfrac{n}{2}\left(2a+\left(n-1\right)d\right)\right] \\ & \left[\text{Here, first term }= a = 7 \; ; \text{common difference }= d = 4 \right] \\ & = 4 + \dfrac{\left(n-1\right) \left(4n+6\right)}{2} \\ & = 4 + \dfrac{4n^2 + 2n - 6}{2} \\ & = 4 + 2n^2 + n - 3 \\ \text{i.e. } t_n & = 2n^2 + n + 1 \end{aligned} $ $ \begin{aligned} \therefore \text{ Sum to n terms} = S_n & = 2 \sum_{k = 1}^{n} k^2 + \sum_{k=1}^{n} k + n \\ & = 2 \times \dfrac{n \left(n+1\right) \left(2n+1\right)}{6} + \dfrac{n \left(n+1\right)}{2} + n \\ & = \dfrac{n}{3} \left(2n^2 + 3n + 1\right) + \dfrac{n \left(n+1\right)}{2} + n \\ & = \dfrac{n}{6} \left(4n^2 + 6n + 2 + 3n + 3 + 6\right) \\ \therefore S_n & = \dfrac{n}{6} \left(4n^2 + 9n + 11\right) \end{aligned} $

Sequences and Series

The product of the third and the eighth terms of a G.P is 243. If the fourth term is 3, find its seventh term.

Third term of G.P $= t_3 = ar^2$
Eighth term of G.P $= t_8 = ar^7$
$\therefore$ $ar^2 \times ar^7 = 243$ $\implies$ $a^2 r^9 = 243$ $\cdots$ (1)
Fourth term of G.P $= t_4 = ar^3 = 3$ $\cdots$ (2)
Now, equation (1) can be rewritten as
$ar^3 \times ar^3 \times r^3 = 243$ $\cdots$ (3)
In view of equation (2), equation (3) becomes
$3 \times 3 \times r^3 = 243$ $\implies$ $r^3 = \dfrac{243}{9} = 27$ $\implies$ $r = \sqrt[3]{27} = 3$
Substituting the value of r in equation (2) gives
$a \times 3^3 = 3$ $\implies$ $a = \dfrac{1}{9}$
$\therefore$ Seventh term $= t_7 = ar^6 = \dfrac{1}{9} \times 3^6 = 81$