CBSE Class 10 Maths HOTs Arithmetic Progressions Set 16

PRACTICE QUESTIONS

Question. The first term of an AP of consecutive integers is \( p^2 + 1 \). The sum of \( (2p + 1) \) terms of this AP is
(a) \( (p + 1)^2 \)
(b) \( (2p + 1) (p + 1)^2 \)
(c) \( (p + 1)^3 \)
(d) \( p^3 + (p + 1)^3 \)
Answer: (d) \( p^3 + (p + 1)^3 \)

Question. If the sum of first n terms of an AP is \( An + Bn^2 \) where A and B are constants, the common difference of AP will be
(a) A + B
(b) A – B
(c) 2A
(d) 2B
Answer: (d) 2B

Question. If the nth term of an AP is \( (2n + 1) \), then the sum of its first three terms is
(a) 6n + 3
(b) 15
(c) 12
(d) 21
Answer: (b) 15

Question. An AP consists of 31 terms. If its 16th term is m, then sum of all the terms of this AP is
(a) 16 m
(b) 47 m
(c) 31 m
(d) 52 m
Answer: (c) 31 m

Question. Three numbers in an AP have sum 24. Its middle term is
(a) 12
(b) 8
(c) 4
(d) 2
Answer: (b) 8

Question. The value of the expression 1 – 6 + 2 – 7 + 3 – 8 + ... to 100 terms is
(a) 250
(b) –250
(c) 500
(d) –500
Answer: (b) –250

Question. In an A.P., if the common difference d = 3 and the eleventh term \( a_{11} = 15 \), then find the first term.
Answer: Given \( d = 3 \) and \( a_{11} = 15 \).
We know \( a_n = a + (n - 1)d \)

\( \implies \) \( 15 = a + (11 - 1)3 \)

\( \implies \) \( 15 = a + 30 \)

\( \implies \) \( a = 15 - 30 = -15 \).
The first term is -15.

Question. If sum of first n terms of an AP is \( 2n^2 + 5n \), then find \( S_{20} \).
Answer: Given \( S_n = 2n^2 + 5n \).
For \( n = 20 \):
\( S_{20} = 2(20)^2 + 5(20) \)

\( \implies \) \( S_{20} = 2(400) + 100 \)

\( \implies \) \( S_{20} = 800 + 100 = 900 \).

Question. In an AP, the first term is – 4, the last term is 29 and the sum of all its terms is 150. Find its common difference.
Answer: Given \( a = -4, l = 29, S_n = 150 \).
We know \( S_n = \frac{n}{2}(a + l) \)

\( \implies \) \( 150 = \frac{n}{2}(-4 + 29) \)

\( \implies \) \( 150 = \frac{n}{2}(25) \)

\( \implies \) \( 300 = 25n \)

\( \implies \) \( n = 12 \).
Now, \( l = a + (n - 1)d \)

\( \implies \) \( 29 = -4 + (12 - 1)d \)

\( \implies \) \( 33 = 11d \)

\( \implies \) \( d = 3 \).
The common difference is 3.

Question. The nth term (\( t_n \)) of an Arithmetic Progression is given by \( t_n = 4n – 5 \). Find the sum of the first 25 terms of the Arithmetic Progression.
Answer: Given \( t_n = 4n - 5 \).
First term \( a = t_1 = 4(1) - 5 = -1 \).
25th term \( l = t_{25} = 4(25) - 5 = 100 - 5 = 95 \).
Sum \( S_{25} = \frac{25}{2}(a + l) \)

\( \implies \) \( S_{25} = \frac{25}{2}(-1 + 95) \)

\( \implies \) \( S_{25} = \frac{25}{2}(94) = 25 \times 47 = 1175 \).

Question. How many terms of the AP 3, 5, 7, ... must be taken so that the sum is 120?
Answer: Given AP: 3, 5, 7, ... where \( a = 3, d = 2, S_n = 120 \).
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)

\( \implies \) \( 120 = \frac{n}{2}[2(3) + (n - 1)2] \)

\( \implies \) \( 120 = \frac{n}{2}[6 + 2n - 2] \)

\( \implies \) \( 120 = \frac{n}{2}[2n + 4] \)

\( \implies \) \( 120 = n(n + 2) \)

\( \implies \) \( n^2 + 2n - 120 = 0 \)

\( \implies \) \( (n + 12)(n - 10) = 0 \).
Since \( n \) cannot be negative, \( n = 10 \).
10 terms must be taken.

Question. Find the sum of all 2-digit positive numbers divisible by 3.
Answer: The 2-digit numbers divisible by 3 are 12, 15, 18, ..., 99.
This is an AP with \( a = 12, d = 3, l = 99 \).
Find \( n \): \( 99 = 12 + (n - 1)3 \)

\( \implies \) \( 87 = (n - 1)3 \)

\( \implies \) \( 29 = n - 1 \)

\( \implies \) \( n = 30 \).
Sum \( S_{30} = \frac{30}{2}(12 + 99) = 15(111) = 1665 \).

Question. The sum of first six terms of an AP is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first and the thirteenth terms of the AP.
Answer: \( S_6 = 42 \implies \frac{6}{2}[2a + 5d] = 42 \implies 2a + 5d = 14 \).
Ratio \( \frac{a + 9d}{a + 29d} = \frac{1}{3} \)

\( \implies \) \( 3a + 27d = a + 29d \)

\( \implies \) \( 2a = 2d \implies a = d \).
Substitute \( a = d \) in first equation:
\( 2a + 5a = 14 \implies 7a = 14 \implies a = 2 \).
So \( d = 2 \).
First term \( a = 2 \).
Thirteenth term \( a_{13} = a + 12d = 2 + 12(2) = 26 \).

Question. Find the sum of first 20 terms of an A.P. whose \( n^{th} \) term is given as \( a_n = 5 – 2n \). 
Answer: Given \( a_n = 5 - 2n \).
\( a_1 = 5 - 2(1) = 3 \).
\( a_{20} = 5 - 2(20) = -35 \).
\( S_{20} = \frac{20}{2}(a_1 + a_{20}) = 10(3 - 35) = -320 \).

Question. If the sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
Answer: \( S_6 = 36 \implies n^2 \) pattern suggested.
Let \( S_n = n^2 \). Checking: \( S_6 = 6^2 = 36 \), \( S_{16} = 16^2 = 256 \).
Then \( S_{10} = 10^2 = 100 \).
(Or solve using \( S_n \) formula to find \( a=1, d=2 \)).

Question. Kanika was given her pocket money on Jan 1st, 2008. She puts Rs. 1 on day 1, Rs. 2 on day 2, Rs. 3 on day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs. 204 of her pocket money, and found that at the end of the month she still had Rs. 100 with her. How much was her pocket money for the month?
Answer: Jan 2008 has 31 days.
Savings in piggy bank = \( 1 + 2 + 3 + ... + 31 = \frac{31 \times 32}{2} = 496 \).
Total pocket money = Piggy bank savings + Spent + Remaining
\( = 496 + 204 + 100 = \text{Rs. } 800 \).

Question. Find the sum of all three-digit numbers each of which leaves the remainder 2, when divided by 3. 
Answer: Numbers are 101, 104, 107, ..., 998.
This is an AP with \( a = 101, d = 3, l = 998 \).
Find \( n \): \( 998 = 101 + (n - 1)3 \implies 897 = (n - 1)3 \implies n - 1 = 299 \implies n = 300 \).
Sum \( S_{300} = \frac{300}{2}(101 + 998) = 150(1099) = 164850 \).

Question. The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.
Answer: \( S_5 + S_7 = 167 \implies \frac{5}{2}(2a+4d) + \frac{7}{2}(2a+6d) = 167 \implies 5a+10d + 7a+21d = 167 \implies 12a+31d = 167 \).
\( S_{10} = 235 \implies \frac{10}{2}(2a+9d) = 235 \implies 5(2a+9d) = 235 \implies 2a+9d = 47 \).
Solving these: Multiply second by 6: \( 12a+54d = 282 \).
Subtracting: \( (12a+54d) - (12a+31d) = 282 - 167 \implies 23d = 115 \implies d = 5 \).
\( 2a + 9(5) = 47 \implies 2a = 2 \implies a = 1 \).
\( S_{20} = \frac{20}{2}[2(1) + 19(5)] = 10[2 + 95] = 970 \).

Question. Deepak repays his total loan of Rs. 1,18,000 by paying every month starting with the first instalment of Rs. 1000. If he increases the instalment by Rs. 100 every month, what amount will be paid as the last instalment of loan? What amount of loan he still have to pay after the 30th instalment?
Answer: Total sum = Rs. 1,18,000. \( a = 1000, d = 100 \).
Find \( n \): \( \frac{n}{2}[2(1000) + (n-1)100] = 118000 \implies n[1000 + 50n - 50] = 118000 \implies 50n^2 + 950n - 118000 = 0 \implies n^2 + 19n - 2360 = 0 \).
Factors: \( (n + 59)(n - 40) = 0 \implies n = 40 \).
Last instalment \( a_{40} = 1000 + 39(100) = 4900 \).
Sum paid after 30th instalment: \( S_{30} = \frac{30}{2}[2(1000) + 29(100)] = 15[2000 + 2900] = 15(4900) = 73500 \).
Remaining amount = \( 1,18,000 - 73,500 = \text{Rs. } 44,500 \).

Question. If the sum of first 7 terms of an AP is 49 and that of first 17 terms is 289, find the sum of its first n terms. 
Answer: \( S_7 = 7^2 = 49 \), \( S_{17} = 17^2 = 289 \).
By observation, \( S_n = n^2 \).
(Or derive \( a=1, d=2 \) from equations).

Question. If \( S_1, S_2, S_3 \) are the sum of n terms of three APs, the first term of each being unity and the respective common difference being 1, 2, 3; prove that \( S_1 + S_3 = 2S_2 \). 
Answer: \( S_1 = \frac{n}{2}[2 + (n-1)1] = \frac{n(n+1)}{2} \).
\( S_2 = \frac{n}{2}[2 + (n-1)2] = n^2 \).
\( S_3 = \frac{n}{2}[2 + (n-1)3] = \frac{n(3n-1)}{2} \).
\( S_1 + S_3 = \frac{n^2+n + 3n^2-n}{2} = \frac{4n^2}{2} = 2n^2 \).
\( 2S_2 = 2n^2 \).
Hence \( S_1 + S_3 = 2S_2 \). Proved.

Question. 150 workers were engaged to finish a piece of work in a certain number of days. Four workers dropped the second day, four more worker dropped the third day and so on. It takes 8 more days to finish the work now. Find the number of days in which the work was completed. 
Answer: Let the original number of days be \( x \).
Total work = \( 150x \) worker-days.
Actual work done: \( 150 + 146 + 142 + ... \) for \( (x + 8) \) days.
\( \frac{x+8}{2}[2(150) + (x+8-1)(-4)] = 150x \)

\( \implies \) \( \frac{x+8}{2}[300 - 4x - 28] = 150x \)

\( \implies \) \( (x+8)(136 - 2x) = 150x \)

\( \implies \) \( 136x - 2x^2 + 1088 - 16x = 150x \)

\( \implies \) \( 120x - 2x^2 + 1088 = 150x \)

\( \implies \) \( 2x^2 + 30x - 1088 = 0 \)

\( \implies \) \( x^2 + 15x - 544 = 0 \).
Factors: \( (x + 32)(x - 17) = 0 \implies x = 17 \).
Work was completed in \( 17 + 8 = 25 \) days.

Question. Interior angles of a polygon are in AP. If the smallest angle is 120° and common difference is 5°, find the number of sides of the polygon.
Answer: Let number of sides be \( n \). Sum of interior angles = \( (n - 2)180 \).
In AP: \( \frac{n}{2}[2(120) + (n-1)5] = (n-2)180 \)

\( \implies \) \( n[240 + 5n - 5] = 360n - 720 \)

\( \implies \) \( 5n^2 + 235n = 360n - 720 \)

\( \implies \) \( 5n^2 - 125n + 720 = 0 \)

\( \implies \) \( n^2 - 25n + 144 = 0 \).
Factors: \( (n - 16)(n - 9) = 0 \implies n = 9, 16 \).
If \( n = 16 \), largest angle = \( 120 + 15(5) = 195 > 180 \), not possible.
So \( n = 9 \).

Question. Raghav buys a shop for Rs. 120000. He pays half of the amount in cash and agrees to pay the balance in 12 annual instalments of Rs. 5000 each. If the rate of interest is 12% and he pays with the instalment the interest due on the unpaid amount. Find the total cost of the shop.
Answer: Remaining balance = Rs. 60,000.
Interest on 1st instalment = \( 12\% \text{ of } 60,000 = 7200 \).
Interest on 2nd instalment = \( 12\% \text{ of } 55,000 = 6600 \).
... Interest on 12th instalment = \( 12\% \text{ of } 5000 = 600 \).
Total interest = \( \frac{12}{2}[7200 + 600] = 6(7800) = 46800 \).
Total cost = Rs. 1,20,000 + Rs. 46,800 = Rs. 1,66,800.

Question. Saurav gets pocket money from his father every day. Out of the pocket money, he saves Rs. 2.75 on first day and on each succeeding day he increases his saving by 25 paise. Find
(i) the amount saved by Saurav on 14th day,
(ii) the amount saved by Saurav on 25th day, and
(iii) the total amount saved by Saurav in 30 days.
Answer: \( a = 2.75, d = 0.25 \).
(i) \( a_{14} = 2.75 + 13(0.25) = 2.75 + 3.25 = \text{Rs. } 6.00 \).
(ii) \( a_{25} = 2.75 + 24(0.25) = 2.75 + 6.00 = \text{Rs. } 8.75 \).
(iii) \( S_{30} = \frac{30}{2}[2(2.75) + 29(0.25)] = 15[5.5 + 7.25] = 15(12.75) = \text{Rs. } 191.25 \).

Question. Find the sum of the first 25 terms of an AP whose nth term is given by \( t_n = 7 – 3n \).
Answer: \( t_1 = 4, t_{25} = 7 - 3(25) = -68 \).
\( S_{25} = \frac{25}{2}(4 - 68) = \frac{25}{2}(-64) = 25(-32) = -800 \).

Question. If \( a_1, a_2, b_3, ... \) are in A.P. such that \( a_1 + a_6 + a_8 + a_{10} + a_{21} + a_{23} + a_{25} + a_{30} = 300 \), find \( a_1 + a_2 + a_3 + ... + a_{30} \).
Answer: In an AP, \( a_k + a_{n-k+1} = a_1 + a_n \).
Here \( a_1+a_{30} = a_6+a_{25} = a_8+a_{23} = a_{10}+a_{21} = x \).
Sum given = \( 4x = 300 \implies x = 75 \).
\( S_{30} = \frac{30}{2}(a_1 + a_{30}) = 15(75) = 1125 \).

Question. How many multiples of 4 lie between 10 and 250? Also find their sum.
Answer: Multiples are 12, 16, ..., 248.
\( 248 = 12 + (n-1)4 \implies 236 = (n-1)4 \implies n-1 = 59 \implies n = 60 \).
Sum \( S_{60} = \frac{60}{2}(12 + 248) = 30(260) = 7800 \).

Question. Deepa has to buy a scooty. She can buy scooty either making cashdown payment of Rs. 25,000 or by making 15 monthly instalment as below.
Ist month – Rs. 3425,
IInd month – Rs. 3225,
IIIrd month – Rs. 3025,
IVth month – Rs. 2825 and so on.
(i) Find amount of 6th instalment.
(ii) Total amount paid in 15 instalments.
Answer: \( a = 3425, d = -200 \).
(i) \( a_6 = 3425 + 5(-200) = 3425 - 1000 = \text{Rs. } 2425 \).
(ii) \( S_{15} = \frac{15}{2}[2(3425) + 14(-200)] = \frac{15}{2}[6850 - 2800] = \frac{15}{2}(4050) = 15(2025) = \text{Rs. } 30,375 \).

INTEGRATED (MIXED) QUESTIONS

Question. If the third term of an AP is 12 and the seventh term is 24, then the 10th term is
(a) 34
(b) 35
(c) 36
(d) 33
Answer: (d) 33

Question. If \( a, b, c, d, e \) and \( f \) are in AP, then \( e – c \) is equal to
(a) \( 2(c – a) \)
(b) \( 2(f – d) \)
(c) \( 2(d – c) \)
(d) \( d – c \)
Answer: (c) \( 2(d – c) \)

Question. If the numbers \( a, b, c, d, e \) form an AP, then the value of \( a – 4b + 6c – 4d + e \) is
(a) 1
(b) 2
(c) 0
(d) None of the options
Answer: (c) 0

Question. If the roots of the equation \( (b – c)x^2 + (c – a)x + (a – b) = 0 \) are equal, then
(a) \( 2b = a + c \)
(b) \( a = b = c \)
(c) \( b^2 = ac \)
(d) None of the options
Answer: (a) \( 2b = a + c \)

Question. Sum of the n terms of the series \( \sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + ... \) is
(a) \( \frac{n(n + 2)}{\sqrt{2}} \)
(b) \( \sqrt{2} (n) (n + 1) \)
(c) \( \frac{n(n + 1)}{\sqrt{2}} \)
(d) 1
Answer: (c) \( \frac{n(n + 1)}{\sqrt{2}} \)

Question. If \( \frac{3 + 5 + 7 + ... \text{ up to n terms}}{5 + 8 + 11... \text{ up to 10 terms}} = 7 \), then the value of n is
(a) 35
(b) 36
(c) 37
(d) 40
Answer: (a) 35

Question. Sum of first n terms of a series is \( 5n^2 + 2n \), its second term is
(a) 15
(b) 17
(c) 19
(d) 21
Answer: (b) 17

Question. If \( \frac{3}{4}, a, 4 \) are three consecutive terms of an A.P., then the value of a is
(a) 2.5
(b) 2.4
(c) 2.3
(d) 2.2
Answer: (c) 2.375

Question. The value of k for which the consecutive terms \( 2k+1, 3k+3 \) and \( 5k–1 \) form an AP is 
(a) 6
(b) 7
(c) 8
(d) 9
Answer: (a) 6

Question. The 9th term from the end (towards the first term) of the A.P. 5, 9, 13, ..., 185 is 
(a) 149
(b) 153
(c) 157
(d) 161
Answer: (b) 153

Question. If the last term of an AP is 119 and the 8th term from the end is 91, then the common difference of the AP is
(a) 2
(b) 4
(c) 3
(d) –3
Answer: (b) 4

Question. In an AP, if \( a = 3.5, d = 0 \) and \( n = 101 \), then \( a_n = \)
(a) 0
(b) 3.5
(c) 103.5
(d) 104.5
Answer: (b) 3.5

Question. If nth term of an AP is \( (2n + 1) \), then the sum of its first three terms is
(a) 7
(b) 15
(c) 17
(d) 21
Answer: (b) 15