AP

$a, a+d, a+2 d \cdots$,

$a_n=n^{\text {th }} \text { term }=a+(n-1) d$

$S_n=\frac{n}{2}[I$ term + Last term]

$S_n = \frac{n}{2}[2a + (n-1)d]$

If $S_1$ is the sum of first " $n$ " terms of an A.P, where $n$ is odd.

and $S_2$ is the sum of terms of this series in odd places. Find the ratio $\frac{S_1}{S_2}$.

Let $a_1, a_2\cdots$ be the A.P.

Let us denote its common difference by “d”

$S_1=a_1+a_2+\ldots+a_n$

$=\frac{n}{2}\left[2 a_1+(n-1) \cdot d\right]$

$S_2=a_1+a_3+a_5+\cdots+a_n$

$a_3-a_1 =a_3-a_2+a_2-a_1$

$=d+d =2 d$

$a_5-a_3 =a_5-a_4+a_4-a_3 =2 d$

$S_1=a_1+a_2+\ldots+a_n =\frac{n}{2}\left[2 a_1+(n-1) \cdot d\right]$

$S_2=a_1+a_3+a_5+\cdots+a_n \\ \ =\frac{n+1}{4}\left[2 a_1+(n-1) \cdot d\right]$

Thus,

$\frac{S_1}{S_2} =\frac{\frac{n}{2}\left[2 a_1+(n-1) d\right]}{\frac{n+1}{4}\left[2 a_1+(n-1) d\right]}$

$=\frac{2 n}{n+1}$

GP

$a, a r, a r^2\cdots$,

$a_n=n^{th} \text { term }=a r^{n-1}$

$S_n=\frac{a\left(r^n-1\right)}{r-1} \text { if } r \neq 1$

$a+a r+a r^2+\cdots$

$=\frac{a}{1-r} \text { if }|r|<1$

$P^{\text {th }}$ term of an AP is $\frac{1}{q}$ and $q^{\text {th }}$ term of the same $A P$ is $\frac{1}{p}$.

Prove that sum of first “$pq$” terms is $\frac{1}{2}(p q+1) , \ p \neq q$

Let “a” be the first term and “d” be the common difference.
$n^{\text {th }} \text { term }=a+(n-1) d$

Sum of “n “terms

$S_n =\frac{n}{2}[2 a+(n-1) d]$

$=\frac{n}{2} [I^{\text{st}} \text { term }+ \text { Last term}]$

$a+(p-1) d=\frac{1}{q} \longrightarrow (1)$

$a+(q-1) d=\frac{1}{p} \longrightarrow (2)$

${[(p-1)-(q-1)] d=\frac{1}{q}-\frac{1}{p}}$

$\Rightarrow(p-q) d=\frac{p-q}{q p}$
$\Rightarrow d=\frac{1}{q p}$

$a =\frac{1}{q}-(p-1) \cdot d$

$=\frac{1}{q}-(p-1) \cdot \frac{1}{q p}$

$=\frac{1}{q}-\frac{1}{q}+\frac{1}{p q}$

$a =\frac{1}{p q}$

$S_{p q} =\frac{p q}{2}\left[2 \cdot \frac{1}{p q}+(p q-1) \cdot \frac{1}{p q}\right]$

$=\frac{p q}{2}\left[\frac{2}{p q}+1-\frac{1}{p q}\right]$

$=\frac{p q}{2}\left[\frac{1}{p q}+1\right]$

$=\frac{p q}{2}\left[\frac{1+p q}{p q}\right]$

$=\frac{1+p q}{2}$

$a, b$ and $c$ are three consecutive terms of a GP.

Further $a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}$

Prove that $x, y, z$ are in A.P

$a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}=k$

Thus $\quad a^{\frac{1}{x}}=k$

$\Rightarrow a=k^x$

$b=k^y$

$c=k^z$

Since $a, b, c$ are in GP

$b=\sqrt{a c}$

$b^2=a c$

$i.e.\quad\left(k^y\right)^2=k^x \cdot k^z$

$\Rightarrow k^{2 y}=k^{x+z}$

$\Rightarrow 2 y=x+z$

$\Rightarrow y=\frac{x+z}{2}$
Thus $y$ is AM of $x$ and $z$.

Let m, n be positive reals.

Assume that arithmetic mean of m, n is A and geometric mean of m, n is G.

Then show that the quadratic whose roots are m, n is $x^2 - 2Ax +G^2 = 0.$

$A M(m, n)=A$

$\text { i.e. } \quad\frac{m+n}{2}=A$

$\Rightarrow m+n=2 A .$

$G M(m, n)=G$

$\sqrt{m n}=G$

$\Rightarrow m n=G^2$

$m+n=2 A$

$m n=G^2$

A quadratic with roots $m, n$ is given by

$(x-m)(x-n)=0.$

$x^2 - (m + n)x + mn = 0.$

If $A$ is Arithmetic mean and $G$ is Geometric mean of two positive numbers then show that numbers are

$A \pm \sqrt{A^2 - G^2}.$

Let the numbers be m and n.
$\frac{m+n}{2} = A \ \Rightarrow m + n = 2A$

$\sqrt{mn} = G$

$\Rightarrow mn = G^2$

$(m-n)^2 = m^2 - 2mn + n^2 = (m+n)^2 -4mn$

$= 4A - 4 G^2$

$(m-n)^2 =4\left(A^2-G^2\right)$

$m-n = \pm \sqrt{4\left(A^2-G^2\right)}$

$= \pm 2 \sqrt{A^2-G^2}$

Thus

$m+n =2 A$

$m-n = \pm 2 \sqrt{A^2-G^2}$

$2 m=2 A \pm 2 \sqrt{A^2-G^2}$

$\Rightarrow m=A \pm \sqrt{A^2-G^2}$

Given $f$ is a function satisfying

$f(x+y)=f(x) \cdot f(y) \ \ \forall x, y \in \mathbb{N} .$

Let $f(1)=3$

If $\sum_{x=1}^n f(x)=120$, find value of $n$.

$f(x+y)=f(x) \cdot f(y)$

$\forall x, y \in \mathbb{N}$

$f(2)=f(1+1)$

$=f(1) \cdot f(1)$

$\text { i.e.. } \quad f(2)=[f(1)]^2$

Similarly

$f(3) =f(2+1)$

$=f(2) \cdot f(1) =[f(1)]^3$

continuing like this

$f(n)=[f(1)]^n$

Given

$\sum_{x=1}^n f(x)=120$

$\text { i.e. }$

$f(1)+f(2)+\cdots+f(n) =120.$

$\text {i.e.}$

$f(1)+f(1)^2+\cdots+[f(1)]^n =120 .$

i.e.

$3+3^2+\cdots+3^n =120$

The LHS represents Sum of terms of a GP with $a=3, \text{and} \ r=3$

$\text{using} \quad S_n=\frac{a\left(r^n-1\right)}{r-1}$

$\frac{3\left(3^n-1\right)}{3-1}=120 .$

$3\left(3^n-1\right)=120 \times 2 =240$

$3^n-1=\frac{240}{3}=80$

$3^n=81$
Thus

$3^n = 81 = 3^4$

$\Rightarrow n = 4$

Find sum of the terms of the following sequence

$7, 77, 777, 7777, \cdots (n$ terms).

$S_n = 7 + 77 + 777 + \cdots (n$ terms).

$= 7 (1 + 11 + 111 + \cdots )$

$= \frac{7}{9} [ 9 + 99 + 999 + \cdots ]$

$= \frac{7}{9} [ (10 - 1) + (100 - 1) + \cdots ]$

$= \frac{7}{9} [ (10 + 100 + 1000 + \cdots -n]$

$=\frac{7}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right]$