Arithmetic Progression
Mathematics Class Tenth
NCERT Exercise 5.2
Question: 1. Fill in the blanks in the following table, given that a is the first term, d the common differnece and `a_n` the nth term of the AP.
| Sl | a | d | n | `a_n` |
|---|---|---|---|---|
| (i) | 7 | 3 | 8 | ... |
| (ii) | `-18` | ... | 10 | 0 |
| (iii) | ... | `-3` | 18 | `-5` |
| (iv) | `-18.9` | 2.5 | ... | 3.6 |
| (v) | 3.5 | 0 | 105 | .... |
Solution:
Answer Table
| Sl | a | d | n | `a_n` |
|---|---|---|---|---|
| (i) | 7 | 3 | 8 | 28 |
| (ii) | `-18` | 2 | 10 | 0 |
| (iii) | 46 | `-3` | 18 | `-5` |
| (iv) | `-18.9` | 2.5 | 10 | 3.6 |
| (v) | 3.5 | 0 | 105 | 3.5 |
Explanation
(i) Given, `a=7, d=3, n=8`
Therefore, `a_n=?`
We know that, `a_n=a+(n-1)d`
`:. a_8 = 7+(8-1)3`
`=7+7xx3=28`
(ii) Given, `a=-18, n=10, a_n=0`
Therefore, `d=?`
We know that, `a_n=a+(n-1)d`
`:. 0 = -18+(10-1)xxd`
`=>0=-18+9d`
`=>9d=18`
`:. d=18/9=2` Answer
(iii) Given, `d=-3, n=18, a_n=-5`
Therefore, `a=?`
We know that, `a_n=a+(n-1)d`
`:. -5=a+(18-1)xx(-3)`
`=>-5=a+17xx(-3)`
`=>-5=a-51`
`=>a=-5+51=46` Answer
(iv) Given, `a=-18.9, d=2.5, a_n=3.6`
Therefore, `n=?`
We know that, `a_n=a+(n-1)d`
`:. 3.6=-18.9+(n-1)xx2.5`
`=>3.6=-18.9+2.5\ n-2.5`
`=>3.6=-21.4+2.5\ n`
`=>2.5\ n = 3.6+21.4`
`=>2.5\ n = 25`
`:. n = 25/2.5=10` Answer
(v) Given, `a=3.5, d=0, n=105`
Therefore, `a_n=?`
Here, since common differnece is equal to zero (0) thus, every terms will come equal to first term i.e. `=3.5`
`:. a_105= 3.5` Answer
Question: (2) Choose the correct choice in the following and justify:
(i) 30th term of AP: 10, 7, 4, ..., is
(A) 97
(B) 77
(C) `-77`
(D)`-87`
Solution:
Given, 10, 7, 4, ...
Here, First term `a_1= 10`
Common differnece (d)
`= a_2-a_1=a_3-a_2`
`=7-10=4-7`
`=>d=-3`
∴ 30th term =?
We know that, `a_n=a+(n-1)d`
`:. a_30=10+(30-1)xx(-3)`
`=10+29xx(-3)=10-87`
`=>a_30=-77`
∴ Option (C)`-77` is the correct answer
(ii) 11th term of the AP: `-3, -1/2, 2, `......, is
(A) 28
(B) 22
(C) `-38`
(D)`-87`
Solution:
Given, `-3, -1/2, 2, `......,
Here, First term `(a_1)=-3`
Common differnece (d)
`a_2-a_1=-1/2-(-3)`
`=-1/2+3=(-1+6)/2`
`d=5/2`
∴ `a_11=?`
We know that, `a_n=a+(n-1)d`
`:. a_11=-3+(11-1)xx(5/2)`
`=-3+10xx5/2`
`=-3+5xx5=-3+25`
`a_11=22`
∴ Option (B) 22 is the correct answer
Question. (3) In the following APs, find the missing terms in the boxes
(i) 2, `square`, 26
Solution:
Here, First term `(a_1)=2`
And Third term `(a_3)=26`
∴ 2nd term `(a_2)=?`
Let, commond differnece =d
We know that, `a_n=a+(n-1)d`
And third term `(a_3)`
`=>26 = 2+(3-1)d`
`=>26=2+2d`
`=>2d=26-2=24`
`:. d = 24/2=12`
Now missing term i.e. second term `(a_2)` can be calculated by following two methods
Method (1)
Since, In and AP Second term = First term + common differnece
`:. a_2 = a_1+d`
`=>a_2 = 2+12=14`
Method (2)
Using` a_n=a+(n-1)d`
`a_2= 2+(2-1)xx12`
`=>a_2=2+12=14`
Thus, number missing in box = 14 Answer
Alternate method
In an AP middle term is the average of other two terms
Or, Middle term `=(text{1st term + 2nd term})/2`
`a_2=(2+26)/2=28/2`
`=>a_2=14`
Thus, number missing in box = 14 Answer
(ii) `square, 13, square, 3`
Solution:
Given, `a_2=13 and a_4=3`
`:. a_1 and a_3=?`
We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece
`:. a_2= a+(2-1)d`
`=>13=a+d` -----(i)
And `a_4 = a+(4-1)d`
`=>3=a+3d` -----(ii)
After subtracting euqation (ii) from (i) we get
`13-3=(a+d)-(a+3d)`
`=>10=a+d-a-3d`
`=>10=-2d`
`:. d = -10/2=-5`
Now, First term = second term – common differnece
`=>a_1= 13-(-5)`
`=>a_1=13+5=18`
And third term = second term + common differnece
`=a_3=13+(-5)`
`=>a_3=13-5= 8`
Thus, missing numbers in boxes are 18 and 8 Answer
Alternate Method
In an AP middle term is the average of other two terms
Or, Middle term `=(text{1st term + 2nd term})/2`
Given, `a_2=13 and a_4 = 3`
`:. a_3 = (a_2+a_4)/2`
`=>a_3 = (13+3)2=16/2`
`=>a_3 = 8`
Here, common difference (d) `= a_4-a_3`
`=>d=3-8=-5`
`:. a_1 = a_2-d=13-(-5)`
`=>a_1=13+5=18`
Thus, missing numbers in boxes are 18 (first term) and 8 (third term) Answer
Question 3 (iii) `5, square, square, 9\1/2`
Solution:
Given, First term `(a_1) = 5`
And the Fourth term `(a_4) = 9\1/2=19/2`
`:. a_2 and a_3 =?`
Let, common difference = d
We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece
`:. a_4 = a + (4-1)d`
`=> 19/2=5+3d`
`=>19/2-5=3d`
`=>(19-10)/2=3d`
`=>3d=9/2`
`=>d=9/(2xx3)=3/2`
Now, `a_2 = a_1+(2-1)d`
`=>a_2 = 5+1xx3/2`
`=>a_2 = (10+3)/2=13/2`
`=>a_2 = 6\1/2 `
And, `a_3 = a + (3-1)d`
`=>a_3 = 5+2xx3/2=5+3`
`=>a_3 = 8`
Thus, `6\1/2 and 8` are the missing numbers in the boxes respectively. Answer
Question. 3 (iv) `-4, square, square, square, square, 6,` .......
Solution:
Given, `-4, square, square, square, square, 6,` .......
Here, `a_1 = -4 and a_6 = 6`
We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece
`:. a_6= a+(6-1)d`
`=> 6 = -4+5d`
`=>5d=6+4=10`
`=>d=10/5=2`
Now, ∵ `a_2 = a+d`
`:. a_2 = -4+2=-2`
And, `a_3 = a+(3-1)d`
`=>a_3 = -4+2xx2`
`=>a_3 = -4+4=0`
And, `a_4 = -4+(4-1)2`
`=>a_4 = -4+3xx2=-4+6`
`=>a_4=2`
And, `a_5 = -4+(5-1)2`
`=>a_5 = -4+4xx2`
`=>a_5 = -4+8=4`
&there4, missing numbers in boxes are, `-2`, 0, 2 and 4 respectively.
Question. 3 (v) `square, 38, square, square, square, -22`
Solution:
Given, Second term `(a_2)=38` and Sixth term `(a_6) = -22`
We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece
`:. a_2 = a+(2-1)d`
` 38=a+d` -----(i)
And, `a_6 = a + (6-1)d`
`:. -22=a+5d` -------(ii)
Now by subtracting equation (ii) from equation (i) we get
`38-(-22)=a+d-(a+5d)`
`=>38+22=a+d-a-5d`
`=> 60 = -4d`
`=>d = 60/(-4)=-15`
Now, first term `(a_1) = a_2-d`
`:. a_1 = 38-(-15)=38+15`
`=> a_1 = 53`
Third term `(a_3)=a+(3-1)d`
`=>a_3 = 53+2xx(-15)`
`=>a_3 = 53-30=23`
Fourth term `(a_4)=a+(4-1)xxd`
`=>a_4 = 53+3xx(-15)`
`=>a_4 = 53-45=8`
Fifth term `(a_5) = a +(5-1)d`
`a_5 = 53+4xx(-15)`
`a_5 = 53-60 = -7`
Thus, missing numbers in the boxes are 53, 23, 8 and `-7` respectively. Answer
Question. (4) Which term of the AP: 3, 8, 13, 18, ..., is 78?
Solution:
Given, First term `(a_1)` = 3
Common difference `(d) = 8-3=13-8=5`
Let, the given term `a_n= 78`
Thus, `n =?`
We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece
` a_n = 3+(n-1)xx5`
`:. 78 = 3 + (n-1)xx5`
`=> 78 = 3+5n-5`
`=> 78 = 5n -2`
`=> 5n = 78+2`
`=> n =80/5 = 16`
Therefore, given term is `16^(th)` term. Answer
Question. (5) Find the number of terms in each of the following APs:
(i) 7, 13, 19, ....., 205
Solution:
Given, First term `(a_1) = 7`
Last term `(l)` = 205
According to the given list of numbers, which is in AP
Common difference `(d) = a_2 - a_1 `
` => d = 13-7=6`
Let last term is `n^(th) term
Thus, `n=?`
Now, `l=a_n = 205`
We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece
`:. 205 = 7 + (n-1) 6`
`=> 205 = 7 + 6n - 6`
`=> 1+6n = 205`
`=> 6n = 205-4= 204`
`:. n = 204/6 = 34`
∴ There are 34 terms in the given Arithmeti Progression (AP)
Question. 5 (ii) `18, 15\1/2, 13, ..., -47`
Solution:
Here, First term `(a_1) = 18`
Second term `(a_2) = 15\1/2=31/2`
Last term `(l) = -47`
there4; Common difference `(d) = a_2-a_1=31/2-18`
` =>d = (31-36)/2`
`=>d = -5/2`
Let, given last term `(l) = nth` term
`:. n = ?`
We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece
`:. a_n= 18+(n-1)xx(-5/2)`
`=> -47 = 18+(-(5n)/2+5/2)`
`=> -47 = 18 - (5n)/2+5/2`
`=> -47-18 -5/2=-(5n)/2`
`=> -(5n)/2=-65-5/2`
`=> -(5n)/2=(-130-5)/2`
`=> -5n=-135`
`:. n = 135/5=27`
∴ There are 27 terms in the given Arithmetic Progression (AP). Answer
Question (6) Check whether `-150` is a term of the AP: 11, 8, 5, 2, ....
Solution:
Here, First term `(a_1) = 11`
And Second term `(a_2) = 8`
Since, the given list of number is in Arithmetic Progression (AP)
Therefore, Common difference `(d) = a_2-a_1`
`d = 8-11 = -3`
Now, let the given term `-150 = n^(th)` term
We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece
`:. -150 = 11 + (n-1) xx (-3)`
`=> -150 = 11 -3n + 3`
`=> -150 = -3n + 14`
`=>-3n = -150 - 14 = -164`
`=> n = 164/3 = 54\2/3`
Since, result of `n` is not a whole number, thus given term is not a term of the given Arithmetic Progression (AP). Answer
Reference: