10th Maths 5.2
Chapter 5
Arithmetic Progressions
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NCERT Class 10th solution of Exercise 5.1
Exercise 5.2
Q1. Fill in the blanks in the following table, given that `a` is the first term, `d` the common difference and `a_n` the `nth` term of the AP ;
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Sol. :
i)
`a_n=a+(n-1)timesd`
`a_n=7+(8-1)times3`
`a_n=7+7times3`
`a_n=7+21`
`a_n=28`
Answer :
`a_n=28`
ii)
`a_n=a+(n-1)timesd`
`0=-18+(10-1)timesd`
`0=-18+9timesd`
`9d=18`
`d=2`
Answer :
`d=2`.
iii)
`a_n=a+(n-1)timesd`
`-5=a+(18-1)(-3)`
`-5=a+17(-3)`
`-5=a-51`
`a=46`
Answer :
`a=46`.
iv)
`a_n=a+(n-1)timesd`
`3.6=-18.9+(n-1)times2.5`
`3.6=-18.9+2.5n-2.5`
`3.6=-21.4+2.5n`
`2.5n=-21.4-3.6`
`n=-25.0/2.5`
`n=-10`
Answer :
`n=-10`.
v)
`a_n=a+(n-1)timesd`
`a_n=3.5+(105-1)times0`
`a_n=3.5+0`
`a_n=3.5`
Answer :
`a_n=3.5`.
Q2. Choose the correct choice is the following and justify :
i) `30^(th)` term of the AP : `10, 7, 4,`...is
A) `97` B) `77` C) `-77` D) `-87`
ii) `11^(th)` term of the AP : `-3, -1/2, 2`,....is :
A) `28` B) `22` C) `-38` D) `-48(1)/2`
Sol. :
i) Given : `a=10, d=-3`, and `n=30`
`a_n=a+(n-1)d`
`a_(10)=10+(30-1)(-3)`
`a_(10)=10+29(-3)`
`a_(10)=10-87`
`a_(10)=-77`
Answer :
(C).
ii) Given :`a=-3, d=2(1)/2`, and `n=11`
`a_n=a+(n-1)d`
`a_(11)=-3+(11-1)(5/2)`
`a_(11)=-3+25`
`a_(11)=22`
Answer :
(B).
Q3. In the following APs, find the missing terms in the boxes :
i) `2, square, 26`
ii) `square, 13, square, 3`
iii) `5, square, square, 9(1)/2`
iv) `-4, square, square, square, square, 6`
v) `square, 38, square, square, square, -22`
Sol. :
i) Given : `a_1=2, a_3=26`, and `n=1, 3`
`a_n=a+(n-1)d`
`26=2+(3-1)d`
`26=2+2d`
`26-2=2d`
`24=2d`
`d=(24)/2`
`d=12`
`a_2=2+(2-1)12`
`a_2=2+12`
`a_2=14`
Answer :
`d=12, square_2=12`
ii) Given : `a_2=13, a_4=3`, and `n=2, 4`
`a_n=a+(n-1)d`
`13=a+(2-1)d`
`13=a+d`_________(1)
`3=a+(4-1)d`
`3=a+3d`___________(2)
By equation (1) - (2)
`13-3=a+d-(a+3d)`
`10=a+d-a-3d`
`10=(-2)d`
`d=(-10)/2`
`d=-5`
Put in (1)
`a+(-5)=13`
`a=18`
`a_3=18+(3-1)(-5)`
`a_3=18+2(-5)`
`a_3=18-10`
`a_3=8`
Answer :
`square_1=18, square_3=8`.
iii) Given : `a_1=5, a_4=9(1)/2=9.5`, and `n=4`
`a_n=a+(n-1)d`
`9.5=5+(4-1)d`
`9.5-5=3d`
`4.5=5d`
`d=(4.5)/5`
`d=1.5`
`a_2=a+(n-1)d`
`a_2=5+(2-1)1.5=6.5`
`a_3=a+(n-1)d`
`a_3=5+(3-1)1.5=8`
Answer :
`square_2=6.5, square_3=8`
iv) Given : `a=-4, a_6=6`, and `n=6`
`a_n=a+(n-1)d`
`a_6=-4+(6-1)d`
`6=-4+5d`
`6+4=5d`
`10=5d`
`d=10/5`
`d=2`
`a_2=-4+(2-1)2=-4+2=-2`
`a_3=-4+(3-1)2=-4+4=0`
`a_4=-4+(4-1)2=-4+6=2`
`a_5=-4+(5-1)2=-4+8=4`
Answer :
`square_2=-2, square_3=0, square_4=2, and square_5=4`.
v) Given : `a_2=38, a_6=-22`, and `n=2, 6`
`a_n=a+(n-1)d`
`38=a+(2-1)d`
`38=a+d`_________(1)
`-22=a+(6-1)d`
`-22=a+5d`________(2)
By equation (2) - (1)
`-22-38=a+5d-(a+d)`
`-60=a+5d-a-d`
`-60=4d`
`d=(-60)/4=-15`
Put in equation (1)
`38=a+(-15)`
`38=a-15`
`a=38+15=53`
`a_3=53+(3-1)(-15)=53-30=23`
`a_4=53+(4-1)(-15)=53-45=8`
`a_5=53+(5-1)(-15)=53-60=-7`
Answer :
`square_1=53, square_3=23, square_4=8`, and `square_5=-7`.
Q4. Which term of the AP: `3, 8, 18,` ...., is `78`?
Sol.
Given: `a=3, d=8-3=5`, and `a_n=78`
`a_n=a+(n-1)d`
`78=3+(n-1)(5)`
`78=3+5n-5`
`78=-2+5n`
`5n=78+2`
`5n=80`
`n=80/5`
`n=16`
Answer :
`n=16`.
Q5. Find the number of terms in each of the following APs :
i) `7, 13, 19, ...,205` ii) `18, 15(1)/5, 13, .... -47`
Sol. :
i) Give: `a=7, d=13-7=6`, and `a_n=205`
`a_n=a+(n-1)d`
`205=7+(n-1)(6)`
`205=7+6n-6`
`205=1+6n`
`6n=205-1`
`6n=204`
`n=204/6`
`n=34`
Answer :
`n=34`.
ii) Given: `a=18, d=15(1)/2-1=31/2-1=15.5-81=-2.5`, and `a_n=-47`
`a_n=a+(n-1)d`
`-47=18+(n-1)(-2.5)`
`-47=18-2.5n+2.5`
`-47=20.5-2.5n`
`2.5n=47+20.5`
`2.5n=67.5`
`n=(67.5)/(2.5)`
`n=(675)/25`
`n=27`
Answer :
`n=27`.
Q6. Check whether `-150` is a term of the AP: `11, 8, 5, 2.... .`
Sol. : Given: `a=11, d=8-11=-3`, and `a_n=-150`
`a_n=a+(n-1)d`
`-150=11+(n-1)(-3)`
`-150=11-3n+3`
`-150=14-3n`
`3n=14+150`
`3n=164`
`n=164/3`
`n=54(2)/3`[ it is not natural number ]
Answer :
No, `-150` is not a term of given AP.
Q7. Find the `31`st term of an AP whose `11`th term is `38` and the `16`th term is `73`.
Sol. :
Given: `a_11=38, a_16=73`, and `n=11, 16`
`a_n=a+(n-1)d`
`38=a+(11-1)d`
`38=a+10d`___________(1)
`73=a+(16-1)d`
`73=a+15d`___________(2)
By equation (2) - (1)
`73-38=a+15d-(a+10d)`
`35=a+15d-a-10d`
`35=5d`
`d=35/5`
`d=7`
Put in equation (1)
`38=a+10(7)`
`38=a+70`
`a=-70+38`
`a=-32`
Now
`a_31=a+(30-1)d`
`a_31=-32+30(7)`
`a_31=-32+210`
`a_31=178`
Answer :
`a_31=178`.
Q8. An AP consists of `50` terms of which `3`rd term is `12` and the last term is `106`. Find the `29`th term.
Sol. :
Given: `a_3=12, a_50=106` and `n=12, 50`
`a_n=a+(n-1)d`
`12=a+(3-1)d`
`12=a+2d`________________(1)
`106=a+(50-1)d`
`106=a+49d`______________(2)
By equation (2) - (1)
`106-12=a+49d-(a+2d)`
`94=a+49d-a-2d`
`94=49d-2d`
`94=47d`
`d=94/47`
`d=2`
Put in equation (1)
`12=a+2(2)`
`12=a+4`
`a=8`
`a_29=a+(29-1)d`
`a_29=8+28(2)`
`a_29=8+56`
`a_29=64`
Answer :
`a_29=64`.
Q9. If the `3`rd and the `9`th terms of an AP are `4` and `-8` respectively, which term of this AP is zero.
Sol. :
Given: `a_3=4, a_9=-8`, and `a_n=0`
`a_n=a+(n-1)d`
`4=a+(3-1)d`
`4=a+2d`______________(1)
`-8=a+(9-1)d`
`-8=a+8d`______________(2)
By equation (2) - (1)
`-8-4=a+8d-(a+2d)`
`-12=a+8d-a-2d`
`-12=6d`
`d=-12/6`
`d=-2`
Put in equation (1)
`4=a+2(-2)`
`4=a-4`
`a=8`
Now
`a_n=8+(n-1)(-2)`
`0=8-2n+2`
`0=10-2n`
`2n=10`
`n=10/2`
`n=5`
Answer :
`n=5`.
Q10. The `17`th term of an AP exceeds its `10`th term by `7`. Find the common difference.
Sol. :
According to question
`a_17-a_10=7`
`a+(17-1)d-(a+(10-1)d=7`
`a+16d-a-9d=7`
`7d=7`
`d=7/7`
`d=1`
Answer :
`d=1`.
Q11. Which term of the AP: `3, 15, 27, 39,...` will be `132` more than its `54`th term?
Sol. :
Given: `a=3, d=15-3=12`
According to question
`a_n-a_54=132`
`a+(n-1)d-[a+(54-1)d]=132`
`3+(n-1)12-[3+53(12)]=132`
`3+12n-12-3-636=132`
`12n=132+636+12`
`12n=780`
`n=780/12`
`n=65`
Answer :
`65`th term.
Q12. Two APs have the same command difference. The difference between their `100`th term is `100`, what is the difference between their `1000`th terms?
Sol. :
Let first term of first AP is `a`,
and first term of second AP is `b`.
According to question
`a_100-b_100=100`
`a+(100-1)d-[b+(100-1)d]=100`
`a+99d-b-99d=100`
`a-b=100___________________(1)
Now
`a_1000-b_1000`
`=a+(1000-1)d-[b+(1000-1)d]`
`=a+999d-b-999d`
`=a-b`
from equation (1)
`a-b=100`
Answer :
The difference between `1000`th terms will also be `100`.
Q13. How many three-digit numbers are divisible by `7`?
Sol. :
AP is `105, 112, 119,.......994`.
where `a=105, d=7,` and `a_n=994`
`a_n=a+(n-1)d`
`994=105+(n-1)7`
`994=105+7n-7`
`994=98+7n`
`7n=994-98`
`7n=896`
`n=896/7`
`n=128`
Answer :
`n=128`.
Q14. How many multiples of `4` lie between `10` and `250`?
Sol. :
AP is `12, 16, 20, 24, .....248`
where `a=12, d=4,` and `a_n=248`
`a_n=a+(n-1)d`
`248=12+(n-1)4`
`248=12+4n-4`
`248=8+4n`
`4n=248-8`
`4n=240`
`n=240/4`
`n=60`
Answer :
`n=60`.
Q15. For what value of `n`, are the `n`th terms of two APs: `63, 65, 67`,....and `3, 10, 17,..... equal?
Sol. :
AP `63, 65, 67,.....
where `a=63, d=65-63=2`
AP `3, 10, 17, ....
where `a=3, d=10-3=7`
According to question
`a_n=b_n`
`63+(n-1)2=3+(n-1)7`
`63+2n-2=3+7n-7`
`7n-2n=63-3-2+7`
`5n=65`
`n=65/5`
`n=13`
Answer :
`n=13`.
Q16. Determine the AP whose third term is `16` and the `7`th term exceeds the `5`th term by `12`.
Sol. :
Let the AP is `a, a+d, a+2d, .....`
According to question
`a_3=a+2d`
`16=a+2d`________(1)
and
`a_7-a_5=12`
`a+(7-1)d-[a+(5-1)d]=12`
`a+6d-a-4d=12`
`2d=12`
`d=6`
Put in equation (1)
`16=a+2(6)`
`16=a+12`
`a=4`
`a_2=4+(2-1)6=4+6=10`
`a_3=4+(3-1)6=4+12=16`
`a_4=4+(4-1)6=4+18=22`
Answer :
The AP is `4, 10, 16, 22, .....`.
Q17. Find the `20`th term from the last term of the AP: `3, 8, 13,...., 253`.
Sol. :
By the given AP take in decending order
`253, 248, ......, 8, 3`
where `a=253, d=248-253=-5`
`a_20=253+(20-1)(-5)`
`a_20=253+19(-5)`
`a_20=253-95`
`a_20=158`
Answer :
`a_20=158`.
Q18. The sum of the `4`th and `8`th terms of an AP is `24` and the sum of the `6`th and `10`th term is `44`. Find the first three terms of the AP.
Sol. :
Let the AP is `a, a+d, a+2d, ......`
According to question
`a_4+a_8=24`
`a+(4-1)d+a+(8-1)d=24`
`a+3d+a+7d=24`
`2a+10d=24`
`a+5d=12`______________(1)
and
`a_6+a_10=44`
`a+(6-1)d+a+(10-1)d=44`
`a+5d+a+9d=44`
`2a+14d=44`
`a+7d=22`_____________(2)
By equation (2) - (1)
`a+7d-(a+5d)=22-12`
`a+7d-a-5d=10`
`2d=10`
`d=10/2`
`d=5`
Put in equation (1)
`a+5(5)=12`
`a+25=12`
`a=-13`
`a_2=-13+(2-1)5=-13+5=-8`
`a_3=-13+(3-1)5=-13+10=-3`
Answer :
The first three terms of AP is `-13, -8, -3, ....`.
Q19. Subba Rao started work in `1995` at an annual salary of `₹5000` and received an increment of `₹ 200` each year. In which year did his income reach `₹ 7000`?
Sol. :
Given: `a=₹ 5000, d=₹ 200,` and `a_n=₹ 7000`.
`a_n=a+(n-1)d`
`7000=5000+(n-1)200`
`7000=5000+200n-200`
`200n=7000-5000+200`
`n=2200/200
`n=11`
Answer :
`11`th year.
Q20. Ramkali served `5` in the first week of a year and then increased her weekly savings by `₹ 1.75`. in the `n`th week, her weekly saving become `₹ 20.75,` find `n`.
Sol. :
Given: `a=₹ 5, d=₹ 1.75` and `a_n=₹ 20.75`
`a_n=a+(n-1)d`
`20.75=5+(n-1)1.75`
`20.75=5+1.75n-1.75`
`1.75n=20.75-5+1.75`
`n=17.50/1.75`
`n=10`
Answer :
`10`th year.
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