Feedback Daily Quiz
1 2 3 4 5 6 7 8 9 10 11 12
InstaCue : Mathematics / Sequences And Series

Sequence


\(\bullet\)  By a Sequence we mean a list of numbers, arranged according to some definite rule.

                                             or

   We definite a sequence as a function whose domain is the set of  natural numbers or some subsets of type \(\{1,\ 2,\ 3,\ ....\ k\}.\)

\(\bullet\)  If \(a_1,\ a_2,\ a_3,\ ...\ a_n, ....\) is a sequence, then the expression \(a_1+\ a_2+\ a_3 +\ ...\ a_n +....\) is called the series.

\(\bullet\)  If the terms of a sequence follow a certain pattern, then it is called a progression.

 

Visit https://www.testbee.in for practice test, concept-based learning
InstaCue : Mathematics / Sequences And Series

Arithmetic Progression (AP)


Arithmetic Progression (AP)

\(\bullet\)  It is a sequence in which the difference between any two consecutive terms is always same

\(\bullet\)  An AP can be represented as \(a,\ a + d,\ a + 2d,\ a + 3d, ...\) where, \(a\) is the first term, \(d\) is the common difference. 

\(\bullet\)  The \(n^{th}\) term, \(t_n= a \ +\ (n-1)d\)

\(\bullet\)  Common difference \(d=t_n-t_{n-1}\)

\(\bullet\)  The \(n^{th}\) term from end, \(t=l\ -\ (n-1)d,\) where \(l\) is the last term.

\(\bullet\)  Sum of first \(n\) terms, \(S_n=\frac{n}{2}\ [2a\ +\ (n-1)d]\ =\ \frac{n}{2}\ [a\ +\ l],\) where \(l\)    is the last term.

\(\bullet\)  If sum of \(n\) terms is \(S_n, \) then \(n^{th}\) term is \(t_n=S_n\ -\ S_{n-1},\ t_n = \frac{1}{2}\ [t_{n-k}\ +\ t_{n\ +\ k}],\) where \(k \lt n\)

Visit https://www.testbee.in for practice test, concept-based learning
InstaCue : Mathematics / Sequences And Series

Important Result Of AP


\(\underline {\overline{NOTE}}\)

            \(\bullet\)  Any three numbers in AP can be taken as

                \(a-d,\ a,\ a\ +\ d.\)

            \(\bullet\)  Any four numbers in AP can be taken as 

                \(a-3d,\ a - d,\ a\ +\ d,\ a\ +\ 3d.\)

            \(\bullet\)   Any five numbers in AP can be taken as

                 \(a-2d,\ a - d,\ a,\ \ a\ +\ d,\ a\ +\ 2d.\)

            \(\bullet\)  Three numbers \(a,\ b,\ c,\ \)are in AP iff \(2b= a\ +\ c\)

An Important Result of AP

\(\bullet\)  In a finite AP, \(a_1, ..., a_n,\) the sum of the terms equidistant from the beginning and end is always same and equal to the sum of first and last term.

\(i.e.\ \ a_1\ +\ a_n= a_k\ +\ a_{n-(k-1)},\ \forall\ k=1, 2, 3, ..., n-1\)

Visit https://www.testbee.in for practice test, concept-based learning
InstaCue : Mathematics / Sequences And Series

Arithmetic Mean(AM)


ARITHMETIC MEAN (AM)

\(\bullet\) If \(a,\ A\ and\ B\) are in AP, then \(A=\frac{a+b}{2}\) is the arithmetic mean of \(a\ and \ b\)

\(\bullet\)  If \(a,\ A_1,\ A_2, ..., A_n,\ b\) are in AP, then \(A_1,\ A_2, ..., A_n\) are the \(n\) arithmetic means between \(a\ and\ b\).

\(\bullet\)  The \(n\) arithmetic means \(A_1,\ A_2, ..., A_n\), between \(a\ and \ b\) are given by the formula \(A_r= a\ +\ \frac{r\ (b-a)}{n+1}\ \forall\ r=1,2,...n\)

\(\bullet\)  Sum of \(n\ \text{AM's}\) inserted between \(a\ and\ b\) is \(n\ \text{A}\) i.e.

   \(A_1\ +\ A_2\ +\ A_3\ +...\ +\ A_n=n\ \big(\frac{a+b}{2}\big)\)

\(\underline{\overline{NOTE}}\ \ \ \bullet\)   The AM of \(n\) numbers \(a_1, a_2, ..., a_n\) is given by

                         \(AM=\frac{(a_1\ +\ a_2\ +\ a_3\ +...\ +\ a_n)}{n}\)

Visit https://www.testbee.in for practice test, concept-based learning
InstaCue : Mathematics / Sequences And Series

Geometric Progression (GP)


Geometric Progression (GP)

\(\bullet\)  It is a sequence in which the ratio of any two consecutive terms is always same.

\(\bullet\)  A GP can be represented as \(a,\ ar,\ ar^2, ...\)

    where, \(a\) is the first ratio and \(r\) is the common ratio.

\(\bullet\)  The \(n^{th}\) term, \(t_n=ar^{n-1}\).

\(\bullet\)  The \(n^{th}\) term from end, \(t'_n=\frac{l}{r_{n-1}},\) where \(l\) is the last term.

\(\bullet\)  First of \(n\) terms, \(S_n= \begin{cases} a\Big(\frac{1-r^n}{1-r}\Big), & r\ne1\\ na,& r=1 \end{cases} \)

\(\bullet\)  If \(|r|\ \lt 1,\) then the sum of the infinite GP is \(S_\infty=\frac{a}{1-r}\)

 

Visit https://www.testbee.in for practice test, concept-based learning
InstaCue : Mathematics / Sequences And Series

Properties Of GP


Properties of GP

\(\underline{\overline{\text{NOTE}}}\)

\(\bullet\)  Any three numbers in GP can be taken as \(\frac{a}{r},\ a,\ ar\).

\(\bullet\)  Any four numbers in GP can be taken as \(\frac{a}{r^3},\ \frac{a}{r},\ ar,\ ar^3\).

\(\bullet\)  Any five numbers in GP can be taken as \(\frac{a}{r^2},\ \frac{a}{r},\ a,\ ar,\ ar^2\).

\(\bullet\)  Three non-zero numbers \(a, b, c\) are in GP iff \(b^2=ac\).

\(\bullet\)  If \(a, b\ and\ c\) are in AP as well as GP then \(a=b=c\).

\(\bullet\)  If \(a \gt 0\ \ and\ \ r \gt 1\ \ or\ \ a \lt 0\ \ and\ \ 0 \lt r \lt 1,\) then the GP will be an increasing GP.

\(\bullet\) If \(a \gt 0\ \ and\ \ 0 \lt r \lt 1,\ or\ \ \ a \lt 1\ \ and\ \ r \gt 1\), then the GP will be decreasing GP.

 

Visit https://www.testbee.in for practice test, concept-based learning
InstaCue : Mathematics / Sequences And Series

Important Results On GP


Important Results on GP

\(\bullet\) If \(a_1, a_2, a_3,... a_n\) is a GP of positive terms, then \(\log a_1, \log a_2, \log a_3, ... \log a_n\) is AP and vice-versa.

\(\bullet\)  In a finite GP, \(a_1, a_2, ... a_n\), the product of the terms equidistant from the beginning and the end is always same and is equal to the product of the first and the last term.

\(\bullet\)  \(i.e.\ \ a_1a_n=a_k \cdot a_{n-(k-1)},\ \forall\ k=1, 2, 3,...n-1.\)

Visit https://www.testbee.in for practice test, concept-based learning
InstaCue : Mathematics / Sequences And Series

Geometric Mean (GM)


Geometric Mean (GM)

\(\bullet\)  If \(a, G\ \ and \ \ b\) are in GP then, \(GP= \sqrt{ab}\) is the geometric mean of \(a\ \ and \ \ b\).

\(\bullet\)  If \(a, G_1, G_2,..., G_n, b\) are in GP, then \(G_1, G_2,..., G_n\) are the \(n\) geometric means between \(a\ \ and \ \ b\).

\(\bullet\)  The \(n\) GM's, \(G_1, G_2,..., G_n\), inserted between \(a\ \ and \ \ b\), are given by the formula, \(G_r=a\Big(\frac{b}{a}\Big)^{\frac{r}{n+1}}\).

\(\bullet\)  Product of \(n\) GM's, inserted between \(a\ \ and \ \ b\), is the \(n^{th}\) power of the single GM between \(a\ \ and \ \ b\).

  \(i.e.\ \ G_1\ \cdot\ G_2\ \cdot\ ...\ \cdot\ G_n=(ab)^{n/2}.\)

\(\underline{\overline{\text{NOTE}}}\)

\(\bullet\)  If \(a\ \ and \ \ b\) are of opposite signs, then their GM cannot exist.

\(\bullet\)  If \(A\ \ and \ \ G\) are respectively the AM and GM between two numbers \(a\ \ and \ \ b\), then \(a,b\) are given by \([A\ \pm\ \sqrt{(A+G)\ (A\ -\ G)}].\)

\(\bullet\) If \(a_1, a_2, a_3, ..., a_n\) are positive numbers, then their GM \(=(a_1, a_2, a_3, ..., a_n)^{1/n}\)

Visit https://www.testbee.in for practice test, concept-based learning
InstaCue : Mathematics / Sequences And Series

Arithmetico-Geometric Progression (AGP)


Arithmetico-Geometric Progression (AGP)

\(\bullet\)  A progression in which every term is a product of a term of AP and corresponding term of GP, is known as arithmetico-geometric progression.

\(\bullet\)  If the series of AGP be a \(a\ +\ (a+d)\ r\ +\ (a+2d)\ r^2\ +\ ...\ +\ \{a\ +\ (n-1)\ d\}r^{n-1}\ +\ ...,\) then

  (i) \(S_n=\frac{a}{1-r}\ +\ \frac{dr(1-r^{n-1)}}{(1-r)^2}\ -\ \frac{\{a+(n-1)d\}r^n}{1-r},\ r \ne 1\)

  (ii) \(S_\infty=\frac{a}{1-r}\ +\ \frac{dr}{(1-r)^2},\ |r| \lt 1\)

Visit https://www.testbee.in for practice test, concept-based learning
InstaCue : Mathematics / Sequences And Series

Method To Find The Sum Of N-terms


N terms of AP

Method to find the sum of n-terms of Arithmetic-Geometric Progression

Usually, we do not use the above formula to find the sum of \(n\) terms.

Infact we use the mechanism by which we derived the formula, shown below:

Let, \(S_n=a\ +\ (a+d)\ r\ +\ (a+2d)\ r^2\ + ... +\ (a\ +(n-1)d)\ r^{n-1}\ ...\ (i)\)

Step I Multiply each term by \(r\) (Common ratio of GP) and obtain a new series 

\(\Rightarrow\ \ r\ S_n= ar\ +\ (a+d)\ r^2\ + ... +\ (a\ +\ (n-2)\ d)\ r^{n-1}\ +\ (a\ +\ (n-1)\ d)\ r^n\ ... \ (ii)\)

Step II Subtract the new series from the original series by shifting the terms of new series by one term.

\(\Rightarrow\ \ (1-r)\ S_n=a\ +\ [dr\ +\ dr^2\ + ...\ +\ dr^{n-1}]\ -\ (a\ +\ (n-1)\ d)\ r^n\\ \Rightarrow\ \ S_n\ (1-r)=\ a\ +\ dr\ \Big(\frac{1-r^{n-1}}{1-r}\Big)\ -\ (a\ +\ (n-1)\ d)\ r^n\\ \Rightarrow\ \ S_n= \frac{a}{1-r}\ +\ dr\ \Bigg(\frac{1-r^{n-1}}{(1-r)^2}\Bigg)\ -\ \frac{(a+(n-1)d)}{1-r}\ r^n\)

Visit https://www.testbee.in for practice test, concept-based learning
InstaCue : Mathematics / Sequences And Series

  Sum Of Special Series


      Sum of Special Series

\(\bullet\)  Sum of first \(n\) natural numbers, 

       \(1\ +\ 2\ + ... +\ n=\ \sum\ n=\frac{n(n+1)}{2}\)

\(\bullet\)  Sum of squares of first \(n\) natural numbers,

      \(1^2\ +\ 2^2\ + ... +\ n^2=\ \sum\ n^2\ =\frac{n(n+1)\ (2n+1)}{6}\)

\(\bullet\)  Sum of cubes of first \(n\) natural numbers,

      \(1^3\ +\ 2^3\ +\ 3^3\ + ... +\ n^3=\ \sum\ n^3=\Big[\frac{n(n+1)}{2}\Big]^2\)

(i)  Sum of first \(n\) even natural numbers

        \(2\ +\ 4\ +\ 6\ + ...\ +2n=\ n(n+1)\)

(ii)  Sum of first \(n\) odd natural numbers

        \(1\ +\ 3\ +\ 5\ + ...\ +(2n-1)=\ n^2\)

 

Visit https://www.testbee.in for practice test, concept-based learning
InstaCue : Mathematics / Sequences And Series

Summation Of Series By The Difference Method


Summation of Series by the Difference Method

If the \(n^{th}\) term of a series cannot be determined by the methods discussed so far. Then, \(n^{th}\) term can be determined by the method of difference, if the difference between successive terms of series are either in AP or in GP, as shown below:

Let, \(T_1\ +\ T_2\ +\ T_3\ +\ ...\) be a given infinite series.

If \(T_2\ -\ T_1,\ \ T_3\ -\ T_2,\ ...\) are in AP or GP, then \(T_n\) can be found by following procedure

Clearly, \(S_n=\ T_1\ +\ T_2\ +\ T_3\ +\ ...\ T_n\ \ \ \ ...\ (i)\)

Again, \(S_n=\ T_1\ +\ T_2\ +\ ...\ T_{n-1}\ +\ T_n\ \ \ \ ...\ (ii)\)

\(\therefore\ \ \ S_n\ -\ S_n=T_1\ +\ (T_2-T_1)\ +\ (T_3-T_2)\ +\ ...\ +\ (T_n-T_{n-1})-\ T_n\\ \Rightarrow\ \ \ T_n=\ T_1\ +\ (T_2-T_1)\ +\ (T_3-T_2)\ +\ ...\ +\ (T_n-T_{n-1})\\ \Rightarrow\ \ \ T_n=\ T_1\ +\ t_1\ +\ t_2\ +\ t_3\ +\ t_{n-1}\)

where, \(t_1,\ t_2,\ t_3, ....\) are terms of the new series \( \Rightarrow\ \ S_n=\sum\limits^n_{r=1}\ T_r \)

Visit https://www.testbee.in for practice test, concept-based learning
1 2 3 4 5 6 7 8 9 10 11 12

Log In to Your Testbee Account!

X

or Forgot password?

Sign up and Start...

X
Please enter valid Mobile No to get OTP
Please enter valid Email Id to get OTP

Lost your password? Please enter your mobile number. You will receive a otp to create a new password.