Progressions - A.P. and G.P. (Part-1)
© ® Written by Sankar Ghosh
Now, if in a sequence the difference of a term and the previous term is always same (i.e. constant) then the sequence is called Arithmetic Progression (A.P.). The constant difference is called Common Difference (C.D.).
In this chapter we will discuss how to construct an A.P. and the different things related to it.
What will we learn in this Chapter?
If we study this chapter, we will
- learn Arithmetic progression and its characteristics.
- determine the common difference and the general term of any A.P.
- learn the definition of Arithmetic Mean (A.M.) and how to determine it.
- learn how to determine the sum to first n terms of any A.P.
- determine the sum to first n natural numbers.
- solve some related problems
We have discussed earlier that the numbers by which an A.P. is formed are called the terms of the A.P. The number in the first position of the A.P. is called first term (or in brief `t_1`), number in the second position is called second term (or `t_2`) and in this way the number in the rth position is called rth term (or `t_r`). Now if the first term `t_1` and common difference of any Progression be denoted by a and d respectively then by the properties of A.P. we can write,
`t_2-t_1=t_3-t_2=t_4-t_3=...=t_t-t_{r-1}=...=d` and as `t_1=a` therefore the terms of the A.P. can be written as
`t_1=a,t_2=t_1+d=a+d,t_3=t_2+b=a+2d,t_4=t_3+d=a+3d,...` i.e., `t_1=a+\left(1-1\right)d,t_2=a+\left(2-1\right)d,t_3=a+\left(3-1\right)d,t_4=a+\left(4-1\right)d,...` and in this way the rth term `t_r` can be written as `t_r=a+\left(r-1\right)d`.
We can get the different terms of an A.P. by putting `r=1, 2, 3, .. ,` etc in the rth term `t_r`. For this reason `t_r` is called general term of A.P. Here, we see that `t_r` is only dependent on the first term and common difference, hence we can say that any term an A.P. can be determine if we know the first term and common difference.
Let us know few terms and definition from the above discussion.
- First term: The numbers by which a progression is formed its first one is called first term ( `t_1`) and is generally denoted by a.
- Last term: If an A.P. is consist of n number of terms (i.e. finite number of terms) the nth term i.e. `t_n` is calld last term.
- Common difference: In any A.P. the difference of a term and the previous term is called common difference. Common difference may be positive as well as negative also and the common difference is generally denoted by d.
- General term: The term from which the other terms of the A.P. can be obtained is called general term. As from `t_r=a+\left(r-1\right)d` we can get `t_1,t_2,t_3...` etc. by putting `r= 1, 2, 3,..`, `t_r` is called general term.
rth Term of an A.P. from the end: Let a be the first term and d the common difference of an A.P. having n terms. Then, rth term from the end is `(n-r+1)` th term from the beginning.
`\therefore t_{n-r+1}=a+\left(n-r+1-1\right)d=a+\left(n-r\right)d`
For finding the rth term from the end, we may take `t_n` as the first term and `-d` as the common difference.
Taking `t_n` as the first term and the common difference `-d`, we find rth term from the end `=t_n+\left(r-1\right)\left(-d\right)`.
Note: If three terms are in A.P. then consider them as `a-d,a,a+d`.
If four terms are in A.P. then consider them as `a-3d,a-d,a+d,a+3d`
ILLUSTRATIVE EXAMPLES
Example 1: Find the A.P. in each of the following cases.
(i) first term is 5 and fourth term is 17
(ii) numbers of terms 15 and second and last term are respectively 9 and 61
Solution (i): Given that, `a=5`,`t_4=17`. Let the common difference be `d`
`\thereforet_4=17\impliesa+3d=17``\implies5+3d=17\implies3d=12` \therefore `d=4`
Hence the required A.P. is,
`5,5+\left(2-1\right)4,5+\left(3-1\right)4,5+\left(4-1\right)4,...`
or `5,9,13,17,....`
Example 2: The first term and the common difference of an A.P. are `-3` and `3` respectively and the last term is 54. How many terms are there in the A.P?
Solution: Here, `a=-3,d=3` and the last term `t_n=54`. `\therefore-3+\left(n-1\right)3=54``\implies n=20`
Thus the number of terms in the A.P. is 20
Example 3: The 7th and 10th term of an A.P. are 20 and 29 respectively. Find the first term and common difference.
Solution: Let the first term be a and common difference be d.
`\thereforea+\left(7-1\right)d=20\impliesa+6d=20` and
`a+\left(10-1\right)d=29\impliesa+9d=29`.
Solving the above two equation we get `a=2` and `d=3`. Thus, the first term and common difference of the A.P. are 2 and 3 respectively.
Example 4: If pth, qth and rth term of an A.P. are a, b, c respectively, show that:
`\left(a-b\right)r+\left(b-c\right)p+\left(c-a\right)q=0`
Solution: Let the first term and the common difference of the A.P. be m and n respectively. Then,
`t_p=a\Rightarrow m+\left(p-1\right)n=a`
`t_q=b\Rightarrow m+\left(q-1\right)n=b`
`t_r=c\Rightarrow m+\left(r-1\right)n=c`
Now, we put these values of a, b, c, in `\left(a-b\right)r+\left(b-c\right)p+\left(c-a\right)q`
`=\left(a-b\right)r+\left(b-c\right)q+\left(c-a\right)q`
`=\left(p-q\right)nr+\left(q-r\right)np+\left(r-p\right)nq`
`=\ntimes0``=0`
Example 5: If the mth term of an A.P. be `\frac{1}{n}`, and nth term be `\frac{1}{m}` then show that its `\left(mn\right)`th term is 1.
Solution: Let the first term and the common difference of the A.P. be `a` and `d` respectively. Then,
`t_m=\frac{1}{n}` and `t_n=\frac{1}{m}`
`impliesa+\left(m-1\right)b=\frac{1}{n}`..................(1)
and `impliesa+\left(n-1\right)b=\frac{1}{m}`....................(2)
Solving (1) and (2) we get, `a=\frac{1}{mn}` and `d=\frac{1}{mn}`
`\therefore` `t_mn=a+\left(mn-1\right)d=\frac{1}{mn}+\left(mn-1\right)\frac{1}{mn}=1`
Example 6: Divide 57 into three parts such that they are in A.P. and product of the first and last is 345.
Solution: Let the three parts are `a-d,a,a+d`
`\therefore` `a-d+a+a+d=57` `\implies3a=57` `\impliesa=19`. and `\left(a-d\right)\left(a+d\right)=345`.
`\implies` `a^2-d^2=345` `\impliesd^2=16\impliesd=\pm4`.
`\therefore` the required three parts are `19-4,19,19+4` or `15,19,23`
[Remark: Same result will come if we consider `d=-4` also.]
Example 7: In an A.P. `2,5,8,...` upto 50 terms, and `3,5,7,9,...` upto 60 terms, find how many terms are identical.
Solution: Let the pth term of the first A.P. be identical to the qth term of the second A.P.
`\therefore` `2+\left(p-1\right)3=3+\left(q-1\right)2`
`\implies` `3p-1=2q+1`
`\implies` `3p=2q+2`
`\implies` `\frac{p}{2}=\frac{q+1}{3}=\lamda` (say)
`\implies` `p=2\lamda` and `q=3\lamda-1`
`\implies` `2\lamda\leq50` and `3\lamda-1\leq60`
`\implies` `\lamda\leq25` and `\lamda\leq20\frac{1}{3}`
`\implies` `\lamda\leq20`
`\implies` `\lamda=1,2,3,...,20`.
Corresponding to each value of `\lamda`, there will be a pair of identical terms.
Hence, there are 20 identical terms in two A.P.'s.
Example 8: If `a_1,a_2,a_3,...,a_n` be an A.P. of non-zero terms, prove that`\frac{1}{a_1a_2}+\frac{1}{a_2a_3}+...+\frac{1}{a_{n-1}\times a_n}=\frac{n-1}{a_1a_n}`.
Solution: Let the common difference of the given A.P. be `d`.
Then, `a_2-a_1=a_3-a_2=...=a_n-a_{n-1}=d`. (say) .................(1)
Now,
`\frac{1}{a_1a_2}+\frac{1}{a_2a_3}+...+\frac{1}{a_{n-1}a_n}`
`=\frac{1}{d}\{\frac{d}{a_1a_2}+\frac{d}{a_2a_3}+\frac{d}{a_3a_4}+...+\frac{d}{a_{n-1}a_n}}`
`=\frac{1}{d}{\frac{a_2-a_1}{a_1a_2}+\frac{a_3-a_2}{a_2a_3}+\frac{a_4-a_3}{a_3a_4}+...+\frac{a_n-a_{n-1}}{a_{n-1}a_n}}`
`=\frac{1}{d}{\left(\frac{1}{a_1}-\frac{1}{a_2}\right)+\left(\frac{1}{a_2}-\frac{1}{a_3}\right)+\left(\frac{1}{a_3}-\frac{1}{a_4}\right)+...+\left(\frac{1}{a_{n-1}}-\frac{1}{a_n}\right)}`
`=\frac{1}{d}{\frac{1}{a_1}-\frac{1}{a_n}}`
`=\frac{1}{d}{\frac{a_n-a_1}{a_1a_n}}``=\frac{1}{d}{\frac{a_1+\left(n-1\right)d-a_1}{a_1a_n}}``=\frac{n-1}{a_1a_n}`
Exercise 1(A)
(MCQ)
A. Chose thee correct Answer
- If the 7th and 13th terms of an A.P. be 34 and 64 respectively. then it its 18th term is
(a) 87 (b) 88 (c) 89 (d) 90. - If the second and sixth term of an A.P. is 1 and 13 respectively then the sum of the last two terms (when the number of terms of the A.P. is 15) is,
(a) 77 (b) 88 (c) 66 (d) 70. - If the pth term of an A.P. is q and the qth term is p, then `\left(p+q\right)`th term is
(a) 0 (b) 1 (c) `pq` (d) none of these. - Tenth term of the progression 62, 55, 48, .... is
(a) `-7` (b) `-1` (c) 0 (d) 7. - rth term of an A.P. is `4p-7`; the common difference of the progression is
(a) 2 (b) `-2` (c) `-4` (d) 4. - The 10th common term between the A.P.s 3, 7, 11, 15, ... and 1, 6, 11, ...is
(a) 191 (b) 193 (c) 221 (d) none of these. - Consider first three terms of a sequence `S_n` `|x-1|,3,|x-3|`, which are in A.P.
Statement-I: The sixth term of `S_n` is 7<third term.
Statement-II: If `a,a+d,a+2d,...` are in A.P. Then sixth term is `\left(a+5d\right)`. `\left(d\ne0\right)`.
(a) Both statement-1 and statement-II are true and statement-II is the correct explanation of statement-I
(b) Both statement-1 and statement-II are true and statement-II is not the correct explanation of statement-I
(c) Statement-I is true but statement-II is false.
(d) Statement-I is false and statement-II is true.
B. Answer the following questions
- Can `-447` be a term of the A.P.: 8, 5, 2, ...? Give reasons for your answer.
[Ans: no] - Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least. [Ans: 5, 10, 15, 20]
- (a) If `a_1,a_2,a_3,.....,a_{2n-+1}` are in A.P., then show that,`\frac{1}{a_1a_3}+\frac{1}{a_3a_5}+....+\frac{1}{a_{2n-1}a_{2n+1}}=\frac{n}{a_1a_{2n+1}}`
(b) If three positive numbers `a,b,c` are in A.P., then show that, `\frac{1}{\sqrtb+\sqrtc},\frac{1}{\sqrtc+\sqrta},\frac{1}{\sqrta+\sqrtb}` are in A.P.
(c) If `a_1,a_2,a_3,...,a_n` are in A.P. where `a_i>0`, for all values of `i`, show that `\frac{1}{\sqrta_1+\sqrta_2}+\frac{1}{\sqrta_2+\sqrta_3}+.....+\frac{1}{\sqrta_{n-1}+\sqrta_n}=\frac{n-1}{\sqrta_1+\sqrta_n}`.
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