Mathematics solution NCERT
Class 10 - Chapter 5: Arithmetic Progressions
EXERCISE 5.2
Question 1 (i)
Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP.
Given:
$$ a=7,\qquad d=3,\qquad n=8 $$Solution:
Use the formula for the nth term of an AP:
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ \begin{aligned} a_8 &=7+(8-1)\times3\\[4pt] &=7+7\times3\\[4pt] &=7+21\\[4pt] &=28 \end{aligned} $$Answer:
$$ \boxed{a_8=28} $$Question 1 (ii)
Given:
$$ a=-18,\qquad n=10,\qquad a_{10}=0 $$Find the common difference d.
Solution:
Use the formula
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 0=-18+(10-1)d $$ $$ 0=-18+9d $$ $$ 9d=18 $$ $$ d=2 $$Answer:
$$ \boxed{d=2} $$Question 1 (iii)
Given:
$$ d=-3,\qquad n=18,\qquad a_{18}=-5 $$Find the first term a.
Solution:
Use the formula for the nth term of an AP:
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ -5=a+(18-1)(-3) $$ $$ -5=a+17(-3) $$ $$ -5=a-51 $$Add 51 to both sides.
$$ a=-5+51 $$ $$ a=46 $$Answer:
$$ \boxed{a=46} $$Question 1 (iv)
Given:
$$ a=-18.9,\qquad d=2.5,\qquad a_n=3.6 $$Find the value of n.
Solution:
Use the formula
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 3.6=-18.9+(n-1)(2.5) $$Add 18.9 to both sides.
$$ 3.6+18.9=(n-1)(2.5) $$ $$ 22.5=2.5(n-1) $$Divide both sides by 2.5.
$$ \frac{22.5}{2.5}=n-1 $$ $$ 9=n-1 $$ $$ n=10 $$Answer:
$$ \boxed{n=10} $$Question 1 (v)
Given:
$$ a=3.5,\qquad d=0,\qquad n=105 $$Find the value of a105.
Solution:
Use the formula for the nth term of an AP:
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ \begin{aligned} a_{105} &=3.5+(105-1)\times0\\[4pt] &=3.5+104\times0\\[4pt] &=3.5+0\\[4pt] &=3.5 \end{aligned} $$Since the common difference is
$$ d=0, $$all the terms of the AP are equal to the first term.
Answer:
$$ \boxed{a_{105}=3.5} $$Question 2 (i)
Choose the correct choice in the following and justify:
(i) 30th term of the AP:
$$ 10,\;7,\;4,\;\ldots $$Options:
(A) 97 (B) 77 (C) −77 (D) −87
Solution:
The given AP is
$$ 10,\;7,\;4,\;\ldots $$Therefore,
$$ a=10,\qquad d=7-10=-3 $$Using the formula for the nth term,
$$ a_n=a+(n-1)d $$Substitute
$$ a=10,\qquad d=-3,\qquad n=30. $$ $$ \begin{aligned} a_{30} &=10+(30-1)(-3)\\[4pt] &=10+29(-3)\\[4pt] &=10-87\\[4pt] &=-77 \end{aligned} $$Answer:
$$ \boxed{a_{30}=-77} $$Correct Option:
$$ \boxed{\textbf{(C) }-77} $$Question 2 (ii)
(ii) 11th term of the AP:
$$ -3,\;-\frac12,\;2,\;\ldots $$Options:
(A) 28 (B) 22 (C) −38 (D) −48
Solution:
The given AP is
$$ -3,\;-\frac12,\;2,\;\ldots $$Therefore,
$$ a=-3 $$The common difference is
$$ d=-\frac12-(-3) $$ $$ =-\frac12+3 $$ $$ =\frac52 $$Using the formula
$$ a_n=a+(n-1)d $$Substitute
$$ a=-3,\qquad d=\frac52,\qquad n=11. $$ $$ \begin{aligned} a_{11} &=-3+(11-1)\times\frac52\\[4pt] &=-3+10\times\frac52\\[4pt] &=-3+25\\[4pt] &=22 \end{aligned} $$Answer:
$$ \boxed{a_{11}=22} $$Correct Option:
$$ \boxed{\textbf{(B) }22} $$Question 3 (i)
Find the missing term in the following AP:
$$ 2,\;\Box,\;26 $$Solution:
Let the missing term be
$$ x. $$Since the numbers are in an Arithmetic Progression, the common difference is the same.
$$ x-2=26-x $$ $$ 2x=28 $$ $$ x=14 $$Verification:
$$ 14-2=12 $$ $$ 26-14=12 $$The common difference is the same.
Answer:
$$ \boxed{14} $$Question 3 (ii)
Find the missing terms in the following AP:
$$ \Box,\;13,\;\Box,\;3 $$Solution:
Let the first term be
$$ a $$and the common difference be
$$ d. $$The four terms are
$$ a,\;a+d,\;a+2d,\;a+3d. $$Given,
$$ a+d=13 $$ $$ a+3d=3 $$Subtracting the two equations,
$$ (a+3d)-(a+d)=3-13 $$ $$ 2d=-10 $$ $$ d=-5 $$Now,
$$ a+d=13 $$ $$ a-5=13 $$ $$ a=18 $$The third term is
$$ a+2d $$ $$ 18+2(-5) $$ $$ 18-10 $$ $$ =8 $$Verification:
$$ 18,\;13,\;8,\;3 $$Common difference:
$$ 13-18=-5 $$ $$ 8-13=-5 $$ $$ 3-8=-5 $$Answer:
$$ \boxed{18,\;13,\;8,\;3} $$Question 3 (iii)
Find the missing terms in the following AP:
$$ 5,\;\Box,\;\Box,\;9\frac12 $$Solution:
Let the common difference be
$$ d. $$Since there are four terms,
$$ a=5 $$ $$ a+3d=9\frac12 $$Therefore,
$$ 5+3d=9\frac12 $$ $$ 3d=9\frac12-5 $$ $$ 3d=4\frac12 $$ $$ 3d=\frac92 $$ $$ d=\frac32 $$Second term:
$$ 5+\frac32 = 6\frac12 $$Third term:
$$ 6\frac12+\frac32 = 8 $$Verification:
$$ 5,\;6\frac12,\;8,\;9\frac12 $$Common difference:
$$ 6\frac12-5=\frac32 $$ $$ 8-6\frac12=\frac32 $$ $$ 9\frac12-8=\frac32 $$Answer:
$$ \boxed{5,\;6\frac12,\;8,\;9\frac12} $$Question 3 (iv)
Find the missing terms in the following AP:
$$ -4,\;\Box,\;\Box,\;\Box,\;\Box,\;6 $$Solution:
There are six terms in the AP.
Let the common difference be
$$ d. $$The first term is
$$ a=-4. $$The sixth term is
$$ a_6=6. $$Using the formula
$$ a_n=a+(n-1)d, $$we get
$$ 6=-4+(6-1)d $$ $$ 6=-4+5d $$ $$ 10=5d $$ $$ d=2 $$Now find the missing terms.
Second term:
$$ -4+2=-2 $$Third term:
$$ -2+2=0 $$Fourth term:
$$ 0+2=2 $$Fifth term:
$$ 2+2=4 $$Verification:
$$ -4,\;-2,\;0,\;2,\;4,\;6 $$Common difference:
$$ 2 $$Answer:
$$ \boxed{-4,\;-2,\;0,\;2,\;4,\;6} $$Question 3 (v)
Find the missing terms in the following AP:
$$ \Box,\;38,\;\Box,\;\Box,\;\Box,\;-22 $$Solution:
Let the first term be
$$ a $$and the common difference be
$$ d. $$The six terms are
$$ a,\;a+d,\;a+2d,\;a+3d,\;a+4d,\;a+5d. $$Given,
$$ a+d=38 $$ $$ a+5d=-22 $$Subtract the first equation from the second.
$$ (a+5d)-(a+d)=-22-38 $$ $$ 4d=-60 $$ $$ d=-15 $$Now,
$$ a+d=38 $$ $$ a-15=38 $$ $$ a=53 $$Hence, the six terms are
$$ 53,\;38,\;23,\;8,\;-7,\;-22 $$Verification:
Common difference:
$$ 38-53=-15 $$ $$ 23-38=-15 $$ $$ 8-23=-15 $$ $$ -7-8=-15 $$ $$ -22-(-7)=-15 $$Answer:
$$ \boxed{53,\;38,\;23,\;8,\;-7,\;-22} $$Question 4
Which term of the AP:
$$ 3,\;8,\;13,\;18,\;\ldots $$is 78?
Solution:
The given AP is
$$ 3,\;8,\;13,\;18,\;\ldots $$Therefore,
$$ a=3,\qquad d=5 $$Let the required term be the nth term.
Then,
$$ a_n=78 $$Using the formula for the nth term of an AP,
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 78=3+(n-1)\times5 $$ $$ 78=3+5(n-1) $$ $$ 78-3=5(n-1) $$ $$ 75=5(n-1) $$ $$ 15=n-1 $$ $$ n=16 $$Verification:
$$ \begin{aligned} a_{16} &=3+(16-1)\times5\\[4pt] &=3+15\times5\\[4pt] &=3+75\\[4pt] &=78 \end{aligned} $$Answer:
78 is the
$$ \boxed{16^{\text{th}}\text{ term}} $$of the given AP.
Question 5 (i)
Find the number of terms in the AP:
$$ 7,\;13,\;19,\;\ldots,\;205 $$Solution:
Here,
$$ a=7,\qquad d=6,\qquad a_n=205 $$Using the formula
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 205=7+(n-1)\times6 $$ $$ 205-7=6(n-1) $$ $$ 198=6(n-1) $$ $$ 33=n-1 $$ $$ n=34 $$Answer:
$$ \boxed{34\text{ terms}} $$Question 5 (ii)
Find the number of terms in the AP:
$$ 18,\;15\frac12,\;13,\;\ldots,\;-47 $$Solution:
Here,
$$ a=18,\qquad d=15\frac12-18=-2\frac12=-\frac52, $$ $$ a_n=-47. $$Using the formula
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ -47=18+(n-1)\left(-\frac52\right) $$ $$ -65=(n-1)\left(-\frac52\right) $$Multiply both sides by 2.
$$ -130=-5(n-1) $$ $$ 26=n-1 $$ $$ n=27 $$Answer:
$$ \boxed{27\text{ terms}} $$Question 6
Check whether −150 is a term of the AP:
$$ 11,\;8,\;5,\;2,\;\ldots $$Solution:
The given AP is
$$ 11,\;8,\;5,\;2,\;\ldots $$Therefore,
$$ a=11,\qquad d=-3 $$Let −150 be the nth term of the AP.
Using the formula
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ -150=11+(n-1)(-3) $$ $$ -150=11-3(n-1) $$Subtract 11 from both sides.
$$ -161=-3(n-1) $$Divide both sides by −3.
$$ n-1=\frac{161}{3} $$ $$ n=\frac{161}{3}+1 $$ $$ n=\frac{164}{3} $$Since
$$ \frac{164}{3}=54\frac23 $$which is not a whole number, there is no such term in the AP.
Answer:
Since the value of n is not a positive integer,
$$ \boxed{-150\text{ is not a term of the given AP.}} $$Question 7
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
Let the first term be
$$ a $$and the common difference be
$$ d. $$Using the formula for the nth term of an AP,
$$ a_n=a+(n-1)d $$Step 1: Form the equations.
The 11th term is 38.
$$ a_{11}=38 $$ $$ a+10d=38 \qquad\cdots(1) $$The 16th term is 73.
$$ a_{16}=73 $$ $$ a+15d=73 \qquad\cdots(2) $$Step 2: Find the common difference.
Subtract equation (1) from equation (2).
$$ (a+15d)-(a+10d)=73-38 $$ $$ 5d=35 $$ $$ d=7 $$Step 3: Find the first term.
Substitute
$$ d=7 $$in equation (1).
$$ a+10(7)=38 $$ $$ a+70=38 $$ $$ a=-32 $$Step 4: Find the 31st term.
$$ a_{31}=a+(31-1)d $$ $$ =-32+30(7) $$ $$ =-32+210 $$ $$ =178 $$Verification:
$$ a_{11}=-32+10(7)=38 $$ $$ a_{16}=-32+15(7)=73 $$Hence, the values of a and d are correct.
Answer:
$$ \boxed{a_{31}=178} $$Question 8
An AP consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
Let the first term be
$$ a $$and the common difference be
$$ d. $$Using the formula
$$ a_n=a+(n-1)d $$Step 1: Form the equations.
The 3rd term is 12.
$$ a_3=12 $$ $$ a+2d=12 \qquad\cdots(1) $$The last term is the 50th term and is equal to 106.
$$ a_{50}=106 $$ $$ a+49d=106 \qquad\cdots(2) $$Step 2: Find the common difference.
Subtract equation (1) from equation (2).
$$ (a+49d)-(a+2d)=106-12 $$ $$ 47d=94 $$ $$ d=2 $$Step 3: Find the first term.
Substitute
$$ d=2 $$in equation (1).
$$ a+2(2)=12 $$ $$ a+4=12 $$ $$ a=8 $$Step 4: Find the 29th term.
$$ a_{29}=a+(29-1)d $$ $$ =8+28(2) $$ $$ =8+56 $$ $$ =64 $$Verification:
$$ a_{50}=8+49(2)=106 $$Hence, the values obtained are correct.
Answer:
$$ \boxed{64} $$Question 9
If the 3rd and the 9th terms of an AP are 4 and −8 respectively, which term of this AP is 0?
Solution:
Let the first term be
$$ a $$and the common difference be
$$ d. $$The 3rd term is 4.
$$ a+2d=4 \qquad\cdots(1) $$The 9th term is −8.
$$ a+8d=-8 \qquad\cdots(2) $$Step 1: Find the common difference.
Subtract equation (1) from equation (2).
$$ (a+8d)-(a+2d)=-8-4 $$ $$ 6d=-12 $$ $$ d=-2 $$Step 2: Find the first term.
Substitute
$$ d=-2 $$in equation (1).
$$ a+2(-2)=4 $$ $$ a-4=4 $$ $$ a=8 $$Step 3: Find the required term.
Let the required term be the nth term.
$$ a_n=0 $$Using the formula
$$ a_n=a+(n-1)d $$ $$ 0=8+(n-1)(-2) $$ $$ 0=8-2(n-1) $$ $$ 2(n-1)=8 $$ $$ n-1=4 $$ $$ n=5 $$Verification:
$$ a_5=8+4(-2)=0 $$Answer:
The
$$ \boxed{5^{\text{th}}\text{ term}} $$of the AP is 0.
Question 10
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Let the first term be
$$ a $$and the common difference be
$$ d. $$Using the formula for the nth term,
$$ a_n=a+(n-1)d $$Step 1: Write the 17th term.
$$ a_{17}=a+16d $$Step 2: Write the 10th term.
$$ a_{10}=a+9d $$According to the question,
$$ a_{17}-a_{10}=7 $$Substitute the values.
$$ (a+16d)-(a+9d)=7 $$ $$ 7d=7 $$ $$ d=1 $$Answer:
$$ \boxed{d=1} $$Question 11
Which term of the AP:
$$ 3,\;15,\;27,\;39,\;\ldots $$will be 132 more than its 54th term?
Solution:
The given AP is
$$ 3,\;15,\;27,\;39,\;\ldots $$Therefore,
$$ a=3,\qquad d=12 $$Step 1: Find the 54th term.
$$ a_{54}=a+(54-1)d $$ $$ =3+53\times12 $$ $$ =3+636 $$ $$ =639 $$Step 2: Find the required term.
The required term is
$$ 639+132=771 $$Let it be the nth term.
$$ a_n=771 $$Using the formula
$$ a_n=a+(n-1)d $$ $$ 771=3+(n-1)\times12 $$ $$ 768=12(n-1) $$ $$ 64=n-1 $$ $$ n=65 $$Verification:
$$ a_{65}=3+64\times12=771 $$Difference from the 54th term:
$$ 771-639=132 $$Answer:
The required term is
$$ \boxed{65^{\text{th}}\text{ term}} $$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?
Solution:
Let the first terms of the two APs be
$$ a_1 \text{ and } a_2, $$and let their common difference be
$$ d. $$The 100th term of the first AP is
$$ a_1+99d. $$The 100th term of the second AP is
$$ a_2+99d. $$According to the question,
$$ (a_1+99d)-(a_2+99d)=100. $$Simplifying,
$$ a_1-a_2=100. $$Now, the 1000th term of the first AP is
$$ a_1+999d. $$The 1000th term of the second AP is
$$ a_2+999d. $$The difference between these terms is
$$ (a_1+999d)-(a_2+999d). $$ $$ =a_1-a_2. $$ $$ =100. $$Answer:
$$ \boxed{100} $$Question 13
How many three-digit numbers are divisible by 7?
Solution:
The smallest three-digit number is
$$ 100. $$The smallest three-digit number divisible by 7 is
$$ 105=7\times15. $$The largest three-digit number is
$$ 999. $$The largest three-digit number divisible by 7 is
$$ 994=7\times142. $$Thus, the required numbers form the AP
$$ 105,\;112,\;119,\;\ldots,\;994 $$Here,
$$ a=105,\qquad d=7,\qquad a_n=994. $$Using the formula
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 994=105+(n-1)\times7 $$ $$ 994-105=7(n-1) $$ $$ 889=7(n-1) $$ $$ 127=n-1 $$ $$ n=128 $$Answer:
The number of three-digit numbers divisible by 7 is
$$ \boxed{128}. $$Question 14
How many multiples of 4 lie between 10 and 250?
Solution:
The smallest multiple of 4 greater than 10 is
$$ 12. $$The largest multiple of 4 less than 250 is
$$ 248. $$Thus, the required multiples form the AP
$$ 12,\;16,\;20,\;\ldots,\;248 $$Here,
$$ a=12,\qquad d=4,\qquad a_n=248. $$Using the formula
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 248=12+(n-1)\times4 $$ $$ 248-12=4(n-1) $$ $$ 236=4(n-1) $$ $$ 59=n-1 $$ $$ n=60 $$Answer:
The number of multiples of 4 between 10 and 250 is
$$ \boxed{60} $$Question 15
For what value of n, are the nth terms of the two APs
$$ 63,\;65,\;67,\;\ldots $$and
$$ 3,\;10,\;17,\;\ldots $$equal?
Solution:
First AP:
$$ 63,\;65,\;67,\;\ldots $$Here,
$$ a_1=63,\qquad d_1=2. $$The nth term is
$$ \begin{aligned} T_n &=63+(n-1)\times2\\[4pt] &=2n+61 \end{aligned} $$Second AP:
$$ 3,\;10,\;17,\;\ldots $$Here,
$$ a_2=3,\qquad d_2=7. $$The nth term is
$$ \begin{aligned} T_n' &=3+(n-1)\times7\\[4pt] &=7n-4 \end{aligned} $$Since the nth terms are equal,
$$ 2n+61=7n-4 $$ $$ 65=5n $$ $$ n=13 $$Verification:
First AP:
$$ 63+12\times2=87 $$Second AP:
$$ 3+12\times7=87 $$Both give the same value.
Answer:
$$ \boxed{n=13} $$Question 16
Determine the AP whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
Let the first term be
$$ a $$and the common difference be
$$ d. $$Using the formula for the nth term of an AP,
$$ a_n=a+(n-1)d $$Step 1: Use the given 3rd term.
The 3rd term is 16.
$$ a_3=16 $$ $$ a+2d=16 \qquad\cdots(1) $$Step 2: Use the condition on the 7th and 5th terms.
The 7th term exceeds the 5th term by 12.
$$ a_7-a_5=12 $$Now,
$$ a_7=a+6d $$ $$ a_5=a+4d $$Therefore,
$$ (a+6d)-(a+4d)=12 $$ $$ 2d=12 $$ $$ d=6 $$Step 3: Find the first term.
Substitute
$$ d=6 $$in equation (1).
$$ a+2(6)=16 $$ $$ a+12=16 $$ $$ a=4 $$Step 4: Write the AP.
The first term is
$$ a=4 $$and the common difference is
$$ d=6. $$Hence, the required AP is
$$ 4,\;10,\;16,\;22,\;28,\;34,\;40,\;\ldots $$Verification:
The 3rd term is
$$ 16. $$The 5th term is
$$ 28. $$The 7th term is
$$ 40. $$Therefore,
$$ 40-28=12, $$which satisfies the given condition.
Answer:
$$ \boxed{4,\;10,\;16,\;22,\;28,\;34,\;40,\;\ldots} $$Question 17
Find the 20th term from the last term of the AP:
$$ 3,\;8,\;13,\;\ldots,\;253 $$Solution:
The given AP is
$$ 3,\;8,\;13,\;\ldots,\;253 $$Here,
$$ a=3,\qquad d=5,\qquad l=253 $$Step 1: Find the total number of terms.
Using the formula
$$ l=a+(n-1)d $$Substitute the given values.
$$ 253=3+(n-1)\times5 $$ $$ 250=5(n-1) $$ $$ 50=n-1 $$ $$ n=51 $$Step 2: Find the 20th term from the last.
The 20th term from the last is the
$$ (51-20+1)=32^{\text{nd}} $$term from the beginning.
Now find the 32nd term.
$$ a_{32}=a+(32-1)d $$ $$ =3+31\times5 $$ $$ =3+155 $$ $$ =158 $$Answer:
$$ \boxed{158} $$Question 18
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
Let the first term be
$$ a $$and the common difference be
$$ d. $$Step 1: Form the equations.
4th term:
$$ a_4=a+3d $$8th term:
$$ a_8=a+7d $$According to the question,
$$ (a+3d)+(a+7d)=24 $$ $$ 2a+10d=24 \qquad\cdots(1) $$6th term:
$$ a_6=a+5d $$10th term:
$$ a_{10}=a+9d $$According to the question,
$$ (a+5d)+(a+9d)=44 $$ $$ 2a+14d=44 \qquad\cdots(2) $$Step 2: Find the common difference.
Subtract equation (1) from equation (2).
$$ (2a+14d)-(2a+10d)=44-24 $$ $$ 4d=20 $$ $$ d=5 $$Step 3: Find the first term.
Substitute
$$ d=5 $$in equation (1).
$$ 2a+10(5)=24 $$ $$ 2a+50=24 $$ $$ 2a=-26 $$ $$ a=-13 $$Step 4: Write the first three terms.
First term:
$$ a=-13 $$Second term:
$$ a+d=-13+5=-8 $$Third term:
$$ a+2d=-13+10=-3 $$Verification:
$$ a_4=2,\qquad a_8=22 $$ $$ 2+22=24 $$ $$ a_6=12,\qquad a_{10}=32 $$ $$ 12+32=44 $$Hence, the obtained values are correct.
Answer:
$$ \boxed{-13,\;-8,\;-3} $$Question 19
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?
Solution:
The annual salaries form an Arithmetic Progression.
Here,
$$ a=5000,\qquad d=200 $$The required salary is
$$ a_n=7000. $$Step 1: Find the value of n.
Using the formula
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 7000=5000+(n-1)\times200 $$ $$ 7000-5000=200(n-1) $$ $$ 2000=200(n-1) $$ $$ 10=n-1 $$ $$ n=11 $$Step 2: Find the year.
The 1st term corresponds to the year
$$ 1995. $$Therefore, the 11th term corresponds to
$$ 1995+(11-1) $$ $$ =1995+10 $$ $$ =2005. $$Answer:
Subba Rao's income reached ₹7000 in the year
$$ \boxed{2005.} $$Question 20
Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.
Solution:
The weekly savings form an Arithmetic Progression.
Here,
$$ a=5,\qquad d=1.75,\qquad a_n=20.75. $$Step 1: Find the value of n.
Using the formula
$$ a_n=a+(n-1)d $$Substitute the given values.
$$ 20.75=5+(n-1)\times1.75 $$ $$ 20.75-5=1.75(n-1) $$ $$ 15.75=1.75(n-1) $$Divide both sides by
$$ 1.75. $$ $$ n-1=\frac{15.75}{1.75} $$ $$ n-1=9 $$ $$ n=10 $$Verification:
$$ a_{10}=5+9(1.75) $$ $$ =5+15.75 $$ $$ =20.75 $$Hence, the obtained value is correct.
Answer:
Ramkali's weekly savings became ₹20.75 in the
$$ \boxed{10^{\text{th}}\text{ week}.} $$