What is the sum of all three digit numbers that leave a remainder of 2 when divided by 3 ?

Let $a, a+1, a+2, a+3$ consecutive positive integers. Further if a is divisible by 3, then $a+1$ leaves a reminder 1 and $a+2$ leaves a remainder 2, when divided by 3 .

First 3 digit number, Viz, 100 leaves reminder 1 when divided by 3

101 leaves re minder 2 when divided by 3.

Thus, the numbers which leaves remainer 2 when divided by 3 are

$101,104,107, \cdots, 9.98$

Note that 101, 104….. 998 is in A.P with

common difference 3 .

$a=101$

$d=3 .$

Sum of first $n$ terms of an AP

$=[\frac{n}{2}\text { first term } + \text { Last term }]$

Let 998 be the $n^{\text {th }}$ term

$998 =a+(n-1) d$

$=101+(n-1) 3$

$n-1 =\frac{998-101}{3}=299$

$n=300.$

Required sum

$=\frac{n}{2}[\text { first term }+ \text { last term }]$

$=\frac{300}{2}[101+998]$

$=164,850$

Obtain the sum of all positive integers up to 1000, which are divisible by 5 and Not divisible by 2.

Positive integers up to 1000 , that are divisible by 5 are

$5, 15,\cdots, 995,1000$

Positive integers that are divisible by 5 , but not by 2 are

$5,15, \ldots . \cdots 995$

Note that the above sequence is an A.P with $a=5$, $d=10$.

Let 995 be the $n^{\text {th }}$ term.

$995 =a+(n-1) d$

$=5+(n-1) \cdot 10$

$\Rightarrow n =\frac{995-5}{10}+1$

$\Rightarrow n =100 .$

Required sum $=\frac{n}{2}[$ first term + last term]

$=\frac{100}{2}[5+995]$

$=50,000 .$

If $A=2^{65}$

and $B=2^{64}+2^{63}+\cdots+2^0$,

then is $A>B$ ?

$B=2^{64}+2^{63}+\cdots+2^0$

$B$ is sum of terms of a G.P with

$a=2^0=1 \quad r=2$

Recall that sum of first " $n$ " terms of GP

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

Let $2^{64}$ be the $n^{\text {th }}$ term.

a $r^{n-1}=2^{64}$

$1 \cdot 2^{n-1}=2^{64}$

$n-1= 64$

$n=65$

$B=2^{64}+2^{63}+\cdots+2^0$

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

$=\frac{1\left(2^{65}-1\right)}{2-1}=2^{65}-1$

$\therefore \quad B=A-1 .$

$A=B+1$

$A>B .$

$\text { yes ! }$

A piece of equipment cost a certain factory Rs. 600,000 . If it depreciates in value;
$15$ %the first year, $13.5$ % the next year $12$ % the third year and so on, what will be its value at the end of 10 years?

(All percentages applying to the Original cost)

Let the cost be Rs. 100

Now % of depreciation at the end of I, II, III years are

$15,13.5,12 \text {……. }$

This is in A.P with

$a=15$

$d=-1.5$

Therefore % of depreciation in $10^{\text {th }}$ year

$=a+9 d$

$=15+9(-1.5)$

$=1.5 .$

Total value depreciated in 10 years

$=15+13.5+\cdots+1.5$

$=\frac{10}{2}[15+1.5]$

$=82.5$

Value of the equipment at the end of 10 years

$=100-82.5$

$=17.5$

Total cost being $600000$, its value $=\frac{600000 \times 17.5}{100}$

$=1,05,000$

If $\log 2, \log \left(2^x-1\right)$, and $\log \left(2^x+3\right)$ are in $A \cdot P$,

find the value of $x$.

Logarithm

• inverse of exponential function.

Logarithm of a positive real number $x$ is the exponent to which another positive real number should be raised to obtain $x$.

$\log _b(x)=y$

$if \quad b^y=x$

$2^3 =8$

$\text { i.e } \quad \log _2 8 =3$

We know

$5^2=25$

$\log _5(25)=2$

We know

$25^1 =25$

$\log _{25} 25 =1$

$\log _3 9=2$

$\left(\because 3^2=9\right)$

Should be raised to obtain $x$.

$\log _b(x) =y$

$\text { if } \quad b^y =x .$

$\log _e(x)$ is called natural logarithm of $x$, and natural logarithm is denoted by $\ln$.

• $\log _b(x y) =\log _b x +\log _b y$

• $\log _b\left(x^p\right) =p \cdot \log _{b } x$

$\log 2, \log \left(2^x-1\right)$ and $\log \left(2^x+3\right)$ are $\text { in } A \cdot P \text {. }$

$\therefore \log \left(2^x-1\right)=\frac{\log 2+\log \left(2^x+3\right)}{2}$

$2 \log \left(2^x-1\right) =\log 2+\log \left(2^x+3\right)$

$=\log \left(2 \cdot\left(2^x+3\right)\right)$

$\log \left(2^x-1\right)^2=\log \left(2 \cdot\left(2^x+3\right)\right)$

$\left(2^x-1\right)^2=2 \cdot\left(2^x+3\right)$

$\left(2^x\right)^2-2 \cdot 2^x +1=2 \cdot 2^x+6$

$\left(2^x\right)^2-4 \cdot 2^x-5=0$

Let $2^x=y$

$y^2-4 y-5=0 \text {. }$

Solving, $y=5$ or $y=-1$

i.e $2^x=5$ or $2^x=-1$

$\therefore 2^x=5$

${{x=\ln 2^5}}$