Progressions-A.P. and G.P. (Part-2)

 

© ® Written by Sankar Ghosh

Hello friends!

I hope you are all doing well and staying happy.

In my last blog, I discussed the concept of a sequence, what an arithmetic progression (A.P.) is, and how to find its general term. We also looked into special cases such as when 3 terms or 4 terms are in A.P., along with different types of problems related to these topics.

In today’s post, I will take the discussion forward and explain how to find the sum of n terms of an A.P. as well as the concept of the arithmetic mean.

I would love to hear your thoughts on the post. Please share your comments and suggestions—they will help me improve further!

Sum to n terms of an A.P.

Let the first term and common difference of an A.P. are `a` and `d` respectively and sum to `n` terms of the A.P. be `S_n`. 

`\therefore` `S_n=a+\left(a+d\right)+\left(a+2d\right)+...+\left(l-2d\right)+\left(l-d\right)+l`     [where `l=a+\left(n-1\right)d`]

`\implies``S_n=l+\left(l-d\right)+\left(l-2d\right)+...+\left(a+2d\right)+\left(a-d\right)+a`

Adding `2S_n=\left(a+l\right)+\left(a+l\right)+....\left(a+l\right)` `\implies 2S_n=n\left(a+l\right)\impliesS_n=\frac{n}{2}\left(a+l\right)` 

Now, Putting the value of `l=a+\left(n-1\right)d` in the above equation, we get--

`S_n=\frac{n}{2}\{2a+\left(n-1\right)d}`

ILLUSTRATIVE EXAMPLES

Example 1: If the first term and common difference of an A.P. are respectively 4 and 2, then find the sum of its first 13 terms.

Solution: We know that sum to `n` terms of an A.P. whose first term and common difference are `a` and `d` respectively and having numbers of term is `n`, is  `S_n=\frac{n}{2}\{2a+\left(n-1\right)d}` .

Here, `a=4`, `d=2` and `n=13`        `\therefore S_13=\frac{13}{2}{2\times4+\left(13-1\right)2}=208.`

Example 2: If the nth term of an A.P. be `\left(2n-1\right), then find the sum of the first 10 terms of the A.P.

Solution: We have, `t_n=2n-1`      

`\therefore` first term `\left(a\right)=t_1=2\times1-1=1` and common difference `d=t_2-t_1=\left(2\times2-1\right)-\left(2\times1-1\right)=3-1=2`.

`\therefore`                `S_10=\frac{10}{2}{2\times1+\left(10-1\right)2}=5\left(2+18\right)=100` 

Example 3: The sum of first `p,q,r` terms of an A.P. are `a,b,c` respectively. Show that --

                    `\frac{a}{p}\left(q-r\right)+\frac{b}{q}\left(r-p\right)+\frac{c}{r}\left(p-q\right)=0`

Solution: Let `m` be the first term and `n` be the common difference of the given A.P. Then,

                `a=S_p\impliesa=\frac{p}{2}{2m+\left(p-1\right)n}\implies\frac{2a}{p}={2m+\left(p-1\right)n}`            ...............(1)

                `b=S_q\impliesb=\frac{q}{2}{2m+\left(q-1\right)n}\implies\frac{2b}{q}={2m+\left(q-1\right)n}`            ................(2)

 and,        `c=S_r\impliesc=\frac{r}{2}{2m+\left(r-1\right)n}\implies\frac{2c}{r}={2m+\left(r-1\right)n}`            .................(3)

Now multiplying (1), (2) and (3) by `\left(q-r\right),\left(r-p\right)` and `\left(p-q\right)` respectively and adding, we get--

                `\frac{2a}{p}\left(q-r\right)+\frac{2b}{q}\(r-p\right)+\frac{2c}{r}\left(p-q\right)`

                `={2m+\left(p-1\right)n}\left(q-r\right)+{2m+\left(q-1\right)n}\left(r-p\right)+{2m+\left(r-1\right)n}\left(p-q\right)`

                `=2m\left(q-r+r-p+p-q\right)+n\{\left(p-1\right)\left(q-r\right)+\left(q-1\right)\left(r-p\right)+\left(r-1\right)\left(p-q\right)}`

                `=2m\times0+n\times0=0`

Example 4: If the sum of first `n` terms of an A.P. be `\left(n^2+3n\right)` then which term of the A.P. is 162?

Solution: Given, `S_n=\left(n^2+3n\right)`        `\thereforeS_{n-1}={\left(n-1\right)^2+3\left(n-1\right)}`

Now,         `t_n=S_n-S_{n-1}`       `\implies`    `\left(n^2+3n\right)-{\left(n-1\right)^2+3\left(n-1\right)}=2n+2`

`\therefore`            `t_n=2n+2\implies2n+2=160\impliesn=80`.

Remember: If sum of `n` terms of a sequence is `S_n`, then nth term of the sequence can be obtained by the formula, `t_n=S_n-S_{n-1}` If the sum of n terms of A.P. is `S_n=an^2+bn+c` then the common difference of the A.P. is `2a`

Example 5: If the middle term of an A.P. consisting of 11 terms is 12, then find the sum of 11terms of that A.P.

Solution:     Given that, an A.P. consisting of 11 terms has its middle term is 12 i.e., `t_6=12\impliesa+5d=12` [taking `a` and `d` as the first term and common difference respectively] 

Now,            `S_11=\frac{11}{2}{2a+\left(11-1\right)d}=\frac{11}{2}\left(2a+10d)=11\left(a+5d\right)=11\times12=132`.

Example 6: The sum of the first four terms of an A.P. is 56. The sum of last four terms is 112. If its first term is 11, then find the number of terms.

Solution: Given that, sum of the first four terms is 56. 

Let the number of terms of the A.P. be `n` with first term `a=11` and common difference `d`

`\therefore`                   `\frac{4}{2}{2\times11+\left(4-1\right)d=56`.

`\implies`                  `\22+3d=28\implies3d=6\impliesd=2`

Also, we have sum of the last four terms`=112`

`\implies`                  `t_{n-3}+t_{n-2}+t_{n-1}+t_n=112`

`\implies`                 `\frac{4}{2}\left(t_{n-3}+t_n\right)=112`

`\implies`                  `t_{n-3}+t_n=56`

`\implies`                  `{11+\left(n-4\right)2}+{11+\left(n-1)2}=56`

`\implies`                  `22+2\left(2n-5\right)=56`

`\implies`                  `4n=44\impliesn=11`

Thus, the number of terms of the A.P. is 11.

Example 7: Find the number of terms in the series `27+24+21+18+.....` must be added to get the sum 126? Explain the double answer.

Solution: Here, the given series is `27+24+21+18+......`

Let `n` number of terms must be added to the series to get the sum 126.

Here,        `a=27,d=24-27=-3`

Thus,       `\frac{n}{2}{2\times27+\left(n-1\right)\times-3}=126`

`\implies`         `n{\left(54-3n+3\right)}=126\times2`  `\impliesn\left(57-3n\right)=252`

`\implies`         `3n^2-57n+252=0\impliesn^2-19n+84=0\implies\left(n-12\right)\left(n-7\right)=0` 

`\therefore`          `n=12,7`.

`therefore` the value of `n` is 7 or 12. Here we have two values of n, since the first term of the series is positive and the common difference is negative, so the sum of the last 5 terms is zero.

[As `t_7=a+6d=27+6\left(-3\right)=27-18=9`. `\therefore` the remaining 5 terms are `9-3,6-3,3-3,0-3,-3-3` i.e., `6,3,0,-3,-6`. Clearly, the sum of the last five terms is zero.] 

Example 8: The ratio of the sum of first `n` terms of two different A.P.'s is `\left(2n+1\right):\left(2n-1\right)`. Find the ratio of their 8 th terms.

Solution: Let the first terms and common differences of the two A.P.'s be `a_1,a_2`and `d_1,d_2`respectively.

Now, by the condition given we have

                         `\frac{\frac{n}{2}{2a_1+\left(n-1\right)d_1}}{\frac{n}{2}{2a_2+\left(n-1\right)d_2}}=\frac{2n+1}{2n-1}`

                        `\implies\frac{2a_1+\left(n-1\right)d_1}{2a_2+\left(n-1\)d_2}=\frac{2n+1}{2n-1}``\implies\frac{a_1+\frac{n-1}{2}d_1}{a_2+\frac{n-1}{2}d_2}=\frac{2n+1}{2n-1}.........(1)`

Now, the ratio of 8th terms of the two A.P.'s is `\frac{a_1+7d_1}{a_2+7d_2}`,

`\therefore` putting `frac{n-1}{2}=7` i.e., `n=15` on both side of (1) we have,

                         `\frac{a_1+7d_1}{a_2+7d_3}=\frac{2\times15+1}{2\times15-1}=\frac{31}{29}`

Thus, the ratio of 8th terms of the two A.P.'s is `31:29`.

Remember Students are advised to observe the solution of the above example carefully. Here, we are given the ratio of the sums of `n` terms of two A.P.'s and then asked to find the ratio of their 8th terms. Suppose we are asked to find the ratio of their kth term, then just put `n=2k-1`and get the required ratio. Let's check the above concept in the given problem. we had, `\frac{S_1}{S_2}=\frac{2n+1}{2n-1}` `\therefore`        `\frac{t_1{8}}{t_2{8}}=\frac{2\times15+1}{2\times15-1}` Here, `n=2k-1=2\times8-1=15`.

Exercise 1(B)

(MCQ)

A. Chose thee correct Answer

1. The sum of all odd integers between 2 and 100 divisible by 3 is
   (a) 862 
   (b) 863 
   (c) 865
   (d) 867

2. If the ratio of the sums of `m` and `n` term of an A.P. is `m^2:n^2`, then the ratio of its `m`th and `n`th terms is given by
   (a) `2m+1:2n+1`
   (b) `2m-1:2n-1`   
   (c) `m:n`             
   (d) `m-1:n-1`
3. If sum to n terms of an A.P. is `5n^2+6n` and rth term is 402, then the value of `r` is
   (a) 40 
   (b) 30 
   (c) 20
   (d) 10
4. If `S_1,S_2,S_3` are the sums of `n,2n,3n` terms respectively of an A.P., then `\frac{S_3}{S_2-S_1}=`
   (a) 1
   (b) 2 
   (c) 3   
   (d) 4
5. The first and last terms of an A.P. are 1and 11. If the sum of its terms is 36, then the number of terms will be
   (a) 5 
   (b) 6 
   (c) 7 
   (d) 8
6. If an A.P. `S_n=n^2q` and `S_m=m^2q`, where `S_r` denotes the sum of `r` terms of the A.P., then `S_q`equals
   (a) `\frac{q^3}{2}` 
   (b) `mnq` 
   (c) `q^3`   
   (d) `\left(m^2+n^2\right)q` 
7. If the sum of n terms of an A.P. is given by `S_n=3n+2n^2`, then the common difference of the A.P. is
   (a) 3 
   (b) 2   
   (c) 6 
   (d) 4 
8. Statement-1: The sum of `n` terms of two arithmetic progressions are in the ratio `\(7n+1\right):\left(4n+17\right)`, then the ratio of their nth term is 7:4.
   Statement-II: If `S_n=ax^2+bx+c`, then `t_n=S_n-S_{n-1}`
   (a) Both the Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement-I
   (b) Both the Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I
   (c) Statement-1 is true but Statement-II is false
   (d) Statement-1 is false and Statement-II is true

B. Answer the following questions

1. The 12th term of an A.P. is 23 and sum of its first 22 terms is 484. Find the sum of first 30 terms of this A.P. [Ans: 900]
2. If the 5th and 12th terms of an A.P. are 30 and 65, then find the sum of its first 20 terms. [Ans: 1150] 
3. If the sum of the first `n` terms of an A.P. is `n\left(2-n\right)` then find its 10th term.  [Ans: -17]
4. If the sum of the first `n` terms of an A.P. be `n^2` then find its common difference. [Ans: 2]                                                                                                     
5. If the nth terms of an A.P. is `p`, then show that the sum of the first `\left(2n-1\right)` terms is `(2n-1)p`.
6. If the sum of `P` terms of an A.P. is equal to the sum of first `Q` terms, then show that the sum of first `\left(P+Q\right)` terms is zero.  
7. The first, second and last terms of an A.P. are `a,b` and `c` respectively. Show that the sum of the terms of the series is `\frac{\left(a+c\right)\left(b+c-2a\right)}{2\left(b-a\right)}`.
8. The sum of first `n` terms of an A.P. is `q`, and the nth term is `p`; show that the first term is `\frac{2q-pn}{n}`.
9. The pth term of an A.P. is `\frac{1}{q}` and qth term is `\frac{1}{p}`. Show that the sum of `pq` number of terms is `\frac{1}{2}\left(pq+1\right)`.
10. Let pth and qth terms of an A.P. are respectively `a` and `b`. Show that the sum of the first `\left(p+q\right)` terms is `\frac{1}{2}\left(p+q\right)\left(a+b+\frac{a-b}{p-q}\right)`.
11. If `S_n` be the sum of the first `n` terms of an A.P., then show that, `S_{3n}=3\left(S_{2n}-S_n\right)`.
12. If `S_1,S_2` and `S_3` be the sums of `n` terms of three A.P.'s, the first term of each A.P. being 1 and the respective common differences being 1, 2, 3, show that, `S_1+S_3=2S_2`.

Arithmetic Mean (or A.M.): Let `a,b,c` are in A.P. `\thereforeb-a=c-b\implies2b=a+c\impliesb=\frac{a+c}{2}` . Here, `b` is called arithmetic mean of `a` and `c`.

If, `a,x_1,x_2,x_3,.....,x_n,b` are in A.P. then `x_1,x_2,x_3,...,x_n` are called `n` arithmetic means between `a` and `b`. 

`\therefore`                `t_1=a` and the `\left(n+2\right)` th term is `b`. Let the common difference is `d`. 

`\therefore`                `b=t_{n+2}=a+{\left(n+2\right)-1}d`  `\implies` `b=a+\left(n+1\right)d`    `\therefored=\frac{b-a}{n+1}`.    

`\therefore`                            `x_1=a+d=a+\frac{b-a}{n+1}`, 

                               `x_2=a+2d=a+\frac{2\left(b-a\right)}{n+1}`,

                                   ....................................................... , 

                               `x_n=a+nd=a+\frac{n\left(b-a\right)}{n+1}` .

Thus the required A.M.'s are respectively,

                    `a+\frac{b-a}{n+1},a+\frac{2\left(b-a\right)}{n+1},...,a+\frac{n\left(b-a\right)}{n+1}`

Now, let us find the sum of these A.M.'s --

            `a+\frac{b-a}{n+1}+a+\frac{2\left(b-a\right)}{n+1}+...+a+\frac{n\left(b-a\right)}{n+1}`

            `=na+\left(b-a\right){\frac{1}{n+1}+\frac{2}{n+1}+...+\frac{n}{n+1}}`

            `=na+\left(b-a\right)\frac{1+2+...+n}{n+1}` `=na+\left(b-a\right)frac{\frac{n\left(n+1}}{2}}{n+1}`

            `=na+\left(b-a\right)\frac{n}{2}=\frac{n}{2}\left(a+b\right)`

Thus, we can conclude sum of the `n` arithmetic means between two numbers is `n` times the single A.M. between them.

ILLUSTRATIVE EXAMPLES

Example 1: Insert 5 arithmetic means between `-5` and `7`

Solution: We have to insert 5 A.M.s between `-5` and `7`. Hence we get an A.P. whose first term `a=-5`, number of terms `n+2=5+2=7` and last term i.e., `t_7=7`. Let the common difference be `d`. `\therefored=\frac{b-a}{5+1}` [Why?]

`\implies`         `d=2` 

`\therefore` the five arithmetic mans are, `a+d,a+2d,a+3d,a+4d,a+5d` 

i.e., `-5+2,-5+2\times2,-5+3\times2,-5+4\times2,-5+5\times2` i.e., `-3,-1,1,3,5`.

Example 2: There are `n` arithmetic means between 4 and 40. If the ratio of first mean and last mean is `2:9`, then find the value of `n`.

Solution: Here, `a=4,t_{n+2}=40` and `d=\frac{40-4}{n+1}=\frac{36}{n+1}`.

Now, according to the problem we have, 

         `\frac{a+d}{a+nd}=\frac{2}{9}\implies\frac{4+\frac{36}{n+1}}{4+\frac{36n}{n+1}}=\frac{2}{9}\implies\frac{4\left(n+1\right)+36}{4\left(n+1\right)+36n}=\frac{2}{9}\implies\frac{n+10}{10n+1}=\frac{2}{9}\impliesn=8`. 

  

Exercise 1(C)

(MCQ)

A. Chose thee correct Answer

1. If `n` A.M.'s are introduced between 3 and 17 such that the ratio of the last mean to the first mean is 3:1, then the value of n is

   (a)

   6

   (b)

   8

   (c)

   4

   (d)

   10

   
   
2. Three arithmetic means between 3 and 19 are

   (a)

   7,11,15

   (b)

   5,10,15

   (c)

   6,11,16

   (d)

   5,7,9

   
   
3. There are `m` A.M.'s between 1 and 31. If the ratio of the 7th and `\left(m-1\right)`th means is 5:9, then `m=`
   

   (a)

   10

   (b)

   12

   (c)

   14

   (d)

   16

   
   
4. If `b` is the arithmetic mean of `a` and `c`, then the arithmetic mean of `ab` and `ac` will be
   

   (a)

   `2b`

   (b)

   `b^2`

   (c)

   `2b^2`

   (d)

   `abc`

B. Answer the following questions

1. Insert 10 arithmetic mean between 2 and 57. [Ans: 7,123,17,22,27,31,37,42,47,52]
2. Insert few arithmetic means between 12 and 52 so that the sum of all the terms of the A.P. becomes 192 [Ans: 20, 28,36,44]
3. If `\frac{1}{2\left(y-a\right)}` is the arithmetic mean between `\frac{1}{y-x}` and `\frac{1}{y-z}` , show that `\left(x-a\right)\left(z-a\right)=|left(y-a\right)^2`.
4. If `\frac{a^{n+1}+b^{n+1}}{a^n+b^n}` is the arithmetic mean between `a` and `b`, find value of `n`. [Ans: 0]
5. Prove that sum of the `n` arithmetic means between two numbers is `n` times the single A.M. between them.
6. The sum of two numbers is `\frac{13}{6}`. An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. find the number of means inserted. [Ans:6]