Progressions-A.P. and G.P. (Part-3)

 

© ® Written by Sankar Ghosh

Hello friends!
I hope you are doing great.

I’ve already shared two blog posts on Arithmetic Progression (A.P.), where we explored its basic concepts, general term, various types of problems, and how to find the sum of n terms and the arithmetic mean.

Today, I’m bringing you the final post on Arithmetic Progression. In this post, we will discuss the important properties of A.P. and also look at some special series related to it.

With this, we will wrap up our journey through A.P. and get ready to move on to an exciting new topic — Geometric Progression (G.P.) in my upcoming fourth blog post.

I would love to hear your feedback and suggestions. Your support keeps me motivated to share more with you!

Important properties of A.P. 
(a) If a non-zero constant is (i) added to or subtracted from each term of an A.P. then the newly form sequence is an A.P. (b) If each term of an A.P.  is multiplied or divided by a nonzero constant then the newly form sequence is also an A.P.

Let us consider an A.P., `a,a+d,a+2d,...`
Now, by adding `\lamda\left(\lamda\ne0\right)` to each term of the given A.P. we have--
            (a) `a+\lamda,a+d+\lamda,a+2d+\lamda,....`  which are also in an A.P. whose first term is `a+\lamda` and common difference is `d`
By subtracting `\lamda` from each term of the given A.P. we have--
            (b) `a-\lamda,a+d-\lamda,a+2d-\lamda.........` which are also in A.P. whose first term is `a-\lamda` and common difference is `d`.
On multiplying each term of the given A.P. by `\lamda`we have--
            (c) `a\lamda,\left(a+d\right)\lamda,\left(a+2d\right)\lamda,.....` which are also in A.P. whose first term is `a\lamda` and common difference is `\left(a+d\right)lamda-a\lamda=d\lamda`.
            (d) On dividing each term of the given A.P. by `\lamda` we have--
`\frac{a}{\lamda},\frac{a+d}{\lamda},\frac{a+2d}{\lamda},....` which are also in A.P. whose first term is `\frac{a}{\lamda}` and common difference is `\frac{a+d}{\lamda}-\frac{a}{\lamda}=\frac{d}{\lamda}`.
 
Some special series

(a)    Sum of first `n` natural numbers: `1+2+3+...+n=\frac{n\left(n+1\right)}{2}`
(b)    Sum of the squares of the first `n` natural numbers: `1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}`
(c)    Sum of the cubes of the first n natural numbers: `1^3+2^3+3^3+...+n^3={\frac{n\left(n+1\right)}{2}}^2`.

Proofs of the above theorems are not given here. We will consider them as questions and questions & answers will be available in 'Illustrative example section' 
 
ILLUSTRATIVE EXAMPLES

Example 1: If `p,q,r` are in A.P., prove that the following are also in A.P. 
(i)    `p^2\left(r+q\right),q^2\left(r+p\right),r^2\left(p+q\right)`, 
(ii)    `\frac{1}{qr},\frac{1}{rp},\frac{1}{pq}`
(iii)    `p\left(\frac{1}{q}+\frac{1}{r}\right),q\left(\frac{1}{r}+\frac{1}{p}\right),r\left(\frac{1}{p}+\frac{1}{q}\right)`
(iv)    `\frac{1}{\sqrtq+\sqrtr},\frac{1}{\sqrtr+sqrtp},\frac{1}{sqrtp+\sqrtq}`
(v)   `\left(q+r\right)^2-p^2`, `\left(r+p\right)^2-q^2`, `\left(p+q\right)^2-r^2`

Solution: (i) Since `p,q,r` are in A.P., 
`\therefore` `p\left(pq+qr+rp\right),q\left(pq+qr+rp\right),r\left(pq+qr+rp\right)` will also be in A.P. 
                                                                    [multiplying each term by `\left(pq+qr+rp\right)`] 
`\implies`    `p\left(pq+rp\right)+pqr,q\left(pq+qr\right)+pqr,r\left(qr+rp\right)+pqr` will also be in A.P.
`\implies`    `p^2\left(q+r\right)+pqr+` `q^2\left(p+r\right)+pqr+` `r^2\left(p+q\right)+pqr` will also be in A.P.
`\implies`    `p^2\left(q+r\right),q^2\left(p+r\right),r^2\left(p+q\right)` will also be in A.P. 
                                                                                               [subtracting `pqr` from each term]
  (ii) Since `p,q,r` are in A.P., 
 `\therefore` `\frac{p}{pqr},\frac{q}{pqr},\frac{r}{pqr}` will also be in A.P. [dividing each term by `pqr`]
`implies`    `\frac{1}{qr},\frac{1}{pr},\frac{1}{pq}` will also be in A.P. 

(iii) Since `p,q,r` are in A.P.
`\therefore`    `\frac{p}{pqr},\frac{q}{pqr},\frac{r}{pqr}`will also be in A.P. [dividing each term by `pqr`]
`implies`    `\frac{1}{qr},\frac{1}{pr},\frac{1}{pq}` will also be in A.P. 
`implies`    `\frac{pq+qr+rp}{qr},\frac{pq+qr+rp}{pr},\frac{pq+qr+rp}{pq}` will also be in A.P. 
                                                                     [multiplying each term by `\left(pq+qr+rp\right)`]
`\implies`    `\frac{pq+qr+rp}{qr}-1,\frac{pq+qr+rp}{pr}-1,\frac{pq+qr+rp}{pq}-1`will also be in A.P.                                                                                                                   [adding `-1` to each term]
`\implies`     `\frac{pq+rp}{qr},\frac{pq+qr}{pr},\frac{qr+rp}{pq}` will also be in A.P.
`\implies`    `p\left(\frac{1}{q}+\frac{1}{r}\right),q\left(\frac{1}{r}+\frac{1}{p}\right),r\left(\frac{1}{p}+\frac{1}{q}\right)` will also be in A.P.

(iv) Since `p,q,r` are in A.P.
`\therefore`    `q-p=r-q`
`\implies`        `\left(\sqrtq+\sqrtp\right)\left(sqrtq-sqrtp\right)` `=\left(\sqrtr+\sqrtq\right)\left(sqrtr-sqrtq\right)` 
 `\implies`    `\frac{\left(sqrtq+\sqrtp\right)\left(sqrtq-\sqrtp\right)}{\left(sqrtp+\sqrtq\right)\left(sqrtq+\sqrtr\right)\left(sqrtr+sqrtp\right)}`  `=\frac{\left(\sqrtr+\sqrtq\right)\left(\sqrtr-\sqrtq\right)}{\left(sqrtp+\sqrtq\right)\left(sqrtq+\sqrtr\right)\left(sqrtr+\sqrtp\right)}`
                                                        [dividing both the sides by `\left(\sqrtp+\sqrtq\right)\left(\sqrtq+\sqrtr\right)\left(\sqrtr+\sqrtp\right)`]   
`\implies`    `\frac{\left(\sqrtq-\sqrtp\right)}{\left(\sqrtq+\sqrtr\right)\left(\sqrtr+\sqrtp\right)}=` `\frac{\left(\sqrtr-\sqrtq\right)}{\left(\sqrtp+\sqrtq\right)\left(\sqrtr+\sqrtp\right)}`
`\implies`    `\frac{\left(\sqrtq+\sqrtr\right)-\left(\sqrtr+\sqrtp\right)}{\left(sqrtq+\sqrtr\right)\left(\sqrtr+\sqrtp\right)}=\frac{\left(\sqrtr+\sqrtp\right)-\left(\sqrtp+\sqrtq\right)}{\left(\sqrtp+\sqrtq\right)\left(\sqrtr+\sqrtp\right)}` 
`\implies`    `\frac{1}{\sqrtr+\sqrtp}-\frac{1}{\sqrtq+\sqrtr}=\frac{1}{\sqrtp+\sqrtq}-\frac{1}{\sqrtr+\sqrtp}`   
Thus, `\frac{1}{\sqrtq+\sqrtr},\frac{1}{\sqrtr+sqrtp},\frac{1}{sqrtp+\sqrtq}` are in A.P.

(v) Since `p,q,r` are in A.P., 
`therefore`    `-2p,-2q,-2r` are in A.P.                  [Multiplying each term by `-2`]
`\implies`    `\left(p+q+r\right)-2p`, `\left(p+q+r\right)-2q`, `\left(p+q+r\right)-2r`  are in A.P. 
                                                                                   [Adding `p+q+r` to each term]
`\implies`    `\left(q+r-p\right),\left(p+r-q\right),\left(p+q-r\right)`  are in A.P.
`\implies`    `\left(p+q+r\right)\left(q+r-p\right),\left(p+q+r\right)\left(p+r-q\right),\left(p+q+r\right)\left(p+q-r\right)` are in A.P.
                                                                                   [Multiplying each term by `\left(p+q+r\right)`]
`\therefore`   `\left(q+r\right)^2-p^2`, `\left(r+p\right)^2-q^2`, `\left(p+q\right)^2-r^2` are in A.P. 

Example 2: Prove that (i) Sum of first `n` natural numbers is `\frac{n\left(n+1\right)}{2}`.

(ii) Sum of the squares of the first `n` natural numbers is `\frac{n\left(n+1\right)\left(2n+1\right)}{6}`.

(iii) Sum of the cubes of first `n` natural numbers is `{\frac{n\left(n+1\right)}{2}}^2`.

Solution: (i) Let        `S_n=1+2+3+...+n=\sumn`

`\therefore`             `S_n=1+2+3+...+n`

                      `S_n=n+\left(n-1\right)+\left(n-2\right)+...+1` 

Now, adding the above equations we get, 

                      `2S_n=n\left(n+1\right)`    `\therefore`    `S_n=` `\frac{n\left(n+1\right)}{2}`.

Hence, `\sumn=1+2+3+...+n=\frac{n\left(n+1\right)}{2}`.

Remark: We can get the same result by using the formula `S_n=\frac{n}{2}\left(a+l\right)`, because, `1,2,3,…,n` is an A.P. whose first term is `1`, last term `n` and number of terms is `n`.

`\therefore`      `S_n=\frac{n}{2}\left(1+n\right)`.

(ii) Let `S_n=1^2+2^2+3^2+...+n^2=\sumn^2`.

Now, putting `n=1,2,3,...,n` in the identity `n^3-\left(n-1\right)^3=3n^2-3n+1`, we get

                        `1^3-0^3=3\.1^2-3\.1+1`

                        `2^3-1^3=3\.2^2-3\.2+1`

                        `3^3-2^3=3\.3^2-3\.3+1`

                        `...................................`

                        `...................................`

                        `n^3-\left(n-1\right)^3=3\.n^2-3\.n+1`

`....................................................................................................................................'

Adding,        `n^3=3\left(1^2+2^2+3^2+...+n^2\right)-3\left(1+2+3+...+n\right)+n`

`implies`        `n^3=3\.S_n-\frac{3n\left(n+1\right)}{2}+n`

`therefore`    `S_n=\frac{n\left(n+1\right)\left(2n+1\right)}{6}`

`\sumn^2=1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}`

(iii) Let `S_n=1^3+2^3+3^3+...+n^3=\sumn^3`

Now, putting `n=1,2,3,...,n` in the identity `n^4-\left(n-1\right)^4=4n^3-6n^2+4n-1`, we get

                       `1^4-0^4=4\.1^3-6\.1^2+4\.1-1` 

                        `2^4-1^4=4\.2^3-6\.2^2+4\.1-1`

                        `3^4-2^4=4\.3^3-6\.3^2+4\.3-1`

                        `................................................`

                        `................................................`

                        `n^4-\left(n-1\right)^4=4\.n^3-6\.n^2+4\.n-1`

`......................................................................................................................................................`

(Adding)    `n^4=4\left(1^3+2^3+3^3+...+n^3\right)-6\left(1^2+2^2+3^2+...+n^2\right)+4\left(1+2+3+...+n\right)-n`

`\implies`    `n^4=4\.S_n-6\.\frac{n\left(n+1\right)\left(2n+1\right)}{6}+4\.\frac{n\left(n+1\right)}{2}-n`

`\therefore`    `S_n={\frac{n\left(n+1\right)}{2}}^2`.

Hence,    `\sumn^3={frac{n\left(n+1\right)}{2}}^2`.    


Example 3: Find the sum of the following series

(i)    `1\.2+2\.3+...+n\.\left(n+1\right)`

(ii)    `1+9+25+49+...` to `n` terms.

(iii)    `1^2+3^2+5^2+...` to `n` terms

(iv)    `3\.1^2+4\.2^2+5\.3^2+......+\left(n+2\right)\.n^2`

Solution (i) Let `S_n=1\.2+2\.3+...+n\.\left(n+1\right)`  and `t_n=n\left(n+1\right)=n^2+n`

`\therefore`    `\sumt_n=\sum\left(n^2+n\right)=\sumn^2+\sumn=\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}`

`\implies`    `S_n=\sumt_n=frac{1}{3}n\left(n+1\right)\left(n+2\right)`

(ii) In this case, the terms of the sequence do not appear to be in A.P. Also, the common difference of the sequence cannot be directly determined from the terms. However, if we observe the terms carefully, it can be noticed that differences between consecutive terms are in A.P. So, by using this concept, let see how we can determine the sum to `n` terms of this given series.

Let the sum to `n` terms of the series be `S_n` and the last term be `t_n`.

`\therefore`    `S_n=1+9+25+49+..................+t_n`

                `S_n=   1+9+25+49+...............+t_{n-1}+t_n` 

`.................................................................................................................`

(Subtract)    `0=1+8+16+24+............+\left(t_n-t_{n-1}\right)-t_n`

 `implies`    `t_n=1+8\left(1+2+3+......+\overline{n-1}\right)=4n^2-4n+1`   

`therefore`    `S_n=\sumt_n=\sum\left(4n^2-4n+1)=4\sumn^2-4\sumn+n`

               `=4frac{n\left(n+1\right)\left(2n+1\right)}{6}-4\frac{n\left(n+1\right)}{2}+n`

                `=\frac{n}{3}\left(4n^2-1\right)`.

(iii) Let `S_n` be the sum of `n`terms of this series and `t_n` be its `n`th term

Now,        `t_n=`  (n th term of `1,3,5,...` to `n` terms)^2 and `S_n=\sumt_n=\sum\left(4n^2-4n+1\right)`

`\implies`        `S_n=4\sumn^2-4\sumn+n`    [Remember `\suma=na` thus, `\sum1=n\times1=n`]

`\implies`        `S_n=4\frac{n\left(n+1\right)\left(2n+1\right)}{6}-{\frac{n\left(n+1\right)}{2}}+n`

`\implies`        `S_n=\frac{n}{3}[2\left(n+1\right)\left(2n+1\right)-6\left(n+1)+3]=\frac{n}{3}\left(4n^2+6n+2-6n-6+3\right)`

`\implies`        `S_n=\frac{n}{3}\left(4n^2-1\right)`

(iv) Here, the `n`th term of the given series is `t_n=\left(n+2\right)\.n^2`.

Let `S_n` be the sum to `n`terms of the given series. Then `S_n=\sum\left(n^3+2n^2\right)`

Now,        `S_n=\sumn^3+2\sumn^2={\frac{n\left(n+1\right)}{2}}^2+2\frac{n\left(n+1\right)\left(2n+1\right)}{6}`

`\implies`        `S_n=\frac{1}{12}n\left(n+1\right)\left(3n^2+11n+4\right)`


Exercise-1(D)

Answer the following Questions:

    1. If `p,q` and `r` are three consecutive terms of an A.P. Show that, (for any nonzero constant `k`)
      (i) `p+k,q+k,r+k` are in A.P.                       (ii) `3k-p,3k-q,3k-r` are in A.P.
      (iii) `4p-5k,4q-5k,4r-5k` are in A.P.      (iv) `p+q,q+r,r+p` are in A.P.
    2. If `\frac{a^2+b^2}{c^2},\frac{b^2+c^2}{a^2},\frac{c^2+a^2}{b^2}` are in A.P. then prove that `\frac{1}{b^2},\frac{1}{a^2},\frac{1}{c^2}` are also in A.P., where (`a^2+b^2+c^2\ne0`).  
    3. Show that for any value of `k`,  `a+2k^2,2c+k^2,2a+c` are in arithmetic progression if `a=c`.
    4. If `\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}` are in A.P., then show that`\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}` are also in A.P., where (`a+b+c\ne0`).  
    5. If `\left(b-c\right)^2,\left(c-a\right)^2,\left(a-b\right)^2` are in A.P., then prove that, `\frac{1}{b-c},\frac{1}{c-a},\frac{1}{a-b}` will be in A.P. 
    6. Find the sum to first `n` terms of the following series
      (i) `2\.4+6\.8+10\.12+......`                             (ii) `1+5+12+22+35+.....`
      (iii) `\left(3^3-2^3\right)+\left(5^3-4^3\right)+\left(7^3-6^3\right)+......`
    7. Find the sum to first `2n` terms of the series `1^2-2^2+3^2-4^2+5^2-6^2+....`
    8. Find the sum to `n` terms 0f the series `\frac{1}{3\.4}+\frac{1}{4\.5}+\frac{1}{5\.6}+....`


Answer

 6.(i) `\frac{4}{3}n\left(n+1\right)\left(4n-1\right)`; (ii) `\frac{1}{2}n^2\left(n+1\right)`; (iii) `\n\left(4n^2+9n+6\right)`; 7.`-n\left(2n+1\right)`. 8. `\frac{n}{3\left(n+3\right)}` 


Let's discuss what we have learned in the last three posts on "Arithmetic Progression"

(i)    If, in a sequence, each term after the first is obtained by adding a fixed number to the preceding term, then the sequence is called an Arithmetic Progression (A.P.). 

(ii)    `t_1,t_2,t_3,...,t_n` will be in A.P. if, `t_2-t_1=t_3-t_2=....,=t_n-t_{n-1}`.

(iii)    If in an A.P., the first term is `a` and common difference is `d` then the `n`th term is `t_n=a+\left(n-1\right)d` and sum to first `n` terms is `S_n=\frac{n}{2}\left(a+l\right)=\frac{n}{2}{2a+\left(n-1\right)d}`  where, `l=a+\left(n-1\right)d`.

(iv)    If sum to first `n` terms of an A.P. is `S_n` then its `n`th term can be obtained by the formula `t_n=S_n-S_{n-1}`.

(v)    If three numbers are in A.P., then the middle number is the arithmetic mean of the other two, and the sum of the two extreme numbers is equal to twice of the middle number. 

(vi)    If each term of an A.P. is multiplied or divided by a fixed non-zero number or if each term is added to or subtracted by same non-zero number, then the new sequence thus obtained will also be an A.P.

(vii)    `\sumn=1+2+3+...+n=\frac{n\left(n+1\right)}{2}`, 
           `\sumn^2=1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}`
            `\sumn^3=1^3+2^3+3^3+...+n^3={\frac{n\left(n+1\right)}{2}}^2`.


            

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