Mathematics solution NCERT
Class 10 - Chapter 5: Arithmetic Progressions
EXERCISE 5.3
Core Formulas Used:
- nth term of an AP: an = a + (n - 1)d
- Sum of first n terms: Sn = n⁄2 [2a + (n - 1)d] or Sn = n⁄2 (a + an)
Question 1 (i)
Find the sum of the AP:
$$ 2,\;7,\;12,\;\ldots,\text{ to }10\text{ terms.} $$Solution:
The given AP is
$$ 2,\;7,\;12,\;\ldots $$Here,
$$ a=2,\qquad d=5,\qquad n=10 $$Use the formula for the sum of n terms of an AP:
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ \begin{aligned} S_{10} &=\frac{10}{2}\left[2(2)+(10-1)(5)\right]\\[4pt] &=5\left[4+45\right]\\[4pt] &=5\times49\\[4pt] &=245 \end{aligned} $$Answer:
$$ \boxed{245} $$Question 1 (ii)
Find the sum of the AP:
$$ -37,\;-33,\;-29,\;\ldots,\text{ to }12\text{ terms.} $$Solution:
The given AP is
$$ -37,\;-33,\;-29,\;\ldots $$Here,
$$ a=-37,\qquad d=4,\qquad n=12 $$Use the formula
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ \begin{aligned} S_{12} &=\frac{12}{2}\left[2(-37)+(12-1)(4)\right]\\[4pt] &=6\left[-74+44\right]\\[4pt] &=6(-30)\\[4pt] &=-180 \end{aligned} $$Verification:
The 12th term is
$$ a_{12}=a+(12-1)d $$ $$ =-37+11(4) $$ $$ =7 $$Using the alternative formula,
$$ S_n=\frac{n}{2}(a+l) $$ $$ =\frac{12}{2}(-37+7) $$ $$ =6(-30) $$ $$ =-180 $$Hence, the answer is correct.
Answer:
$$ \boxed{-180} $$Question 1 (iii)
Find the sum of the AP:
$$ 0.6,\;1.7,\;2.8,\;\ldots,\text{ to }100\text{ terms.} $$Solution:
The given AP is
$$ 0.6,\;1.7,\;2.8,\;\ldots $$Here,
$$ a=0.6,\qquad d=1.1,\qquad n=100 $$Use the formula for the sum of the first n terms of an AP:
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ \begin{aligned} S_{100} &=\frac{100}{2}\left[2(0.6)+(100-1)(1.1)\right]\\[4pt] &=50\left[1.2+108.9\right]\\[4pt] &=50(110.1)\\[4pt] &=5505 \end{aligned} $$Answer:
$$ \boxed{5505} $$Question 1 (iv)
Find the sum of the AP:
$$ \frac{1}{15},\;\frac{1}{12},\;\frac{1}{10},\;\ldots,\text{ to }11\text{ terms.} $$Solution:
The given AP is
$$ \frac{1}{15},\;\frac{1}{12},\;\frac{1}{10},\;\ldots $$Here,
$$ a=\frac{1}{15} $$The common difference is
$$ d=\frac{1}{12}-\frac{1}{15} =\frac{5-4}{60} =\frac{1}{60} $$Also,
$$ n=11 $$Use the formula
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ \begin{aligned} S_{11} &=\frac{11}{2}\left[2\left(\frac{1}{15}\right)+10\left(\frac{1}{60}\right)\right]\\[4pt] &=\frac{11}{2}\left[\frac{2}{15}+\frac{1}{6}\right]\\[4pt] &=\frac{11}{2}\left[\frac{4+5}{30}\right]\\[4pt] &=\frac{11}{2}\times\frac{9}{30}\\[4pt] &=\frac{99}{60}\\[4pt] &=\frac{33}{20} \end{aligned} $$Answer:
$$ \boxed{\frac{33}{20}} $$Question 2 (i)
Find the sum:
$$ 7+10\frac12+14+\cdots+84 $$Solution:
The given series is an Arithmetic Progression.
Here,
$$ a=7,\qquad d=10\frac12-7=3\frac12=\frac72,\qquad l=84 $$Step 1: Find the number of terms.
Using the formula
$$ l=a+(n-1)d $$Substitute the given values.
$$ 84=7+(n-1)\times\frac72 $$ $$ 77=(n-1)\times\frac72 $$ $$ 154=7(n-1) $$ $$ 22=n-1 $$ $$ n=23 $$Step 2: Find the sum.
Using the formula
$$ S_n=\frac{n}{2}(a+l) $$ $$ S_{23}=\frac{23}{2}(7+84) $$ $$ =\frac{23}{2}\times91 $$ $$ =\frac{2093}{2} $$ $$ =1046.5 $$Answer:
$$ \boxed{1046.5} $$Question 2 (ii)
Find the sum:
$$ 34+32+30+\cdots+10 $$Solution:
The given series is an Arithmetic Progression.
Here,
$$ a=34,\qquad d=-2,\qquad l=10 $$Step 1: Find the number of terms.
Using the formula
$$ l=a+(n-1)d $$ $$ 10=34+(n-1)(-2) $$ $$ -24=-2(n-1) $$ $$ 12=n-1 $$ $$ n=13 $$Step 2: Find the sum.
Using the formula
$$ S_n=\frac{n}{2}(a+l) $$ $$ S_{13}=\frac{13}{2}(34+10) $$ $$ =\frac{13}{2}\times44 $$ $$ =13\times22 $$ $$ =286 $$Answer:
$$ \boxed{286} $$Question 2 (iii)
Find the sum:
$$ -5+(-8)+(-11)+\cdots+(-230) $$Solution:
The given series is an Arithmetic Progression.
Here,
$$ a=-5,\qquad d=-3,\qquad l=-230 $$Step 1: Find the number of terms.
Using the formula
$$ l=a+(n-1)d $$ $$ -230=-5+(n-1)(-3) $$ $$ -225=-3(n-1) $$ $$ 75=n-1 $$ $$ n=76 $$Step 2: Find the sum.
Using the formula
$$ S_n=\frac{n}{2}(a+l) $$ $$ S_{76}=\frac{76}{2}\left(-5-230\right) $$ $$ =38(-235) $$ $$ =-8930 $$Answer:
$$ \boxed{-8930} $$Question 3 (i)
Given:
$$ a=5,\qquad d=3,\qquad a_n=50 $$Find n and S_n.
Solution:
Using the formula for the nth term,
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 50=5+(n-1)\times3 $$ $$ 45=3(n-1) $$ $$ 15=n-1 $$ $$ n=16 $$Now, use the formula for the sum of the first n terms.
$$ S_n=\frac{n}{2}(a+l) $$Here,
$$ n=16,\qquad a=5,\qquad l=50 $$ $$ S_{16}=\frac{16}{2}(5+50) $$ $$ =8\times55 $$ $$ =440 $$Answer:
$$ \boxed{n=16,\qquad S_{16}=440} $$Question 3 (ii)
Given:
$$ a=7,\qquad a_{13}=35 $$Find d and S_{13}.
Solution:
Using the formula
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 35=7+(13-1)d $$ $$ 35=7+12d $$ $$ 28=12d $$ $$ d=\frac{7}{3} $$Now find
$$ S_{13}. $$Using the formula
$$ S_n=\frac{n}{2}(a+l) $$Here,
$$ n=13,\qquad a=7,\qquad l=35 $$ $$ S_{13}=\frac{13}{2}(7+35) $$ $$ =\frac{13}{2}\times42 $$ $$ =13\times21 $$ $$ =273 $$Answer:
$$ \boxed{d=\frac73,\qquad S_{13}=273} $$Question 3 (iii)
Given:
$$ a_{12}=37,\qquad d=3 $$Find a and S_{12}.
Solution:
Using the formula
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 37=a+(12-1)\times3 $$ $$ 37=a+33 $$ $$ a=4 $$Now find
$$ S_{12}. $$Using the formula
$$ S_n=\frac{n}{2}(a+l) $$Here,
$$ n=12,\qquad a=4,\qquad l=37 $$ $$ S_{12}=\frac{12}{2}(4+37) $$ $$ =6\times41 $$ $$ =246 $$Answer:
$$ \boxed{a=4,\qquad S_{12}=246} $$Question 3 (iv)
Given:
$$ a_3=15,\qquad S_{10}=125 $$Find d and a_{10}.
Solution:
Using the formula for the 3rd term,
$$ a_n=a+(n-1)d $$we get
$$ a+2d=15 \qquad\cdots(1) $$Also,
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$For
$$ n=10, $$ $$ 125=\frac{10}{2}\left[2a+9d\right] $$ $$ 125=5(2a+9d) $$ $$ 2a+9d=25 \qquad\cdots(2) $$Multiply equation (1) by 2.
$$ 2a+4d=30 \qquad\cdots(3) $$Subtract equation (2) from equation (3).
$$ (2a+4d)-(2a+9d)=30-25 $$ $$ -5d=5 $$ $$ d=-1 $$Substitute
$$ d=-1 $$in equation (1).
$$ a+2(-1)=15 $$ $$ a-2=15 $$ $$ a=17 $$Now find the 10th term.
$$ a_{10}=a+9d $$ $$ =17+9(-1) $$ $$ =8 $$Answer:
$$ \boxed{d=-1,\qquad a_{10}=8} $$Question 3 (v)
Given:
$$ d=5,\qquad S_9=75 $$Find a and a_9.
Solution:
Using the formula
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ 75=\frac{9}{2}\left[2a+8(5)\right] $$ $$ 75=\frac92(2a+40) $$Multiply both sides by 2.
$$ 150=9(2a+40) $$ $$ 150=18a+360 $$ $$ 18a=-210 $$ $$ a=-\frac{35}{3} $$Now find the 9th term.
$$ a_9=a+8d $$ $$ =-\frac{35}{3}+8(5) $$ $$ =-\frac{35}{3}+40 $$ $$ =\frac{85}{3} $$Answer:
$$ \boxed{a=-\frac{35}{3},\qquad a_9=\frac{85}{3}} $$Question 3 (vi)
Given:
$$ a=2,\qquad d=8,\qquad S_n=90 $$Find n and a_n.
Solution:
Using the formula
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ 90=\frac{n}{2}\left[4+8(n-1)\right] $$ $$ 180=n(8n-4) $$ $$ 45=n(2n-1) $$ $$ 2n^2-n-45=0 $$Factorising,
$$ (2n+9)(n-5)=0 $$ $$ n=5 $$(The value
$$ n=-\frac92 $$is rejected since the number of terms cannot be negative.)
Now find the 5th term.
$$ a_5=a+4d $$ $$ =2+4(8) $$ $$ =34 $$Answer:
$$ \boxed{n=5,\qquad a_n=34} $$Question 3 (vii)
Given:
$$ a=8,\qquad a_n=62,\qquad S_n=210 $$Find n and d.
Solution:
Using the formula for the sum of the first n terms,
$$ S_n=\frac{n}{2}(a+l) $$Here,
$$ a=8,\qquad l=62,\qquad S_n=210 $$Substitute the given values.
$$ 210=\frac{n}{2}(8+62) $$ $$ 210=\frac{n}{2}(70) $$ $$ 210=35n $$ $$ n=6 $$Now use the formula for the nth term.
$$ a_n=a+(n-1)d $$ $$ 62=8+(6-1)d $$ $$ 62=8+5d $$ $$ 54=5d $$ $$ d=\frac{54}{5} $$Answer:
$$ \boxed{n=6,\qquad d=\frac{54}{5}} $$Question 3 (viii)
Given:
$$ a_n=4,\qquad d=2,\qquad S_n=-14 $$Find n and a.
Solution:
Using the formula
$$ S_n=\frac{n}{2}(a+l) $$Here,
$$ l=4,\qquad S_n=-14 $$Therefore,
$$ -14=\frac{n}{2}(a+4) \qquad\cdots(1) $$Also, using the formula for the nth term,
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 4=a+2(n-1) $$ $$ 4=a+2n-2 $$ $$ a=6-2n \qquad\cdots(2) $$Substitute equation (2) into equation (1).
$$ -14=\frac{n}{2}\left[(6-2n)+4\right] $$ $$ -14=\frac{n}{2}(10-2n) $$ $$ -14=n(5-n) $$ $$ n^2-5n-14=0 $$Factorising,
$$ (n-7)(n+2)=0 $$ $$ n=7 $$(Reject
$$ n=-2 $$since the number of terms cannot be negative.)
Now substitute
$$ n=7 $$in equation (2).
$$ a=6-2(7) $$ $$ a=-8 $$Verification:
$$ a_7=-8+6(2)=4 $$ $$ S_7=\frac72(-8+4) =\frac72(-4) =-14 $$Hence, the values are correct.
Answer:
$$ \boxed{n=7,\qquad a=-8} $$Question 3 (ix)
Given:
$$ a=3,\qquad n=8,\qquad S=192 $$Find d.
Solution:
Using the formula for the sum of the first n terms of an AP,
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ 192=\frac{8}{2}\left[2(3)+(8-1)d\right] $$ $$ 192=4(6+7d) $$ $$ 48=6+7d $$ $$ 42=7d $$ $$ d=6 $$Verification:
Using
$$ a=3,\qquad d=6,\qquad n=8, $$ $$ S_8=\frac{8}{2}\left[2(3)+7(6)\right] $$ $$ =4(6+42) $$ $$ =4\times48 $$ $$ =192 $$Hence, the value obtained is correct.
Answer:
$$ \boxed{d=6} $$Question 3 (x)
Given:
$$ l=28,\qquad S=144,\qquad n=9 $$Find the first term a.
Solution:
Using the formula for the sum of the first n terms,
$$ S_n=\frac{n}{2}(a+l) $$Substitute the given values.
$$ 144=\frac{9}{2}(a+28) $$Multiply both sides by 2.
$$ 288=9(a+28) $$Divide both sides by 9.
$$ 32=a+28 $$ $$ a=4 $$Verification:
$$ S_9=\frac{9}{2}(4+28) $$ $$ =\frac{9}{2}\times32 $$ $$ =9\times16 $$ $$ =144 $$Hence, the obtained value is correct.
Answer:
$$ \boxed{a=4} $$Question 4
How many terms of the AP
$$ 9,\;17,\;25,\;\ldots $$must be taken to give a sum of 636?
Solution:
The given AP is
$$ 9,\;17,\;25,\;\ldots $$Here,
$$ a=9,\qquad d=8,\qquad S_n=636 $$Using the formula for the sum of the first n terms,
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ 636=\frac{n}{2}\left[2(9)+(n-1)\times8\right] $$ $$ 636=\frac{n}{2}(18+8n-8) $$ $$ 636=\frac{n}{2}(8n+10) $$ $$ 1272=n(8n+10) $$ $$ 4n^2+5n-636=0 $$Factorising,
$$ (4n+53)(n-12)=0 $$ $$ n=12 $$The value
$$ n=-\frac{53}{4} $$is rejected since the number of terms cannot be negative.
Verification:
$$ S_{12}=\frac{12}{2}\left[18+11(8)\right] $$ $$ =6(18+88) $$ $$ =6\times106 $$ $$ =636 $$Answer:
$$ \boxed{12} $$Question 5
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
Given,
$$ a=5,\qquad l=45,\qquad S_n=400 $$Step 1: Find the number of terms.
Using the formula
$$ S_n=\frac{n}{2}(a+l) $$Substitute the given values.
$$ 400=\frac{n}{2}(5+45) $$ $$ 400=\frac{n}{2}(50) $$ $$ 400=25n $$ $$ n=16 $$Step 2: Find the common difference.
Using the formula
$$ l=a+(n-1)d $$Substitute the given values.
$$ 45=5+(16-1)d $$ $$ 45=5+15d $$ $$ 40=15d $$ $$ d=\frac{40}{15} $$ $$ d=\frac{8}{3} $$Verification:
$$ 45=5+15\left(\frac83\right) $$ $$ 45=5+40 $$ $$ 45=45 $$Hence, the values obtained are correct.
Answer:
$$ \boxed{n=16,\qquad d=\frac83} $$Question 6
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
Given,
$$ a=17,\qquad l=350,\qquad d=9 $$Step 1: Find the number of terms.
Using the formula
$$ l=a+(n-1)d $$Substitute the given values.
$$ 350=17+(n-1)\times9 $$ $$ 333=9(n-1) $$ $$ 37=n-1 $$ $$ n=38 $$Step 2: Find the sum of all terms.
Using the formula
$$ S_n=\frac{n}{2}(a+l) $$Substitute the values.
$$ S_{38}=\frac{38}{2}(17+350) $$ $$ =19\times367 $$ $$ =6973 $$Verification:
$$ 17+37\times9=350 $$Hence, the number of terms is correct.
Answer:
$$ \boxed{n=38,\qquad S_{38}=6973} $$Question 7
Find the sum of the first 22 terms of an AP in which
$$ d=7 $$and the 22nd term is
$$ 149. $$Solution:
Given,
$$ d=7,\qquad a_{22}=149 $$Step 1: Find the first term.
Using the formula
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 149=a+(22-1)\times7 $$ $$ 149=a+147 $$ $$ a=2 $$Step 2: Find the sum of the first 22 terms.
Using the formula
$$ S_n=\frac{n}{2}(a+l) $$Here,
$$ n=22,\qquad a=2,\qquad l=149 $$ $$ S_{22}=\frac{22}{2}(2+149) $$ $$ =11\times151 $$ $$ =1661 $$Verification:
$$ 2+21\times7=149 $$Hence, the obtained values are correct.
Answer:
$$ \boxed{S_{22}=1661} $$Question 8
Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Let the first term be
$$ a $$and the common difference be
$$ d. $$Step 1: Form the equations.
The second term is 14.
$$ a+d=14 \qquad\cdots(1) $$The third term is 18.
$$ a+2d=18 \qquad\cdots(2) $$Step 2: Find the common difference.
Subtract equation (1) from equation (2).
$$ (a+2d)-(a+d)=18-14 $$ $$ d=4 $$Step 3: Find the first term.
Substitute
$$ d=4 $$in equation (1).
$$ a+4=14 $$ $$ a=10 $$Step 4: Find the sum of the first 51 terms.
Using the formula
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ \begin{aligned} S_{51} &=\frac{51}{2}\left[2(10)+(51-1)(4)\right]\\[4pt] &=\frac{51}{2}(20+200)\\[4pt] &=\frac{51}{2}\times220\\[4pt] &=51\times110\\[4pt] &=5610 \end{aligned} $$Answer:
$$ \boxed{S_{51}=5610} $$Question 9
If the sum of the first 7 terms of an AP is 49 and that of the first 17 terms is 289, find the sum of the first n terms.
Solution:
Using the formula
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Step 1: Use the given sums.
For the first 7 terms,
$$ 49=\frac{7}{2}\left[2a+6d\right] $$ $$ 14=2a+6d $$ $$ a+3d=7 \qquad\cdots(1) $$For the first 17 terms,
$$ 289=\frac{17}{2}\left[2a+16d\right] $$ $$ 34=2a+16d $$ $$ a+8d=17 \qquad\cdots(2) $$Step 2: Find the values of a and d.
Subtract equation (1) from equation (2).
$$ (a+8d)-(a+3d)=17-7 $$ $$ 5d=10 $$ $$ d=2 $$Substitute
$$ d=2 $$in equation (1).
$$ a+3(2)=7 $$ $$ a=1 $$Step 3: Find the sum of the first n terms.
Using
$$ a=1,\qquad d=2, $$we get
$$ \begin{aligned} S_n &=\frac{n}{2}\left[2(1)+(n-1)(2)\right]\\[4pt] &=\frac{n}{2}(2+2n-2)\\[4pt] &=\frac{n}{2}(2n)\\[4pt] &=n^2 \end{aligned} $$Verification:
$$ S_7=7^2=49 $$ $$ S_{17}=17^2=289 $$Hence, the required formula is correct.
Answer:
$$ \boxed{S_n=n^2} $$Question 10 (i)
Show that
$$ a_n=3+4n $$forms an Arithmetic Progression. Also find the sum of the first 15 terms.
Solution:
Given,
$$ a_n=3+4n $$Find the first few terms.
$$ a_1=3+4(1)=7 $$ $$ a_2=3+4(2)=11 $$ $$ a_3=3+4(3)=15 $$ $$ a_4=3+4(4)=19 $$Thus, the sequence is
$$ 7,\;11,\;15,\;19,\;\ldots $$Find the common difference.
$$ 11-7=4 $$ $$ 15-11=4 $$ $$ 19-15=4 $$Since the common difference is constant, the given sequence is an Arithmetic Progression.
Here,
$$ a=7,\qquad d=4,\qquad n=15 $$Using the formula
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ \begin{aligned} S_{15} &=\frac{15}{2}\left[2(7)+(15-1)(4)\right]\\[4pt] &=\frac{15}{2}(14+56)\\[4pt] &=\frac{15}{2}\times70\\[4pt] &=15\times35\\[4pt] &=525 \end{aligned} $$Answer:
The given sequence is an Arithmetic Progression with
$$ \boxed{d=4} $$and
$$ \boxed{S_{15}=525.} $$Question 10 (ii)
Show that
$$ a_n=9-5n $$forms an Arithmetic Progression. Also find the sum of the first 15 terms.
Solution:
Given,
$$ a_n=9-5n $$Find the first few terms.
$$ a_1=9-5(1)=4 $$ $$ a_2=9-5(2)=-1 $$ $$ a_3=9-5(3)=-6 $$ $$ a_4=9-5(4)=-11 $$Thus, the sequence is
$$ 4,\;-1,\;-6,\;-11,\;\ldots $$Find the common difference.
$$ -1-4=-5 $$ $$ -6-(-1)=-5 $$ $$ -11-(-6)=-5 $$Since the common difference is constant, the given sequence is an Arithmetic Progression.
Here,
$$ a=4,\qquad d=-5,\qquad n=15 $$Using the formula
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ \begin{aligned} S_{15} &=\frac{15}{2}\left[2(4)+(15-1)(-5)\right]\\[4pt] &=\frac{15}{2}(8-70)\\[4pt] &=\frac{15}{2}(-62)\\[4pt] &=15(-31)\\[4pt] &=-465 \end{aligned} $$Answer:
The given sequence is an Arithmetic Progression with
$$ \boxed{d=-5} $$and
$$ \boxed{S_{15}=-465.} $$Question 11
A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than the preceding prize, find the value of each of the prizes.
Solution:
Let the first prize be
$$ a. $$Since each prize is ₹20 less than the previous prize, the common difference is
$$ d=-20. $$The number of prizes is
$$ n=7. $$The total amount distributed is
$$ S_7=700. $$Using the formula for the sum of the first n terms of an AP,
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ 700=\frac{7}{2}\left[2a+(7-1)(-20)\right] $$ $$ 700=\frac{7}{2}(2a-120) $$Multiply both sides by 2.
$$ 1400=7(2a-120) $$Divide both sides by 7.
$$ 200=2a-120 $$ $$ 2a=320 $$ $$ a=160 $$Now find the remaining prizes.
Second prize:
$$ 160-20=140 $$Third prize:
$$ 140-20=120 $$Fourth prize:
$$ 120-20=100 $$Fifth prize:
$$ 100-20=80 $$Sixth prize:
$$ 80-20=60 $$Seventh prize:
$$ 60-20=40 $$Verification:
$$ 160+140+120+100+80+60+40 $$ $$ =700 $$Hence, the prizes are correct.
Answer:
$$ \boxed{₹160,\;₹140,\;₹120,\;₹100,\;₹80,\;₹60,\;₹40} $$Question 12
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The positive integers divisible by 6 are
$$ 6,\;12,\;18,\;24,\;\ldots $$This is an Arithmetic Progression with
$$ a=6,\qquad d=6,\qquad n=40. $$Using the formula for the sum of the first n terms of an AP,
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ \begin{aligned} S_{40} &=\frac{40}{2}\left[2(6)+(40-1)(6)\right]\\[4pt] &=20(12+234)\\[4pt] &=20(246)\\[4pt] &=4920 \end{aligned} $$Verification:
The 40th term is
$$ a_{40}=6+(40-1)\times6 $$ $$ =6+234 $$ $$ =240. $$Using the formula
$$ S_n=\frac{n}{2}(a+l), $$ $$ S_{40}=\frac{40}{2}(6+240) $$ $$ =20\times246 $$ $$ =4920. $$Answer:
$$ \boxed{4920} $$Question 13
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 are
$$ 8,\;16,\;24,\;32,\;\ldots $$This is an Arithmetic Progression with
$$ a=8,\qquad d=8,\qquad n=15. $$Using the formula
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ \begin{aligned} S_{15} &=\frac{15}{2}\left[2(8)+(15-1)(8)\right]\\[4pt] &=\frac{15}{2}(16+112)\\[4pt] &=\frac{15}{2}(128)\\[4pt] &=15\times64\\[4pt] &=960 \end{aligned} $$Verification:
The 15th term is
$$ a_{15}=8+(15-1)\times8 $$ $$ =8+112 $$ $$ =120. $$Using
$$ S_n=\frac{n}{2}(a+l), $$ $$ S_{15}=\frac{15}{2}(8+120) $$ $$ =\frac{15}{2}\times128 $$ $$ =960. $$Answer:
$$ \boxed{960} $$Question 14
Find the sum of the odd numbers between 0 and 50.
Solution:
The odd numbers between 0 and 50 are
$$ 1,\;3,\;5,\;7,\;\ldots,\;49. $$This is an Arithmetic Progression with
$$ a=1,\qquad d=2,\qquad l=49. $$Step 1: Find the number of terms.
Using the formula
$$ l=a+(n-1)d $$Substitute the given values.
$$ 49=1+(n-1)\times2 $$ $$ 48=2(n-1) $$ $$ 24=n-1 $$ $$ n=25 $$Step 2: Find the sum.
Using the formula
$$ S_n=\frac{n}{2}(a+l) $$ $$ S_{25}=\frac{25}{2}(1+49) $$ $$ =\frac{25}{2}\times50 $$ $$ =25\times25 $$ $$ =625 $$Answer:
$$ \boxed{625} $$Question 15
A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows:
₹200 for the first day, ₹250 for the second day, ₹300 for the third day, and so on. The penalty for each succeeding day is ₹50 more than for the preceding day.
How much money does the contractor have to pay as penalty, if the work is delayed by 30 days?
Solution:
The daily penalties form an Arithmetic Progression.
Here,
$$ a=200,\qquad d=50,\qquad n=30. $$Using the formula for the sum of the first n terms,
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ \begin{aligned} S_{30} &=\frac{30}{2}\left[2(200)+(30-1)(50)\right]\\[4pt] &=15\left[400+1450\right]\\[4pt] &=15(1850)\\[4pt] &=27750 \end{aligned} $$Verification:
The penalty on the 30th day is
$$ a_{30}=200+29(50) $$ $$ =1650. $$Using
$$ S_n=\frac{n}{2}(a+l), $$ $$ S_{30}=\frac{30}{2}(200+1650) $$ $$ =15\times1850 $$ $$ =27750. $$Answer:
$$ \boxed{\text{The contractor has to pay }₹27,\!750\text{ as penalty.}} $$Question 16
A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let the first prize be
$$ a. $$Since each prize is ₹20 less than the previous prize,
$$ d=-20. $$The number of prizes is
$$ n=7. $$The total amount distributed is
$$ S_7=700. $$Using the formula
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ 700=\frac{7}{2}\left[2a+(7-1)(-20)\right] $$ $$ 700=\frac{7}{2}(2a-120) $$Multiply both sides by 2.
$$ 1400=7(2a-120) $$Divide both sides by 7.
$$ 200=2a-120 $$ $$ 2a=320 $$ $$ a=160 $$Now find the remaining prizes.
First prize:
$$ ₹160 $$Second prize:
$$ 160-20=₹140 $$Third prize:
$$ 140-20=₹120 $$Fourth prize:
$$ 120-20=₹100 $$Fifth prize:
$$ 100-20=₹80 $$Sixth prize:
$$ 80-20=₹60 $$Seventh prize:
$$ 60-20=₹40 $$Verification:
$$ 160+140+120+100+80+60+40=700 $$Hence, the prizes are correct.
Answer:
$$ \boxed{₹160,\;₹140,\;₹120,\;₹100,\;₹80,\;₹60,\;₹40} $$Question 17
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant will be the same as the class in which they are studying. For example, a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on up to Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
The number of trees planted by one section of each class forms the sequence
$$ 1,\;2,\;3,\;\ldots,\;12. $$This is an Arithmetic Progression with
$$ a=1,\qquad d=1,\qquad n=12. $$Step 1: Find the total number of trees planted by one section of each class.
Using the formula
$$ S_n=\frac{n}{2}(a+l) $$Here,
$$ l=12. $$ $$ S_{12}=\frac{12}{2}(1+12) $$ $$ =6\times13 $$ $$ =78 $$Step 2: Find the total number of trees planted by all sections.
There are three sections in each class.
$$ \text{Total trees}=3\times78 $$ $$ =234 $$Answer:
$$ \boxed{234} $$Therefore, the students will plant 234 trees.
Question 18
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take
$$ \pi=\frac{22}{7} $$)
Solution:
The length of a semicircle of radius r is
$$ \pi r. $$The radii of the thirteen semicircles are
$$ 0.5,\;1,\;1.5,\;2,\;\ldots $$These radii form an Arithmetic Progression with
$$ a=0.5,\qquad d=0.5,\qquad n=13. $$Step 1: Find the sum of the radii.
Using the formula
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ \begin{aligned} S_{13} &=\frac{13}{2}\left[2(0.5)+(13-1)(0.5)\right]\\[4pt] &=\frac{13}{2}(1+6)\\[4pt] &=\frac{13}{2}\times7\\[4pt] &=\frac{91}{2}\\[4pt] &=45.5 \end{aligned} $$Hence, the sum of the radii is
$$ 45.5\text{ cm.} $$Step 2: Find the total length of the spiral.
Total length
$$ =\pi\times(\text{sum of the radii}) $$ $$ =\frac{22}{7}\times45.5 $$Since
$$ 45.5=\frac{91}{2}, $$ $$ =\frac{22}{7}\times\frac{91}{2} $$ $$ =\frac{22\times13}{2} $$ $$ =11\times13 $$ $$ =143 $$Answer:
$$ \boxed{143\text{ cm}} $$Therefore, the total length of the spiral made up of thirteen consecutive semicircles is 143 cm.
Question 19
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
Solution:
The number of logs in each row forms an Arithmetic Progression.
$$ 20,\;19,\;18,\;\ldots $$Here,
$$ a=20,\qquad d=-1,\qquad S_n=200. $$Step 1: Find the number of rows.
Using the formula for the sum of the first n terms,
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ 200=\frac{n}{2}\left[2(20)+(n-1)(-1)\right] $$ $$ 200=\frac{n}{2}(40-n+1) $$ $$ 200=\frac{n}{2}(41-n) $$Multiply both sides by 2.
$$ 400=n(41-n) $$ $$ n^2-41n+400=0 $$Factorising,
$$ (n-16)(n-25)=0 $$ $$ n=16 \quad\text{or}\quad n=25 $$Since the number of logs in the last row cannot be zero or negative, check both values.
For
$$ n=25, $$the last row contains
$$ a_{25}=20+(25-1)(-1) $$ $$ =20-24 $$ $$ =-4, $$which is not possible.
Therefore,
$$ n=16. $$Step 2: Find the number of logs in the top row.
Using the formula for the last term,
$$ a_n=a+(n-1)d $$ $$ a_{16}=20+(16-1)(-1) $$ $$ =20-15 $$ $$ =5. $$Verification:
The rows are
$$ 20,\;19,\;18,\;\ldots,\;5. $$Their sum is
$$ S_{16} =\frac{16}{2}(20+5) $$ $$ =8\times25 $$ $$ =200. $$Hence, the obtained values are correct.
Answer:
Number of rows:
$$ \boxed{16} $$Number of logs in the top row:
$$ \boxed{5} $$Question 20
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up each potato one by one, returns to the bucket after picking each potato, and repeats this until all the potatoes are collected. What is the total distance the competitor has to run?
Solution:
The distances of the potatoes from the bucket are
$$ 5,\;8,\;11,\;14,\;17,\;20,\;23,\;26,\;29,\;32\text{ m.} $$These distances form an Arithmetic Progression with
$$ a=5,\qquad d=3,\qquad n=10. $$Step 1: Find the sum of the distances from the bucket to all potatoes.
Using the formula
$$ S_n=\frac{n}{2}\left[2a+(n-1)d\right] $$Substitute the given values.
$$ \begin{aligned} S_{10} &=\frac{10}{2}\left[2(5)+(10-1)(3)\right]\\[4pt] &=5(10+27)\\[4pt] &=5\times37\\[4pt] &=185\text{ m} \end{aligned} $$Step 2: Find the total distance run.
For each potato, the competitor runs to the potato and returns to the bucket.
Therefore, the total distance is
$$ 2\times185 $$ $$ =370\text{ m} $$Verification:
The total distance can also be written as
$$ 2(5+8+11+14+17+20+23+26+29+32) $$ $$ =2(185) $$ $$ =370\text{ m} $$Hence, the answer is correct.
Answer:
$$ \boxed{370\text{ m}} $$