Arithmetic Progression

Mathematics Class Tenth

10th-Math-home


NCERT Exercise 5.2 Q7-20

Question. (7) Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution:

Given, 11th term `(a_11)=38`

And, 16th term `(a_16) = 73`

there4; 31st term `(a_31)=?`

We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece

Let, the First term `(a_1)=a` and Common difference = d

`:. a_11 = a + (11-1)d`

`=>38 = a+10d` -------(i)

And, `a_16 = a +(16-1)d`

`=> 73 = a+15d` -----(ii)

Now, by subtracting equation (ii) from equation (i) we get

`38-73 = a+10d -(a+15d)`

`=> -35 = a+10d-a-15d`

`=>-35=-5d`

`=> d = 35/5 =7`

By substituting the value of `d` in equation (i) we get

` 38 = a + 10 xx7`

` a = 38-70=-32`

∴ `a_31 = a + (31-1)d`

By substituting the the value of `a` and `d` we get

`a_31 = -32+30xx7`

`=>a_31 = -32 + 210`

`=>a_31 = 178`

Thus, 31st term is equal to 178. Answer

Question. (8) An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:

Given, Total number of terms = 50

And the Last term = 106

∴ `a_50 = 106`

Third term `a_3 = 12`

`:. 29^(th)` term `(a_29) =?`

Let, the common difference `= d` and First term `=a`

We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece

`:. a_3 = a + (3-1)d`

` => 12 = a + 2d ` ------(i)

And, `a_50 = a + (50-1)d`

`=> 106 = a + 49d` ------(ii)

Now after subtracting equation (ii) from equation (i) we get

`106-12 = a + 49d -(a+2d)`

`=> 94 = a + 49d - a -2d`

`=>94= 47d`

`:. d = 94/47 = 2`

By substituting the value of `d` in equation (i), we get

`12 = a + 2 xx 2`

`a = 12-4=8`

Now, `a_29 = a + (29-1)d`

After substituting value of first term `(a)` and common difference `(d)` we get

`a_29 = 8 + (29-1) xx 2`

`= 8 + 28 xx 2 = 8 + 56`

`=> a_29 = 64`

Thus, 29th term = 64 Answer

Question. (9) If the 3rd and the 9th term of an AP are 4 and `-8` respectively, which term of this AP is zero?

Solution:

Given, 3rd term `(a_3) = 4`

And, 9th term `(a_9) = -8`

`:. 0 =` which term

Let, `0 = n(th) term

Let, first term `(a_1) =a` and 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_3 = a + (3-1)d`

` => 4 = a + 2d` ----(i)

And, `a_9 = a + (9-1)d`

`=>-8 = a + 8d` ------(ii)

After subtracting equation (ii) from equation (i) we get

`4-(-8) = a+2d-(a+8d)`

`=>4+8 = a + 2d - a - 8d`

`=> 12 = - 6d`

`:. d = -12/6 = -2`

Now, after substituting the value of `d` in euqation (i) we get

`4 = a + 2 xx (-2)`

`=> 4 = a-4`

`:. a = 4+4 = 8`

Now, ∵ `a_n = a + (n-1)d`

`:. 0 = 8 + (n-1) xx (-2)`

[∵ we assumed that 0 = nth term]

`=> 0 = 8 -2n +2`

`=>0 = -2n +10`

`=> - 2n = 0 - 10`

`=> -2n = -10`

`:. n = 10/2 = 5`

Thus, 5th term of the given AP is equal to zero. Answer

Question. (10) The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution:

Given, `a_17-a_10 = 7`

Let, First term `a_1 = a` and 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_17 = a + (17-1)d`

`=>a_17 = a + 16d` -------(i)

And, `a_10 = a + (10-1)d`

`=>a_10 = a + 9d` ----(ii)

Now, by subtracting equation (ii) from equation (i), we get

`a_17 - a_10 = a + 16d -(a+9d)`

`=>7 = a+ 16d-a-9d`

`=> 7 = 16d-9d = 7d`

`:. d = 7/7=1`

Thus, common difference = 1. Answer

Question. (11) Which term of the AP: 3, 15, 27, 39, .... will be 132 more than its 54th term?

Solution:

Given, First term `(a_1) = 3`

Common difference `d = 15-3 = 12`

Term which is 132 more than 54th =?

Let, this term = nth

We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece

`:. a_54 = a+(54-1)d`

`=>a_54 = 3 + 53 xx 12`

`=>a_54 = 3 + 636=639`

∴ required nth term = 639+132= 771

`=>a_n = 771`

Now, `a_n = 3 + (n-1) 12`

`=>771 = 3 + 12n-12 `

`=> 771 = 12n - 9`

`=> 12n = 771+9=780`

`:. n = 780/12 = 65`

∴ 65th term is 132 more than 54th term. Answer

Question. (12) Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Question. (13) How many three digit numbers are divisible by 7?

Solution:

100 is the the first three-digits number, which is when divided by 7 gives a remainder = 2

Thus, 100+5 = 105 is the first three-digits number divisible by 7

Now, the largest three-digits number = 999

When 999 is divided by 7 it gives a remainer = 5

This means, `999-5 = 994` is the largest three-digits number divisible by 7

Thus, first term of the list of three-digits numbers which is in AP = 105

i.e. `a_1 = 105`

And the last term of the series of three digits numbers divisible by 7 = 994

This, means that `a_n = 994` where `n` = number of terms

Here, since all numbers are divisible by 7, thus common difference `(d)=7`

Now, we have

First term `(a_1)=105`

Common difference `(d)=7`

And `a_n=994`

We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece

`:. a_n = 105+(n-1)7`

`=>994=105+7n-7`

`=>994=7n+98`

`=>7n = 994-98 = 896`

`:. n = 896/7= 128`

 

Thus, there are total 128 three-digits numbers which are divisible by 7. Answer

Question. (14) How many multiples of 4 lie between 10 and 250?

Solution:

When 10 is divided by 4, it gives a remainder = 2

Thus, 10+2 = 12 is the first digit lie between 10 and 250 which is divisible by 4

This means firt term of the AP which is divisible by 4 `(a_1) = 12`

Now, when 250 is divided by 4, it gives a remainder = 2

This means 250 – 2 = 248 is the last number lie between 10 and 250 that is divisible by 4

Thus, for the list of numbers in AP formed,

First term `(a_1)=12`

And the last term, `(l) = 248`

Let total number of terms `=n`

`:. a_n = 248`

Here, common difference `(d)=4`

We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece

`:. a_n = 12 + (n-1)4`

`=> 248 = 12 + 4n -4`

`=> 248 = 4n + 8`

`=> 4n = 248-8`

`=> 4n = 240`

`:. n = 240/4 = 60`

Thus, there are 60 numbers divisible by 4 lie between 10 and 250. Answer

Question. (15) For what value of `n`, are the `n^(th)` terms of two APs: 63, 65, 67, ..... and 3, 10, 17, ... equal?

Solution:

Given, Ist AP: 63, 65, 67, ......

And 2nd AP: 3, 10, 17, .......

For Ist AP:

First term `(a) = 63`

Common difference `(d) = 65-63=2`

For 2nd AP:

First term `(a) = 3`

Common difference `(d) = 10-3=7`

Let for `n` value of `n(th)` term of two APs are equal

We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece

For Ist AP:

`a_n = 63 + (n-1)2` ------(i)

For 2nd AP:

`a_n = 3 + (n-1)7` -------(ii)

Since, `a_n` is equal for both the given APs

Therefore, from (i) and (ii) we get

`63+(n-1)2 = 3 + (n-1)7`

`=>63+2n-2=3+7n-7`

`=>61+2n=-4+7n`

`=>61+4=7n-2n`

`=>5n = 65`

`:. n = 65/5 =13`

This means that 13th term of both the AP will be equal

Thus, value of n = 13 Answer

Proof

For Ist AP:

`a_13 = 63 + (13-1)2`

`=63+12xx2`

`=63+24`

`=>a_13 = 87`

For 2nd AP:

`a_13 = 3 + (13-1)7`

`=3+12xx7`

`=3+84`

`a_13 = 87`

Here it is clear that 13th term of both the APs is equal to 87. Proved

Question. (16) Determine the AP whose third term is 16 and 7th term exceeds the 5th term by 12.

Solution:

Given, 3rd term `(a_3)=16`

And 7th term `(a_7)` – 5th term `(a_5)=12`

Let, first term `=a`

And 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_3 = a + (3-1)d`

`=> 16 = a + 2d` ----(i)

And, `a_5 = a + (5-1)d`

`=> a_5 = a + 4d` -----(ii)

And, `a_7 = a + (7-1)d`

`=a_7 = a+6d` ------(iii)

Now, by subtracting euqation (ii) from equation (iii) we get

`a_7 - a_5 = a+6d - (a+4d)`

`12 = a+6d-a-4d`

`=>12=2d`

`:. d = 12/2= 6`

By substituting the value of `d` in equation (i) we get

`16= a+2xx6`

`=> 16 = =a+12`

`:. a = 16-12 = 4`

Now, we have, `a=4 and d = 6`

`:. a_2 = a+d = 4+6=10`

And, `a_3 = 16` (As given in question)

And, `a_4 = a + (4-1)d`

`a_4 = 4 + 3xx6 = 4+18`

`=>a_4 =22`

And, `a_5 = a + (5-1)d`

`=>a_5 = 4 + 4xx6=4+24`

`=> a_5 = 28`

And, `a_6 = a + (6-1)d`

`=> a_6 = 4 + 5 xx 6`

`=> a_ 6 = 4 +30`

`=> a_ 6 = 34`

Thus, AP is 4, 10, 16, 22, 28, 34, ......

Question. (17) Find the 20th term from the last term of AP: 3, 8, 13, ......, 253.

Solution:

Here, given First term `(a) = 3`

Common difference `(d) = 8-3=5`

Last term `(a_n) = 253`

20th term from last `(n-(20-1))= (n-19)^(th)` term =?`

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)5`

`=>253 =3+5n-5`

`=> 253 = 5n - 2`

`=>5n = 253+2=255`

`:. n = 255/5= 51`

Now, 20th term from last `=51-19= 32`

i.e. `a_32 = 3+(32-1)5`

`=>a_32=3+31xx5=3+155`

`=>a_32 = 158`

Thus, 20th term from last = 158 Answer

Question. (18) The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and the 10th terms is 44. Find the first three terms of the AP.

Solution:

Given, `a_4 + a_8 = 24`

And, `a_6 + a_10 = 44`

Let, First term `=a`

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`

`=>a_4 = a+3d` -------(i)

And, `a_8 = a+(8-1)d`

`=>a_8 = a+7d` -------(ii)

And, `a_6 = a + (6-1)d`

`=> a_6 = a+5d` --------(iii)

And, `a_10 = a + (10-1)d`

`a_10 = a + 9d` --------(iv)

Now, since, `a_4 + a_8 = 24`

∴ from equation (i) and (ii)

`a_4+a_8=a+3d+a+7d`

`=>24 = 2a+10d` ------(v)

`=>2(a+5d)=24`

`=>a+5d=24/2`

`=>a+5d=12` ------(vi)

Now, since `a_6+a_10=44`

∴ From equation (iii) and (iv)

`a_6+a_10 = a+5d+a+9d`

`=>44= 2a+14d` ------(vii)

Now, after equation (vii) - equation (v), we get

`44-24=2a+14d-(2a+10d)`

`=>20 = 2a+14d-2a-10d`

`=>20 = 4d`

`:. d = 20/4=5`

Now, by putting the value of `d` in equation (vii), we get

`44=2a+14xx5=2a+70`

`=>2a=44-70= -26`

`=>a = -26/2=-13`

Now, ∵ ` a_2 = a + d`

`:. a_ 2 = -13+5=8`

And, `a_3 = a + (3-1)d`

`=> a_3 = -13+2xx5=-13+10`

`=>a_3 = -3`

Thus, `-13, -8, -3,` .... are the first three terms of the given AP. Answer

Question. (19) Subba Rao started work in 1995 at the annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000

Solution:

Here, Given, First term `(a) = 5000`

Common difference `(d) = 200`

`a_n = 7000`

Therefore, `n=?`

We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece

`:. 7000 = 5000 + (n-1)200`

`=>7000-5000=200n -200`

`=>2000 = 200n -200`

`=> 2000+200 = 200n`

`:. n = 2200/200`

`=> n = 11`

∴ 1995 + 11 = 2006

Thus, in 2006 salary of Subba Rao reached at Rs 7000

In 11th year or in 2006 Answer

Question. (20) Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the nth week her weekly savings become RS 20.75, find `n`.

Solution:

Given, First term `(a)=5`

Common difference `(d) = 1.75`

nth term, i.e. `a_n = 20.75`

`:. n =?`

We know that, `a_n=a+(n-1)d` where, a = first term, n = number of term and d = common differnece

`:. 20.75 = 5 + (n-1)1.75`

`=> 20.75-5=1.75n-1.75`

`=>1.75n -1.75 = 15.75`

`=>1.75n = 15.75+1.75`

`=>1.75n = 17.70`

`:. n = (17.75)/(1.75)`

`=> n = 10`

Thus, in 10th week Ramkali's salary will become Rs 20.75

Thus, 10th week Answer

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