Subtopic Archive - Page 10 of 10

Sum of n terms of Geometric Sequence

August 6, 2019/by admin_ct

Sum of n terms of Geometric Sequence

Sum of n terms of a finite geometric sequence is given by two formulas depending on the values of common ratio(r).

Let ‘a’ be the first term of geometric sequence and ‘r’ be its common ratio then sum of n terms, is given as,

$\dpi{150} \mathbf{S_{n}= a\left ( \frac{r^{n}-1}{r-1} \right )} for \left | r \right |>1$

$\dpi{150} \mathbf{S_{n}= a\left ( \frac{1-r^{n}}{1-r} \right )} for \left | r \right |< 1$

Example1. Consider the geometric series -90+30+(-10)+10/3 +….

Find the sum of first 16 terms .
Find n such that $\dpi{120} {\color{Red} S_{n}= -66.67}$

Solution: a) First term(a) = -90

Common ratio(r) = -1/3

Here |r| < 1 so we use second formula.

Sum of first 16 terms $\dpi{120} S_{16}= -90\left ( \frac{1-\left ( \frac{-1}{3} \right )^{16}}{1-\left ( \frac{-1}{3} \right )} \right )$

$\dpi{120} S_{16}= -90\left ( \frac{1-\frac{1}{3^{16}}}{1+\frac{1}{3}} \right )$

$\dpi{120} S_{16}= -90\left ( \frac{\frac{3^{16}-1}{3^{16}}}{\frac{4}{3}} \right )$

$\dpi{120} S_{16}= -90\left ( \frac{3^{16}-1}{3^{16}}*\frac{3}{4} \right )$

$\dpi{120} S_{16}= -90\left ( \frac{43046720}{3^{15}*4} \right )$

$\dpi{120} S_{16}= \left ( \frac{-107616800}{1594323} \right ) \Rightarrow 67.4999 \approx 67.5$

b) Given that $\dpi{120} S_{n}= -66.67$

$\dpi{120} S_{n}= -\frac{200}{3}$

$\dpi{120} -90\left ( \frac{1-\left ( \frac{-1}{3} \right )^{n}}{1-\left ( \frac{-1}{3} \right )} \right ) = \frac{-200}{3}$

$\dpi{120} \left ( \frac{1-(\frac{-1}{3})^{n}}{1+\frac{1}{3}} \right )= \frac{-200}{-90*3}$

$\dpi{120} \left ( \frac{1-(\frac{-1}{3})^{n}}{\frac{4}{3}} \right )= \frac{20}{27}$

$\dpi{120} 1-\left ( \frac{-1}{3} \right )^{n} = \frac{20}{27}*\frac{4}{3}= \frac{80}{81}$

$\dpi{120} 1-\frac{80}{81}= \left ( \frac{-1}{3} \right )^{n}$

$\dpi{120} \frac{1}{81}=\left ( \frac{-1}{3} \right )^{n}$

$\dpi{120} \left ( \frac{-1}{3} \right )^{4}=\left ( \frac{-1}{3} \right )^{n}$ [ For same bases , exponents can be equated]

4 = n

Example2: Find sum of given finite geometric series.

$\dpi{120} {\color{Red} \sum_{n=1}^{10}8\left ( \frac{3}{4} \right )^{n-1}}$

Solution: Here we are going to find $\dpi{120} s_{10}$ for the given series

$\dpi{120} 8\left [ 1+\frac{3}{4}+\left ( \frac{3}{4} \right )^{2}+............+\left ( \frac{3}{4} \right )^{9} \right ]$

$\dpi{120} 8+6+\frac{9}{2} +......+ \left ( \frac{3}{4} \right )^{9}$

First term(a) = 8

Common ratio(r) = 3/4

$\dpi{120} S_{10}= 8\left ( \frac{1-\left ( \frac{3}{4} \right )^{10}}{1-\left ( \frac{3}{4} \right )} \right )$

$\dpi{120} = 8\left ( \frac{0.943686}{0.25} \right )=30.20$

Example3: Find the sum of following series.

5+55+555+….n terms

Solution: Let $\dpi{120} S_{n}$ be the sum of given series then,

$\dpi{120} S_{n}$ = 5+55+555+……..n terms

= 5[1+11+111+…..n terms]

= 5/9[9+99+999+……. n terms]

= 5/9 [(10-1)+(100-1)+(1000-1) +…….n terms]

= $\dpi{120} \frac{5}{9}\left [ (10-1)+(10^{2}-1)+\left ( 10^{3}-1 \right )+.....n terms \right ]$

= $\dpi{120} \frac{5}{9}\left [\left (10+10^{2}+10^{3}+.....+10^{n} \right )-(1+1+1+....n times) \right ]$

= $\dpi{120} \frac{5}{9}\left [ 10\left ( \frac{10^{n}-1}{10-1} \right )-n \right ]$

= $\dpi{120} \frac{5}{9}\left [ 10 \frac{10^{n}-1}{9} -n \right ]$

= $\dpi{120} \frac{5}{81}\left [ 10(10^{n}-1) -9n\right ]$

= $\dpi{120} \frac{5}{81}\left [ 10^{n+1}-10-9n \right ]$

Example4: Sum of first three numbers in a G.P. is 16 and sum of next three terms is 128. Find the sum of n terms .

Solution: Let a be the first term and r be the common ratio of G.P. Then,

$\dpi{120} a+ar+ar^{2}=16$

$\dpi{120} a\left (1+r+r^{2} \right )=16$ ———(i)

Also $\dpi{120} ar^{3}+ar^{4}+ar^{5}=128$

$\dpi{120} ar^{3}\left (1+r+r^{2} \right )=128$ ——–(ii)

Dividing equation (ii) by equation(i) we get,

$\dpi{120} \frac{ar^{3}\left ( 1+r+r^{2} \right )}{a(1+r+r^{2})} = \frac{128}{16}$

$\dpi{120} \frac{ar^{3}}{a}= 8$

$\dpi{120} r^{3}=8$

r = 2

Substitute r=2 into equation (i) we get,

$\dpi{120} a\left (1+2+2^{2} \right )=16$

a(7) = 16

a = 16/7

And sum is given as, $\dpi{120} S_{n}= a\left ( \frac{r^{n}-1}{r-1} \right )$

$\dpi{120} S_{n}= \frac{16}{7}\left ( \frac{2^{n}-1}{2-1} \right )= \frac{16}{7}\left ( 2^{n}-1 \right )$

Example 5: A G.P. consists of even number of terms. If the sum of all terms is 5 times the sum of the terms occupying the odd places. Find the common ratio r.

Solution: Let there be 2n(even) terms in G.P with first term as a and common ratio as r.

Given that,

Sum of all terms = 5(sum of terms occupying the odd places)

$\dpi{120} a_{1}+a_{2}+a_{3}+.....+a_{2n}=5\left ( a_{1}+a_{3}+a_{5}+a_{2n-1} \right )$

$\dpi{120} a+ar+ar^{2}+....+ar^{2n-1}=5\left ( a+ar^{2}+ar^{4}+....+ar^{2n-2} \right )$

$\dpi{120} a\left ( \frac{1-r^{2n}}{1-r} \right )=5a\left ( \frac{1-\left ( r^{2} \right )^{n}}{1-r^{2}} \right )$

Note: if there are total 2n terms then there would be n even terms and n odd terms.

$\dpi{120} a\left ( \frac{1-r^{2n}}{1-r} \right )=5a\left ( \frac{1-r^{2n}}{\left (1-r \right )\left ( 1+r \right )} \right )$

Canceling $\dpi{120} \left ( 1-r^{2n} \right )$ from numerators and (1-r) from denominators on both sides, we get

$\dpi{120} a=\frac{5a}{1+r}$

a(1+r) = 5a

1+r = 5

r = 4

Sum of Infinite G.P.

Sum of infinite G.P with a as first term and r as common ratio is given as,

$\dpi{150} \mathbf{S_{\infty }= \frac{a}{1-r}}$

Example6: Find the sum of following geometric infinite series

${\color{Red} \frac{1}{2}+\frac{1}{3^{2}}+\frac{1}{2^{3}}+\frac{1}{3^{4}}+\frac{1}{2^{5}}+......\infty }$

Solution: We can separate these terms to get two infinite geometric series.

$\left ( \frac{1}{2}+\frac{1}{2^{3}}+\frac{1}{2^{5}}+...... \right )+\left ( \frac{1}{3^{2}}+\frac{1}{3^{4}}+\frac{1}{3^{6}}...... \right )$

(Infinite G.P with a=1/2, r=1/4) + (infinite G.P with a=1/9 , r=1/9)

$\dpi{120} \left ( \frac{\frac{1}{2}}{1-\frac{1}{2^{2}}} \right )+\left ( \frac{\frac{1}{3^{2}}}{1-\frac{1}{3^{2}}} \right )$

$\dpi{120} \left ( \frac{\frac{1}{2}}{1-\frac{1}{4}} \right )+\left ( \frac{\frac{1}{9}}{1-\frac{1}{9}} \right )$

$\dpi{120} \frac{1}{2}*\frac{4}{3}+\frac{1}{9}*\frac{9}{8}= \frac{2}{3}+\frac{1}{8}= \frac{19}{24}$

Example 7: prove that $\dpi{120} {\color{Red} 6^{\frac{1}{2}}*6^{\frac{1}{4}}*6^{\frac{1}{8}}......\infty =6}$

Solution: $\dpi{120} 6^{\frac{1}{2}}*6^{\frac{1}{4}}*6^{\frac{1}{8}}......\infty = 6^{\left [ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}.....\infty \right ]}$

Here $\dpi{120} \left [ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....\infty \right ]$ is a infinite geometric series so we can use formula $S_{\infty }= \frac{a}{1-r}$

$\dpi{120} 6^{\frac{1}{2}}*6^{\frac{1}{4}}*6^{\frac{1}{8}}......\infty = 6^{\left [ \frac{\frac{1}{2}}{1-\frac{1}{2}} \right ]}$

$= 6^{\left [ \frac{\frac{1}{2}}{\frac{1}{2}} \right ]}=6^{1}=6$

Example8: In 1990 the average monthly bill for cellular telephone service in the United States was $80.90. From 1990 through 1997, the average monthly bill decreased by about 8.6% per year.

On average, what did a person pay for cellular telephone service during 1990–1997?

Solution: Here we have,

Common ratio(r) = 1-0.086=0.914

Average monthly bill =$80.90

Average annual bill = 12*80.90=970.8

So first term(a) =970.8

Average cost for cellular telephone service during 8 years is given as, $\dpi{120} S_{n}= a\left ( \frac{1-r^{n}}{1-r} \right )$

$\dpi{120} S_{8}= 970.8\left ( \frac{1-0.914^{8}}{1-0.914} \right )= 5790$

On average a person pay 5790 dollars for cellular telephone service.

Practice problems:

Find the sum of the first 8 terms of the geometric series 1 + 8 + 64 + 512 + . . .
Find sum of following series. $\sum_{n=1}^{6}4 \left ( \frac{3}{2} \right )^{n}$
Find the sum of following series 0.7+0.77+0.777+….n terms
How many terms of the geometric series 3, 3/2 , 3/4 ….be taken together to make 3069/512
Find a G.P. for which sum of first two terms is -4 and the fifth term is 4 times the third term.
A person writes a letter to four of his friends. He asks each of them to copy the letter and mail it to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 cents to mail one letter. Find the amount spent on the postage when 8^th set of letter is mailed.

Answers:

1) 2396745

2) 124.6875

3) $\frac{7}{8}\left [ 9n-1 +\frac{1}{10^{n}} \right ]$

4) 10

5) -4/3 ,-8/3 , -16/3…… or 4,-8,16….

6) $43690

General term (nth term rule) Geometric Sequence

August 3, 2019/by admin_ct

General term (nth term rule)

A sequence of non zero numbers is called a geometric sequence if the ratio of a term and the term preceding to it, is always a constant. This constant is called the common ratio denoted by ‘r ’. let $\dpi{120} a_{n}$ denotes the nth term of geometric sequence then,

$\dpi{150} \mathbf{r = \frac{a_{n+1}}{a_{n}}}$ = constant

Nth term of a geometric sequence(whether finite or infinite) with first term as ‘a’ and common ratio as ‘r’ is given by

$\dpi{150} \mathbf{a_{n}= ar^{n-1}}$

1.Example: Decide whether the given sequence is geometric or not. 1,2,6,24,120….

Solution : we check the common ratio(r) of consecutive terms.

r = 2/1 =2 r = 6/2 = 3 r = 24/6 = 4

Since r is not same so this is NOT a geometric sequence.

2.Example: Write a rule for the nth term of the sequence 1,-6, 36, -216 . . . . Then find a8.

Solution: Given sequence: 1, -6, 36, -216….

First term(a) = 1

Common ratio(r)= -6

Nth term $\dpi{120} a_{n}= ar^{n-1}$

$\dpi{120} a_{n}= 1(-6)^{n-1}$

8^th term $\dpi{120} a_{8}= 1(-6)^{8-1}$ = -279936

3.Example: Two terms of a geometric sequence are $\dpi{120} {\color{Red} a_{2}=45}$ and $\dpi{120} {\color{Red} a_{5}=-1215}$ . Find a rule for the nth term.

Solution: Using nth term formula, we find system of equations.

$\dpi{120} a_{n}= ar^{n-1}$

$\dpi{120} a_{2}= ar^{2-1} = ar$

45 = ar ——- (i)

$\dpi{120} a_{5}= ar^{5-1}= ar^{4}$

$\dpi{120} -1215 = ar^{4}$ = ——-(ii)

Dividing eq. (ii) by (i) we get,

$\frac{-1215}{45}=\frac{ar^{n}}{ar}$ ( eliminating a from numerator and denominator)

$-27 = r^{3}$

-3 = r

Substitute r=-3 into equation (i) to get value of a.

45 = ar

45 = a(-3)

-15 = a

So the general rule is given as, $\dpi{120} a_{n}= (-15)(-3)^{n-1}$

4.Example: Which term of geometric sequence 2, 1, ½, ¼….is $\dpi{100} {\color{Red} \frac{1}{128}}$ ?

Solution: Here first term(a) = 2

Common ratio(r) = ½

$a_{n}= \frac{1}{128}$

$ar^{n-1} = \frac{1}{128}$

$2{\left ( \frac{1}{2} \right )}^{n-1} = \frac{1}{128}$

${\left ( \frac{1}{2} \right )}^{n-1} = \frac{1}{128}*\frac{1}{2}$ [multiply both sides with ½ ]

${\left ( \frac{1}{2} \right )}^{n-1} = \frac{1}{256}$

${\left ( \frac{1}{2} \right )}^{n-1} = \left ( \frac{1}{2} \right )^{8}$ [When bases are same, exponents get equated]

n-1 = 8 => n = 9

5.Example: One term of a geometric sequence is ${\color{Red} a_{3}=5}$ . The common ratio is r = 2.

Write a rule for the nth term.
Graph the sequence

Solution: First we need to find value of a.

$a_{3}=5$

$ar^{2}=5$

$a(2)^{2}=5 \Rightarrow a= \frac{5}{4}$

So nth term rule is given as,

$\dpi{120} a_{n}= (\frac{5}{4})(2)^{n-1}$

b) Since $\dpi{120} a_{n}= (\frac{5}{4})(2)^{n-1}$ is an exponential expression with r= 2

So this is an increasing curve as shown. This is true for any geometric sequence with r >1

6.Example: First term of a geometric sequence is 1. The sum of 3^rd and 5^th terms is 90. Find all possible values of r.

Solution: Given that $a_{3}+a_{5} = 90$

$ar^{2}+ar^{4} = 90$

$a(r^{2}+r^{4}) = 90$

$1(r^{2}+r^{4}) = 90$ [substituted a=1]

$r^{4}+r^{2}-90 = 0$

Lets assume $r^{2}=y$ Then we have the quadratic equation as,

$y^{2}+y-90=0$

(y+10)(y-9) = 0

y+10 =0 y-9=0

y = -10 y = 9

$r^{2}= -10$ $r^{2}= 9$

$r^{2}= -10$ is not possible to solve for real values but $r^{2}= 9$ gives two real values of r. r=-3 and r=3

7.Example: Find four numbers forming a geometric sequence in which the third term is greater than the first term by 9, second term is greater than 4^th by 18.

Solution: Let the four numbers be $\dpi{120} a,ar,ar^{2},ar^{3}$

Given that , Third term =first term + 9

$\dpi{120} ar^{2}= a+9 \Rightarrow a(r^{2}-1)=9$ ——-(i)

Also, second term = 4^th term + 18

$\dpi{120} ar=ar^{3}+18\Rightarrow ar(1-r^{2})=18$ ——(ii)

Dividing equation (ii) by (i) we get,

$\dpi{120} \frac{ar(1-r^{2})}{a(r^{2}-1)}= \frac{18}{9}$

$\dpi{120} \frac{-ar(r^{2}-1)}{a(r^{2}-1)}= \frac{18}{9}$

$\dpi{120} \frac{-ar}{a}=2$

-r = 2

r = -2

Substitute r=-2 into equation (i),

$\dpi{120} a((-2)^{2}-1)=9$

a(4-1)=9 => a = 3

Hence the numbers are 3,-2,12,-24.

Nth term from the end of G.P.

Let k be the last term of given Geometric sequence and r be the common ratio then nth term from the end ( $\dpi{120} a_{n}$ ) of that G.P. is given as,

$\dpi{150} \mathbf{a_{n}=k \left ( \frac{1}{r} \right )^{n-1}}$

8.Example: Find the 6th term from the end of the geometric sequence 8,4,2…..1/1024.

Solution: Here last term(k) = 1/1024

Common ratio(r) = ½

Using formula $\dpi{120} a_{n}=k \left ( \frac{1}{r} \right )^{n-1}$

$\dpi{120} a_{6}=\frac{1}{1024}\left ( 2 \right )^{6-1}$

$\dpi{120} a_{6}=\frac{1}{1024}*32 = \frac{1}{32}$

Selection of terms in GP:

Sometimes it is required to select a finite number of terms in G.P. it is always convenient if we select the terms in following manner.

Number of terms	Terms	Common ratio
3	$\dpi{150} \mathbf{\frac{a}{r} , a , ar}$	r
4	$\dpi{150} \mathbf{\frac{a}{r^{3}},\frac{a}{r},ar,ar^{3}}$	$\dpi{150} \mathbf{r^{2}}$
5	$\dpi{150} \mathbf{\frac{a}{r^{2}},\frac{a}{r},a,ar,ar^{2}}$	r

9.Example : The sum of first three terms of a G.P. is 13/12 and their product is -1. Find the G.P.

Solution: Let the three terms are a/r, a, ar

Their product = -1

$\dpi{120} \frac{a}{r}*a*ar =-1$ [ cross canceling r]

$\dpi{120} a^{3}= -1 \Rightarrow a=-1$

Their sum = 13/12

$\dpi{120} \frac{a}{r}+a+ar = \frac{13}{12}$

$\dpi{120} a\left (\frac{1}{r}+1+r \right ) = \frac{13}{12}$

$\dpi{120} (-1)\left (\frac{1+r+r^{2}}{r} \right ) = \frac{13}{12}$ [substitute a=-1]

$\dpi{120} \frac{1+r+r^{2}}{r}=\frac{-13}{12}$

$\dpi{120} 12(1+r+r^{2})= -13r$ [cross multiplying ]

$\dpi{120} 12r^{2}+12r+13r+12=0$

$\dpi{120} 12r^{2}+25r+12=0$

(4r+3)(3r+4) = 0

Solving these two factors we get two values of r.

r = -3/4 and r= -4/3

Using these two values of r, following two geometric sequences are possible,

$\dpi{120} \frac{3}{4},-1,\frac{4}{3}$ ……………… OR

$\dpi{120} \frac{4}{3},-1,\frac{3}{4}$ ………………

10.Example: The sum of three numbers in G.P. is 56.If we subtract 1,7,21 from these numbers in that order, we obtain an arithmetic sequence. Find the numbers.

Solution: let the three numbers are a , ar and $\dpi{120} ar^{2}$

Given that a-1 , ar-7 and $\dpi{120} ar^{2}-21$ form an arithmetic sequence. Therefore their common difference will be same.

$\dpi{120} (ar-7)-(a-1)= \left (ar^{2}-21 \right ) -(ar-7)$

$\dpi{120} ar-7-a+1= ar^{2}-21-ar+7$

$\dpi{120} 0= ar^{2}+a-2ar-8$ ——– (i)

Given that sum of three numbers = 56

$\dpi{120} a+ar+ar^{2}=56$

$\dpi{120} ar^{2}=56-a-ar$ ————–(ii)

Substitute value of $\dpi{120} ar^{2}$ from equation(ii) into equation (i) We get,

0 = 56-a-ar+a-2ar-8

0 = 48 -3ar

$\dpi{120} \frac{16}{a}=r$

Substitute $\dpi{120} \frac{16}{a}=r$ into $\dpi{120} a+ar+ar^{2}=56$ , we get

$\dpi{120} a+16+\frac{256}{a}= 56$

$\dpi{120} a^{2}+16a+256= 56a$

$\dpi{120} a^{2}-40a+256=0$

(a-32)(a-8) = 0

When a= 8, we get r = 16/8 =2 and numbers are: 8,16,32

When a=32, we get r= 16/32 = 1/2 and numbers are : 32,16 and 8

11.Example: In 1990 the average monthly bill for cellular telephone service in the United States was $80.90. From 1990 through 1997, the average monthly bill decreased by about 8.6% per year.

Write a rule for the average monthly cellular telephone bill (in dollars) in terms of the year. Let n = 1 represent 1990.

What was the average monthly cellular telephone bill in 1993?

When did the average monthly cellular telephone bill fall to $50?

Solution: a)Because the average monthly bill decreased by the same percent each year, the average monthly bills from year to year form a geometric sequence.

Since geometric sequence start from year 1990 and monthly bill in year 1990 was $80.90 so,

first term (a)=$80.90

common ratio(r) = 1- 0.086 = 0.914

So the rule for average monthly bill is given as,

$\dpi{120} a_{n}= 80.90\left ( 0.914 \right )^{n-1}$

b) If n=1 represent 1990 then year 1993 is represented by n=4.

So average monthly bill for year 1993 is:

$\dpi{120} a_{4}= 80.90\left ( 0.914 \right )^{4-1}= 61.77$
c) We have to find n such that $\dpi{120} a_{n}= 50$

$\dpi{120} a_{n}= 80.90\left ( 0.914 \right )^{n-1}$

$\dpi{120} 50 = 80.90\left ( 0.914 \right )^{n-1}$

$\dpi{120} \frac{50}{80.90 } = \left ( 0.914 \right )^{n-1}$

$\dpi{120} \ln \left ( 0.618 \right )= \ln (0.914)^{n-1}$ [Taking ln of both sides]

$\dpi{120} \ln \left ( 0.618 \right )= \ln (0.914)^{n-1}$

ln(0.618) = (n-1)ln(0.914)

$\dpi{120} \frac{ln(0.618)}{ln(0.914)} = n-1$

5.3 = n-1

6.3 = n

Rounding to nearest integer we get n=6

That means in year 1995, bill reached to $50.

Practice problems:

Which term of G.P. 18,-12,8… is 512/729.
Two terms of a geometric sequence are $\dpi{120} a_{3}=24$ and $\dpi{120} a_{6}=192$ . Find a rule for the nth term and then find 10^th
If the sum of three numbers in G.P. is 38 and their product is 1728, then find the numbers.
Find the 4^th term from the end of G.P. $\dpi{120} \frac{2}{27},\frac{2}{9},\frac{2}{3}......162$
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2^nd hour and nth hour.

Answers:

9
3072
8,12,18
6
120 , 30 $\dpi{120} \left ( 2 \right )^{n}$

Operations with Scientific Notation Worksheet

August 1, 2019/by admin_ct

Addition and subtraction in scientific notation:

While performing basic arithmetic operations like addition, subtraction, multiplication or division on numbers written in scientific notation, we need to be careful and should follow some rules to reach the correct answer.

While adding and subtracting the numbers in scientific notation, exponent part of all the numbers must be same. If it is not same , then first make it same and then combine the numbers.

Don’t forget to write the final answer in scientific notation.

When exponent part is same-

Example1: Combine $\dpi{120} 4.52\times 10^{3}$ and $\dpi{120} 8.3\times 10^{3}$

Solution: Here we see that exponent part of both numbers is same ( $\dpi{120} 10^{3}$ ). So we just need to combine the coefficients of these two numbers.

$\dpi{120} 4.52\times 10^{3}$ + $\dpi{120} 8.3\times 10^{3}$ = $\dpi{120} (4.52+8.3)\times 10^{3}$

= $\dpi{120} 12.82\times 10^{3}$

As the coefficient part can’t exceed 10, so we need to convert this result into scientific notation.

= $\dpi{120} 1.282\times 10^{4}$

When exponent part is not same-

Example2: Simplify the following expression: $\dpi{120} 8\times 10^{8}- 5\times 10 ^{6}$

Solution: As the exponent part is not same, we make them same first.

Again you can do it two ways. Either lower the exponent $\dpi{120} 10^{8}$ to $\dpi{120} 10^{6}$ or you can raise the exponent $\dpi{120} 10^{6}$ to $\dpi{120} 10^{8}$

First way: raise the exponent $\dpi{120} 10^{6}$ to $\dpi{120} 10^{8}$

As we know, when decimal is moved to left, exponent is positive (increased). We need to raise the exponent by 2 so decimal is moved towards left by 2 places and we get $\dpi{120} 5\times 10^{6} = 0.05\times 10^{2}\times 10 ^{6}$

= $\dpi{120} 0.05\times 10^{8}$

$\dpi{120} 8\times 10^{8}- 5\times 10 ^{6}$ = $\dpi{120} 8\times 10^{8}- 0.05\times 10 ^{8}$

= $\dpi{120} (8-0.05)\times 10^{8}$

= $\dpi{120} 7.95\times 10^{8}$

Second way: Either lower the exponent $\dpi{120} 10^{8}$ to $\dpi{120} 10^{6}$

When decimal is moved to right, exponent is negative (decreased). We need to lower the exponent by 2 so decimal is moved towards right by 2 places and we get $\dpi{120} 8\times 10 ^{8} = 8\times 10^{2}\times 10^{6} = 800\times 10^{6}$

$\dpi{120} 8\times 10^{8}- 5\times 10 ^{6}$ = $\dpi{120} 800\times 10^{6}- 5\times 10 ^{6}$

= $\dpi{120} 795\times 10^{6}$

But this result is not in scientific notation. To get it in scientific notation decimal should be placed after first significant digit.

So final answer is $\dpi{120} 7.95\times 10^{8}$

Example3. Simplify the expression $\dpi{120} 9\times 10^{5} + 6\times 10 ^{7}$

Solution: $\dpi{120} 9\times 10^{5} + 6\times 10 ^{7}$ = $\dpi{120} 9\times 10^{5} + 6\times 10 ^{2}\times 10^{5}$

= $\dpi{120} 9\times 10^{5} + 600\times 10^{5}$

= $\dpi{120} (9+600)\times 10^{5}$

= $\dpi{120} 609 \times 10^{5}$

Then we convert it into scientific notation and get answer as $\dpi{120} 6.09\times 10^{7}$

Example4. Simplify the expression $\dpi{120} 9\times 10^{5} - 6\times 10 ^{7}$

Solution: $\dpi{120} 9\times 10^{5} - 6\times 10 ^{7}$ = $\dpi{120} 9\times 10^{5} - 6\times 10 ^{2}\times 10^{5}$

= $\dpi{120} 9\times 10^{5} - 600\times 10^{5}$

= $\dpi{120} (9-600)\times 10^{5}$

= $\dpi{120} -591 \times 10^{5}$

Rewriting it as scientific notation we get the final answer as $\dpi{120} -5.91\times 10^{-7}$

Practice problems:

Simplify each problem and write the answer in scientific notation.

$\dpi{120} 1\times 10^{6} - 8\times 10 ^{4}$
$\dpi{120} 7\times 10^{7} + 3\times 10 ^{4}$
$\dpi{120} 4\times 10^{8} - 9\times 10 ^{7}$
$\dpi{120} 3\times 10^{-2} - 6\times 10 ^{-5}$

Answers:

1) $\dpi{120} 9.2\times 10^{5}$

2) $\dpi{120} 7.003\times 10 ^{7}$

3) $\dpi{120} 3.1\times 10^{8}$

4) $\dpi{120} 2.994\times 10^{-2}$

Sum of Finite Arithmetic Series

July 21, 2019/by admin_ct

The expression formed by adding the terms of an arithmetic sequence is called an arithmetic series.

The sum of first n terms of an arithmetic series, denoted by $\dpi{120} S_{n}$ , is given by the formula:

$\dpi{120} \mathbf{S_{n} = \mathbf{\frac{n}{2}\left [ 2a+(n-1)d \right ]}}$

If first and last terms are known then we can use formula,

$\dpi{120} \mathbf{S_{n}= \frac{n}{2}\left [ a_{1}+a_{n} \right ]}$

If the sum of n terms of a sequence is given then nth term can be determined by the following formula.

$\dpi{150} \mathbf{a_{n}= s_{n}-{s_{n-1}}}$

Example1: Consider the given arithmetic series 3+8+13+18+……

Find sum of first 20 terms.
Find n such that $\dpi{120} {\color{Red} S_{n}}$ = 366.

Solution: First term(a) = 3

Common difference(d) = 5

We have all the three quantities a, d and n so we can easily use first formula to find $\dpi{120} S_{n}$

$\dpi{120} S_{n}= \frac{n}{2}\left [ 2a+(n-a)d \right ]$

$\dpi{120} S_{n} = \frac{20}{2}\left [ 2(3))+(20-1)5 \right ]$

= 10[ 6+(19)5]

= 10 (101) = 1010

2. $\dpi{120} S_{n}= \frac{n}{2}\left [ 2a+(n-a)d \right ]$

$\dpi{120} 366 = \frac{n}{2}\left [ 2(3))+(n-1)5 \right ]$

$\dpi{120} 366 = \frac{n}{2}\left [ 5n+1 \right ]$

$\dpi{120} 366(2) = 5n^{2}+n$

$\dpi{120} 0 = 5n^{2}+n -732$

We can solve this quadratic equation by factoring. If it is not easily factorable then we can use quadratic formula too

$\dpi{120} 0 = 5n^{2}-60n+61n -732$

0 = 5n(n -12) +61 (n-12)

0 = (n-12)(5n+61)

n-12=0 (Since n, number of terms can’t be in fractions)

n=12

So $\dpi{120} S_{n}$ = 366 when n=12.

Example2: Find the sum of given finite arithmetic series.

$\dpi{120} {\color{Red} \sum_{n=1}^{90} 3-2n}$

Solution: Here we are given $\dpi{120} a_{n}$ = 3-2n

$\dpi{120} a_{1}$ = 3-2(1) = 1

$\dpi{120} a_{90}$ = 3-2(90) = -177

Number of terms(n) = 90

Sum( $\dpi{120} S_{n}$ ) = $\dpi{120} \frac{n}{2}\left [ a_{1}+a_{n} \right ]$

Sum ( $\dpi{120} S_{n}$ ) = $\dpi{120} \frac{90}{2}\left [ 1 + (-177) \right ]$

= 45 (-176)

= -7920

Example3: Find x such that 2x, 3x+2 and 5x+3 are consecutive terms of arithmetic sequence.

Solution: Since these three terms are consecutive terms of arithmetic sequence so their common difference must be same.

Therefore, 3x+2 –(2x) = (5x+3)-(3x+2)

X+2 = 2x+1

2-1 = 2x-x

1 = x

Example4: Find the sum of all odd integers from 1 to 2001.

Solution: Odd integers from 1 to 2001 are : 1,3,5,7…..2001.

To find sum , first we need to know value of n.

So, $\dpi{120} a_{n}$ = a+(n-1)d

2001 = 1+(n-1)2

2001= 2n-1

(2001+1)/2 = n

1001 = n

Sum , $\dpi{120} S_{n}= \frac{n}{2}\left [ a_{1}+a_{n} \right ]$

$\dpi{120} S_{n}= \frac{1001}{2}\left [ 1 + 2001 \right ]$

= $\dpi{120} \frac{1001}{2}(2002)$ = 1002001

Example5: Sum of n terms of an arithmetic series is $\dpi{120} {\color{Red} S_{n} = 3n^{2}+5n}$

And its mth term is 164, find the value of m.

Solution: we have , $\dpi{120} S_{n} = 3n^{2}+5n$

$\dpi{120} S_{n-1} = 3(n-1)^{2}+5(n-1)$

$\dpi{120} = 3n^{2}-n-2$

We know that, $\dpi{150} a_{n}= s_{n}-{s_{n-1}}$

= $\dpi{120} 3n^{2}+5n-(3n^{2}-n-2)$

= 6n+2

Give that,

$\dpi{120} a_{m}$ = 164

6m+2 = 164

6m = 162

m = 27

Example6: Sum of first 4 terms of an arithmetic sequence is 56. Sum of last 4 terms is 112. If first term is 11 then find number of terms.

Solution: Sum of first 4 terms = 56

$\dpi{120} \frac{4}{2}\left [ a_{1}+a_{4} \right ]$ = 56

2[ a +a+3d] = 56

2a+ 3d = 28

2(11) +3d = 28

3d = 28-22= 6

d = 2

Sum of last 4 terms =112

$\dpi{120} a_{n}+a_{n-1}+a_{n-2}+a_{n-3}$ = 112

$\dpi{120} \frac{4}{2}\left [ a_{n}+a_{n-3} \right ]$ = 112

2[a+(n-1)d +a+(n-3-1)d] =112

2a+ (2n-5)d = 56

2(11) +(2n-5)2 = 56

22 + 4n-10 = 56

4n = 56-12

4n = 44

n= 11

There are 11 terms in this Arithmetic sequence.

Example7: Logs are stacked in a pile. The bottom row has 21 logs and the top row has 15 logs. Each row has one less log than the row below it. How many logs are in the pile?

Solution: Based on the given information we get the sequence as, 21,20,…….,15

First term(a) = 21

Common difference(d) = -1

We have last term as 15. If we assume there are n terms(rows) then last term would be $\dpi{120} a_{n}$ , Using nth term rule, first we find n.

$\dpi{120} a_{n}= a+(n-1)d)$

15 = 21 +(n-1)(-1)

15 = 21 –n+1

n = 22-15 = 7

So there are 7 rows in the pile. To find number of logs we use sum formula.

$\dpi{120} S_{n}= \frac{n}{2}\left [ 2a+(n-a)d \right ]$

$\dpi{120} S_{n} = \frac{7}{2}\left [ 2(21))+(7-1)(-1) \right ]$

= $\dpi{120} \frac{7}{2}(42-6)$ = 126

Example8: The first row of a concert hall has 25 seats, and each row after the first has one more seat than the row before it. There are 32 rows of seats.

What is the total number of seats in the concert hall?
Suppose 12 more rows of seats are built (where each row has one more seat than the row before it). How many additional seats will the concert hall have?

Solution:

1. Using a=25 and d= 1 we get nth term formula as,

$\dpi{120} a_{n}$ = 24+n

$\dpi{120} a_{32}$ = 24+32 = 56

Total number of seats $\dpi{120} S_{n}= \frac{n}{2}\left [ a_{1}+a_{n} \right ]$

$\dpi{120} S_{32}= \frac{32}{2}\ [25+56]$

= 16(81) = 1296

2. The expanded concert hall has =32+12= 44 rows

Seats in 44^th row = 24+44 = 68

Total seats in expanded hall $\dpi{120} S_{44}= \frac{44}{2}\ [25+68]$

= 22[25 +68]

= 22(93)= 2046

Number of additional seats = 2046 -1296 = 750

Practice problems:

1.Consider the given arithmetic series 2+9+16+23+……

Find sum of first 68 terms.

Find n such that $\dpi{120} S_{n}$ = 166

2. Find the sum of given finite arithmetic

$\dpi{120} \sum_{n=1}^{15} (-10-3n)$

3. Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

4. The sums of n terms of two arithmetic series are in ratio 5n+4 : 9n+6. Find the ratio of their 18th terms.

5. A farmer buys a used tractor for Rs12,000. He pays Rs6000 cash and agrees to pay the balance in annual installments of Rs500 plus 12% interest on the unpaid amount. How much the tractor cost him.

Answers:

a) 16082 b)22
-510
1210
179 : 321
Rs 16680

Nth term rule (Arithmetic Sequence)

July 17, 2019/by admin_ct

A sequence is an ordered list of numbers whether finite or infinite. Any sequence is called Arithmetic sequence if the difference of a term and the previous term is always same. This constant difference , generally denoted by ‘d’ is called common difference.

Let ‘a’ be the first term and ‘d’ be the common difference of an arithmetic sequence, then its nth term is represented as:

$\dpi{150} a_{n}= a+(n-1)d$

Example: Show that the sequence 9,12,15,18…… is an A.P. (arithmetic progression). Find its general term and 16^th term.

Solution: Given sequence – 9,12,15,18………

Lets check the difference of terms first.

(12-9)=(15-12)=(18-15) =3. Since each time difference is same (3) so this is Arithmetic sequence.

Here we have, first term(a) = 9

Common difference(d)= 3

So general nth term is given as: $\dpi{150} a_{n}= a+(n-1)d$

$\dpi{150} a_{n}= 9+(n-1)3$

$\dpi{150} a_{n}= 9+3n-3 = 3n+6$

16^th term is given as $\dpi{150} a_{16}=3(16)+6 = 54$

Finding number of terms(n) of given sequence.

Example: How many terms are there in the sequence 3,6,9,12….111.

Solution: First term(a)= 3

Common difference(d)= 3

Let there be n terms in the sequence. Then,

nth term $\dpi{150} a_{n}= a+(n-1)d$

111 = 3 +(n-1)3

111 = 3+3n-3

111/3 =n

37 = n

There are 37 terms in this sequence.

Example: Which term of the sequence 4,9,14,19….. is 124?

Solution: First term(a) = 4

Common difference(d) = 5

$\dpi{150} a_{n}= a+(n-1)d$

124 = 4+(n-1)5

124 = 4+5n-5

124 = 5n-1

125 = 5n

25 = n

So 25^th term would be 124 in the given sequence.

To find nth term from end.

To find any nth term from end , we just reverse the sequence such that last term become its first term and common difference get opposite sign .

Example: Consider the sequence 3,7,11,……407. Find 20^th term from its end.

Solution: First we reverse the sequence like this,

407,………11,7,3.

First term(a) = 407

Common difference(d) = -4

20^th term from end, $\dpi{150} a_{n}= a+(n-1)d$

$\dpi{120} a_{20}= 407+(20-1)(-4)$

$\dpi{120} a_{20}= 407-76 = 331$

Finding nth term, given two terms.

Example: Two terms of an arithmetic sequence are $\dpi{120} a_{6}= 10$ and $\dpi{120} a_{21}= 55$ .

Find a rule for the nth term
Find the value of n for which $\dpi{120} a_{n}= 40$

Solution: using rule for nth term , $\dpi{150} a_{n}= a+(n-1)d$

We have,

=> $\dpi{120} a_{6}= a+(6-1)d$ => 10 = a+5d

=> $\dpi{120} a_{21}= a+(21-1)d$ => 55 = a+20d

Solving system of linear equations,

55 = a+20d

10 = a+5d

45 = 15d [subtracting second equation from first ]

3 = d

Substitute value of d into second equation, we get

10 = a +5(3)

10 = a+15

-5 = a

Substitute value of a and d into nth term formula.

$\dpi{120} a_{n}= a+(n-1)d$ = -5+(n-1)3

= -5+3n-3 = -8+3n

b) Given $a_{n}= 40$ , so substitute $a_{n}= 40$ into general rule found in part a.

40 = -8+3n

40+8 =3n

16=n

Example: How many numbers of two digits are divisible by 3.

Solution: Two digits numbers ranges from 10 to 99 and first two digit number divisible by 3, is 12. This sequence will have a common difference of 3 and sequence would be like this. 12,15,18…….99

Assuming there are n terms, we have the last term (nth term) as 99 and we find value of n using nth term formula.

$\dpi{150} a_{n}= a+(n-1)d$

99 = 12+(n-1)3

99 = 12+3n-3

99 = 9+3n

90 = 3n

30 = n

Example: A man starts repaying a loan as first installment of $100. If he increase the installments by $5 each month, what amount we will pay in 30^th installment.

Solution: Based on the given information, we have

First term(a) = $100

Common difference(d) =$5

We need to find 30^th term,

$\dpi{150} a_{n}= a+(n-1)d$

$\dpi{120} a_{30}= 100+(30-1)5$

= 100 + 145 = $245

Selection of terms in arithmetic sequence: sometimes we require certain number of terms in arithmetic sequence. The following ways of selecting terms are generally very convenient.

Number of terms	Terms	Common difference
3	a-d , a, a+d	d
4	a-3d, a-d, a+d , a+2d	2d
5	a-2d,a-d, a,a+d,a+2d	d

In case of odd number of terms, middle term is a and the common difference is d while in case of even number of terms, middle terms are a-d and a+d and common difference is 2d.

Example: Three consecutive terms of an arithmetic progression has a product of 4760 and a sum of 51. Find all the three terms.

Solution:

Let the three numbers be a-d, a and a+d. Then,

Sum = 51 =(a-d)+a+(a+d)

51 = 3a

17 = a

Product =4760

a(a-d)(a+d) = 4760

a $\dpi{120} {(a^2-d^2)}$ = 4760

17 $\dpi{120} {(17^2-d^2)}$ =4760

$\dpi{120} {(289-d^2)}$ =280

289-280 = $\dpi{120} d^{2}$

9 = $\dpi{120} d^{2}$

-3 , 3 = d

Using any value of d and a=17 we get the three numbers as

14,17 and 20

Example: In a certain arithmetic sequence 24^th term is twice the 10^th term. Prove that 72^nd term is twice the 34^th term.

Solution: As per given information,

24^th term = twice* 10^th term

a + 23d = 2(a+9d)

a + 23d = 2a +18d

5d = a

To prove:

72^nd term = twice 34^th term

a + 71d = 2(a+33d)

5d +71d = 2( 5d+33d)

76d = 2(38d)

76d = 76d

Hence proved !

Practice problems:

1) Write a rule for the nth term of arithmetic sequence.

d=4 , $\dpi{120} a_{14}= 46$

2) Graph the arithmetic sequence $\dpi{120} a_{n}= -3+5n$

3) Two terms of an arithmetic sequence are $\dpi{120} a_{6}= 19$ and $\dpi{120} a_{17}= 41$ . a) Find a rule for the nth term. b) Find the 40^th term.

4) How many numbers of two digits are divisible by 7?

5) Is 302 a term of the arithmetic sequence 3,8,13…..?

6) If 9^th term of an AP is 0, prove that its 29^th term is double the 19^th term.

7) The first row of a concert hall has 25 seats, and each row after the first has one more seat than the row before it. There are 32 rows of seats.

Write a rule for the number of seats in the nth row.
Thirty-five students from a class want to sit in the same row. How close to the front can they sit ?

Answers:

4n-10
Linear graph with positive slope (Rising line)
a) 2n+7 b) 87
13
NO
a) 24+n b) 11^th row

Archives

Sum of n terms of Geometric Sequence

Sum of Infinite G.P.

Selection of terms in GP:

Number of terms

Terms

Common ratio

3

r

4

5

r

Addition and subtraction in scientific notation:

When exponent part is same-

When exponent part is not same-

Finding number of terms(n) of given sequence.

To find nth term from end.