Practice Set 3.1
1Which of the following sequences are A.P.? If they are A.P. find the common difference.
(1) 2, 4, 6, 8, …
t1 = 2, t2 = 4, t3 = 6, t4 = 8
t2 − t1 = 4 − 2 = 2
t3 − t2 = 6 − 4 = 2
t4 − t3 = 8 − 6 = 2
The difference is constant.
It is an A.P.; common difference d = 2
(2) 2, 52, 3, 73, …
t1 = 2, t2 = 52, t3 = 3, t4 = 73
t2 − t1 = 52 − 2 = 12
t3 − t2 = 3 − 52 = 12
t4 − t3 = 73 − 3 = −23
t4 − t3 ≠ t3 − t2, so the difference is not constant.
It is not an A.P.
The fourth term is printed as 73 in the textbook. If it were 72, the sequence would be an A.P. with d = 12. As printed, it is not an A.P.
(3) −10, −6, −2, 2, …
t1 = −10, t2 = −6, t3 = −2, t4 = 2
t2 − t1 = −6 − (−10) = 4
t3 − t2 = −2 − (−6) = 4
t4 − t3 = 2 − (−2) = 4
The difference is constant.
It is an A.P.; common difference d = 4
(4) 0.3, 0.33, 0.0333, …
t1 = 0.3, t2 = 0.33, t3 = 0.0333
t2 − t1 = 0.33 − 0.3 = 0.03
t3 − t2 = 0.0333 − 0.33 = −0.2967
The difference is not constant.
It is not an A.P.
Even if the third term is read as 0.333, the differences are 0.03 and 0.003, which are not equal. So the sequence is not an A.P. either way.
(5) 0, −4, −8, −12, …
t1 = 0, t2 = −4, t3 = −8, t4 = −12
t2 − t1 = −4 − 0 = −4
t3 − t2 = −8 − (−4) = −4
t4 − t3 = −12 − (−8) = −4
The difference is constant.
It is an A.P.; common difference d = −4
(6) −15, −15, −15, …
t1 = −15, t2 = −15, t3 = −15
t2 − t1 = −15 − (−15) = 0
t3 − t2 = 0
The difference is constant (zero).
It is an A.P.; common difference d = 0
(7) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …
t1 = 3, t2 = 3 + √2, t3 = 3 + 2√2, t4 = 3 + 3√2
t2 − t1 = (3 + √2) − 3 = √2
t3 − t2 = (3 + 2√2) − (3 + √2) = √2
t4 − t3 = √2
The difference is constant.
It is an A.P.; common difference d = √2
(8) 127, 132, 137, …
t1 = 127, t2 = 132, t3 = 137
t2 − t1 = 132 − 127 = 5
t3 − t2 = 137 − 132 = 5
The difference is constant.
It is an A.P.; common difference d = 5
2Write an A.P. whose first term is a and common difference is d in each of the following.
(1) a = 10, d = 5
t1 = a = 10
t2 = t1 + d = 10 + 5 = 15
t3 = t2 + d = 15 + 5 = 20
t4 = t3 + d = 20 + 5 = 25
A.P.: 10, 15, 20, 25, …
(2) a = −3, d = 0
t1 = a = −3
t2 = t1 + d = −3 + 0 = −3
t3 = t2 + d = −3 + 0 = −3
t4 = t3 + d = −3 + 0 = −3
A.P.: −3, −3, −3, −3, …
(3) a = −7, d = 12
t1 = a = −7
t2 = t1 + d = −7 + 12 = −132
t3 = t2 + d = −132 + 12 = −122 = −6
t4 = t3 + d = −6 + 12 = −112
A.P.: −7, −13/2, −6, −11/2, …
(4) a = −1.25, d = 3
t1 = a = −1.25
t2 = t1 + d = −1.25 + 3 = 1.75
t3 = t2 + d = 1.75 + 3 = 4.75
t4 = t3 + d = 4.75 + 3 = 7.75
A.P.: −1.25, 1.75, 4.75, 7.75, …
(5) a = 6, d = −3
t1 = a = 6
t2 = t1 + d = 6 + (−3) = 3
t3 = t2 + d = 3 + (−3) = 0
t4 = t3 + d = 0 + (−3) = −3
A.P.: 6, 3, 0, −3, …
(6) a = −19, d = −4
t1 = a = −19
t2 = t1 + d = −19 + (−4) = −23
t3 = t2 + d = −23 + (−4) = −27
t4 = t3 + d = −27 + (−4) = −31
A.P.: −19, −23, −27, −31, …
3Find the first term and common difference for each of the A.P.
(1) 5, 1, −3, −7, …
a = t1 = 5
d = t2 − t1 = 1 − 5 = −4
a = 5, d = −4
(2) 0.6, 0.9, 1.2, 1.5, …
a = t1 = 0.6
d = t2 − t1 = 0.9 − 0.6 = 0.3
a = 0.6, d = 0.3
(3) 127, 135, 143, 151, …
a = t1 = 127
d = t2 − t1 = 135 − 127 = 8
a = 127, d = 8
(4) 14, 34, 54, 74, …
a = t1 = 14
d = t2 − t1 = 34 − 14 = 24 = 12
a = 1/4, d = 1/2
Practice Set 3.2
1Write the correct number in the given boxes from the following A.P.
(i) 1, 8, 15, 22, …
Here a = 1, t1 = 1, t2 = 8, t3 = 15
t2 − t1 = 8 − 1 = 7
t3 − t2 = 15 − 8 = 7 ∴ d = 7
(ii) 3, 6, 9, 12, …
Here t1 = 3, t2 = 6, t3 = 9, t4 = 12
t2 − t1 = 3, t3 − t2 = 3 ∴ d = 3
(iii) −3, −8, −13, −18, …
Here t3 = −13, t2 = −8, t4 = −18, t1 = −3
t2 − t1 = −5, t3 − t2 = −5 ∴ a = −3, d = −5
(iv) 70, 60, 50, 40, …
Here t1 = 70, t2 = 60, t3 = 50, …
∴ a = 70, d = −10
2Decide whether the following sequence is an A.P.; if so, find the 20th term of the progression: −12, −5, 2, 9, 16, 23, 30, …
t2 − t1 = −5 − (−12) = 7, t3 − t2 = 2 − (−5) = 7, t4 − t3 = 9 − 2 = 7
The difference is constant, so it is an A.P. with a = −12, d = 7.
tn = a + (n − 1)d
t20 = −12 + (20 − 1) × 7 = −12 + 133 = 121
It is an A.P.; t₂₀ = 121
3Given Arithmetic Progression 12, 16, 20, 24, … Find the 24th term of this progression.
a = 12, d = 16 − 12 = 4
tn = a + (n − 1)d
t24 = 12 + (24 − 1) × 4 = 12 + 92 = 104
t₂₄ = 104
4Find the 19th term of the following A.P.: 7, 13, 19, 25, …
a = 7, d = 13 − 7 = 6
t19 = 7 + (19 − 1) × 6 = 7 + 108 = 115
t₁₉ = 115
5Find the 27th term of the following A.P.: 9, 4, −1, −6, −11, …
a = 9, d = 4 − 9 = −5
t27 = 9 + (27 − 1) × (−5) = 9 − 130 = −121
t₂₇ = −121
6Find how many three digit natural numbers are divisible by 5.
Three digit numbers divisible by 5: 100, 105, 110, …, 995
a = 100, d = 5, tn = 995
995 = 100 + (n − 1) × 5
∴ 895 = (n − 1) × 5
∴ n − 1 = 179 ∴ n = 180
180 three digit numbers are divisible by 5.
7The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.
t11 = a + 10d = 16 … (I)
t21 = a + 20d = 29 … (II)
Subtracting (I) from (II):
10d = 13 ∴ d = 1.3
From (I): a = 16 − 10 × 1.3 = 16 − 13 = 3
t41 = a + 40d = 3 + 40 × 1.3 = 3 + 52 = 55
t₄₁ = 55
811, 8, 5, 2, … In this A.P. which term is number −151?
a = 11, d = 8 − 11 = −3, tn = −151
−151 = 11 + (n − 1) × (−3)
∴ −162 = −3(n − 1)
∴ n − 1 = 54 ∴ n = 55
−151 is the 55th term.
9In the natural numbers from 10 to 250, how many are divisible by 4?
Numbers from 10 to 250 divisible by 4: 12, 16, 20, …, 248
a = 12, d = 4, tn = 248
248 = 12 + (n − 1) × 4
∴ 236 = 4(n − 1)
∴ n − 1 = 59 ∴ n = 60
60 numbers are divisible by 4.
10In an A.P. 17th term is 7 more than its 10th term. Find the common difference.
t17 − t10 = 7
[a + 16d] − [a + 9d] = 7
∴ 7d = 7 ∴ d = 1
d = 1
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Practice Set 3.3
1First term and common difference of an A.P. are 6 and 3 respectively; find S₂₇.
a = 6, d = 3, S27 = ?
Sn = n2[2a + (n − 1)d]
S27 = 272[12 + (27 − 1) × 3]
= 272 × 90
= 27 × 45 = 1215
S₂₇ = 1215
2Find the sum of first 123 even natural numbers.
The even natural numbers are 2, 4, 6, …
a = 2, d = 2, n = 123
Sn = n2[2a + (n − 1)d]
S123 = 1232[2 × 2 + (123 − 1) × 2]
= 1232[4 + 244] = 1232 × 248
= 123 × 124 = 15252
Sum = 15252
3Find the sum of all even numbers between 1 and 350.
Even numbers between 1 and 350: 2, 4, 6, …, 348
a = 2, d = 2, tn = 348
348 = 2 + (n − 1) × 2 ∴ n − 1 = 173 ∴ n = 174
Sn = n2[t1 + tn]
S174 = 1742[2 + 348] = 87 × 350 = 30450
Sum = 30450
4In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.
t19 = a + 18d = 52 … (I)
t38 = a + 37d = 128 … (II)
Subtracting (I) from (II):
19d = 76 ∴ d = 4
From (I): a = 52 − 72 = −20
S56 = 562[2 × (−20) + (56 − 1) × 4]
= 28[−40 + 220] = 28 × 180 = 5040
S₅₆ = 5040
5Complete the following activity to find the sum of natural numbers between 1 and 140 which are divisible by 4.
Between 1 and 140, natural numbers divisible by 4: 4, 8, 12, …, 136
a = 4, d = 4, tn = 136
tn = a + (n − 1)d → 136 = 4 + (n − 1) × 4 → 132 = 4(n − 1) → n − 1 = 33
How many numbers? ∴ n = 34
Sn = n2[2a + (n − 1)d]
S34 = 342[2 × 4 + (34 − 1) × 4] = 17[8 + 132] = 17 × 140 = 2380
Sum of numbers from 1 to 140 which are divisible by 4 = 2380
6Sum of first 55 terms in an A.P. is 3300, find its 28th term.
S55 = 552[2a + (55 − 1)d] = 3300
∴ 552 × 2[a + 27d] = 3300
∴ 55(a + 27d) = 3300
∴ a + 27d = 60
But t28 = a + (28 − 1)d = a + 27d
∴ t28 = 60
t₂₈ = 60
7In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms. (Assume the terms are a − d, a, a + d.)
(a − d) + a + (a + d) = 27
∴ 3a = 27 ∴ a = 9
(a − d) × a × (a + d) = 504
∴ 9(9 − d)(9 + d) = 504
∴ 81 − d2 = 56
∴ d2 = 25 ∴ d = 5 or d = −5
If d = 5: terms are 4, 9, 14. If d = −5: terms are 14, 9, 4.
Check: 4 + 9 + 14 = 27 and 4 × 9 × 14 = 504
The terms are 4, 9, 14 (or 14, 9, 4).
8Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14. (Assume the terms are a − d, a, a + d, a + 2d.)
(a − d) + a + (a + d) + (a + 2d) = 12
∴ 4a + 2d = 12 ∴ 2a + d = 6 … (I)
(a + d) + (a + 2d) = 14 ∴ 2a + 3d = 14 … (II)
Subtracting (I) from (II):
2d = 8 ∴ d = 4
From (I): 2a = 2 ∴ a = 1
Terms: a − d = −3, a = 1, a + d = 5, a + 2d = 9
The terms are −3, 1, 5, 9.
9If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term.
t9 = a + 8d = 0 ∴ a = −8d
t19 = a + 18d = −8d + 18d = 10d
t29 = a + 28d = −8d + 28d = 20d
∴ t29 = 20d = 2 × 10d = 2 × t19
Hence the 29th term is twice the 19th term.
Practice Set 3.4
1On 1st Jan 2016, Sanika decides to save ₹ 10, ₹ 11 on second day, ₹ 12 on third day. If she decides to save like this, then on 31st Dec 2016 what would be her total saving?
2016 is a leap year, so from 1st Jan to 31st Dec there are 366 days.
a = 10, d = 1, n = 366
Sn = n2[2a + (n − 1)d]
S366 = 3662[2 × 10 + (366 − 1) × 1]
= 183[20 + 365] = 183 × 385 = 70455
Total saving = ₹ 70455
2A man borrows ₹ 8000 and agrees to repay with a total interest of ₹ 1360 in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 40. Find the amount of the first and last instalment.
Total amount to be repaid = 8000 + 1360 = ₹ 9360
n = 12, d = −40, S12 = 9360
9360 = 122[2a + (12 − 1)(−40)]
∴ 9360 = 6[2a − 440]
∴ 1560 = 2a − 440
∴ 2a = 2000 ∴ a = 1000
Last instalment = t12 = 1000 + (12 − 1)(−40) = 1000 − 440 = 560
First instalment = ₹ 1000, last instalment = ₹ 560
3Sachin invested in a national saving certificate scheme. In the first year he invested ₹ 5000, in the second year ₹ 7000, in the third year ₹ 9000 and so on. Find the total amount that he invested in 12 years.
a = 5000, d = 2000, n = 12
S12 = 122[2 × 5000 + (12 − 1) × 2000]
= 6[10000 + 22000] = 6 × 32000 = 192000
Total investment in 12 years = ₹ 1,92,000
4There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th row and also find how many total seats are there in the auditorium.
a = 20, d = 2
t15 = 20 + (15 − 1) × 2 = 20 + 28 = 48
S27 = 272[2 × 20 + (27 − 1) × 2] = 272[40 + 52] = 272 × 92 = 27 × 46 = 1242
15th row: 48 seats; total seats: 1242
5Kargil’s temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was 5 °C more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was −30 °C then find the temperature on the other five days.
Let the temperatures from Monday to Saturday be t1, t2, …, t6.
t1 + t6 = t2 + t6 + 5
∴ t1 − t2 = 5 ∴ −d = 5 ∴ d = −5
Wednesday: t3 = a + 2d = −30
∴ a − 10 = −30 ∴ a = −20
Monday −20, Tuesday −25, Wednesday −30, Thursday −35, Friday −40, Saturday −45
Monday −20 °C, Tuesday −25 °C, Thursday −35 °C, Friday −40 °C, Saturday −45 °C
6On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.
a = 1, d = 1, n = 25
S25 = 252[t1 + t25] = 252[1 + 25] = 25 × 13 = 325
Total number of trees = 325
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Problem Set 3
1Choose the correct alternative answer for each of the following sub questions.
(1) The sequence −10, −6, −2, 2, … (A) is an A.P., Reason d = −16 (B) is an A.P., Reason d = 4 (C) is an A.P., Reason d = −4 (D) is not an A.P.
−6 − (−10) = 4, −2 − (−6) = 4, 2 − (−2) = 4
(B) is an A.P., Reason d = 4
(2) First four terms of an A.P. are …, whose first term is −2 and common difference is −2. (A) −2, 0, 2, 4 (B) −2, 4, −8, 16 (C) −2, −4, −6, −8 (D) −2, −4, −8, −16
−2, −2 + (−2) = −4, −4 + (−2) = −6, −6 + (−2) = −8
(C) −2, −4, −6, −8
(3) What is the sum of the first 30 natural numbers? (A) 464 (B) 465 (C) 462 (D) 461
n(n + 1)2 = 30 × 312 = 465
(B) 465
(4) For a given A.P. t7 = 4, d = −4 then a = … (A) 6 (B) 7 (C) 20 (D) 28
a + 6 × (−4) = 4 ∴ a = 28
(D) 28
(5) For a given A.P. a = 3.5, d = 0, n = 101, then tn = … (A) 0 (B) 3.5 (C) 103.5 (D) 104.5
t101 = 3.5 + 100 × 0 = 3.5
(B) 3.5
(6) In an A.P. first two terms are −3, 4 then 21st term is … (A) −143 (B) 143 (C) 137 (D) 17
d = 4 − (−3) = 7
t21 = −3 + 20 × 7 = 137
(C) 137
(7) If for any A.P. d = 5 then t18 − t13 = … (A) 5 (B) 20 (C) 25 (D) 30
t18 − t13 = (a + 17d) − (a + 12d) = 5d = 25
(C) 25
(8) Sum of first five multiples of 3 is … (A) 45 (B) 55 (C) 15 (D) 75
(A) 45
(9) 15, 10, 5, … In this A.P. sum of first 10 terms is … (A) −75 (B) −125 (C) 75 (D) 125
a = 15, d = −5
S10 = 102[30 + 9 × (−5)] = 5 × (−15) = −75
(A) −75
(10) In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is 399, then n = … (A) 42 (B) 38 (C) 21 (D) 19
399 = n2(1 + 20)
∴ n = 399 × 221 = 38
(B) 38
2Find the fourth term from the end in an A.P. −11, −8, −5, …, 49.
a = −11, d = 3, last term = 49
49 = −11 + (n − 1) × 3 ∴ n − 1 = 20 ∴ n = 21
The 4th term from the end is the (21 − 4 + 1) = 18th term.
t18 = −11 + 17 × 3 = −11 + 51 = 40
Check: from the end the terms are 49, 46, 43, 40.
The fourth term from the end is 40.
3In an A.P. the 10th term is 46, sum of the 5th and 7th term is 52. Find the A.P.
t10 = a + 9d = 46 … (I)
t5 + t7 = (a + 4d) + (a + 6d) = 52
∴ 2a + 10d = 52 ∴ a + 5d = 26 … (II)
Subtracting (II) from (I):
4d = 20 ∴ d = 5
From (II): a = 26 − 25 = 1
The A.P. is 1, 6, 11, 16, …
4The A.P. in which 4th term is −15 and 9th term is −30. Find the sum of the first 10 numbers.
t4 = a + 3d = −15 … (I)
t9 = a + 8d = −30 … (II)
Subtracting (I) from (II):
5d = −15 ∴ d = −3
From (I): a = −15 + 9 = −6
S10 = 102[2 × (−6) + 9 × (−3)] = 5[−12 − 27] = 5 × (−39) = −195
S₁₀ = −195
5Two A.P.’s are given 9, 7, 5, … and 24, 21, 18, … If nth term of both the progressions are equal then find the value of n and nth term.
First A.P.: a = 9, d = −2 ∴ tn = 9 + (n − 1)(−2)
Second A.P.: a = 24, d = −3 ∴ tn = 24 + (n − 1)(−3)
9 − 2(n − 1) = 24 − 3(n − 1)
∴ (n − 1) = 15 ∴ n = 16
t16 = 9 + 15 × (−2) = 9 − 30 = −21 (also 24 + 15 × (−3) = −21)
n = 16, and the 16th term of both is −21.
6If sum of 3rd and 8th terms of an A.P. is 7 and sum of 7th and 14th terms is −3 then find the 10th term.
t3 + t8 = (a + 2d) + (a + 7d) = 7 ∴ 2a + 9d = 7 … (I)
t7 + t14 = (a + 6d) + (a + 13d) = −3 ∴ 2a + 19d = −3 … (II)
Subtracting (I) from (II):
10d = −10 ∴ d = −1
From (I): 2a − 9 = 7 ∴ a = 8
t10 = 8 + 9 × (−1) = −1
t₁₀ = −1
7In an A.P. the first term is −5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?
Sn = n2[t1 + tn]
120 = n2[−5 + 45] = 20n
∴ n = 6
t6 = a + 5d
∴ 45 = −5 + 5d ∴ d = 10
There are 6 terms; common difference d = 10.
8Sum of 1 to n natural numbers is 36, then find the value of n.
n(n + 1)2 = 36
∴ n2 + n − 72 = 0
∴ (n + 9)(n − 8) = 0
∴ n = −9 or n = 8
n is a natural number. ∴ n = 8
n = 8
9Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.
Let the parts be a − d, a, a + d (with d > 0).
(a − d) + a + (a + d) = 207 ∴ 3a = 207 ∴ a = 69
Product of the two smaller parts: (a − d) × a = 4623
∴ (69 − d) × 69 = 4623
∴ 69 − d = 67 ∴ d = 2
Parts: 67, 69, 71
Check: 67 + 69 + 71 = 207 and 67 × 69 = 4623
The parts are 67, 69 and 71.
10There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.
With 37 terms, the middle term is t19, so the three middle terms are t18, t19, t20.
t18 + t19 + t20 = (a + 17d) + (a + 18d) + (a + 19d) = 3a + 54d = 225
∴ a + 18d = 75 … (I)
The last three terms are t35, t36, t37: 3a + 105d = 429
∴ a + 35d = 143 … (II)
Subtracting (I) from (II):
17d = 68 ∴ d = 4
From (I): a = 75 − 72 = 3
The A.P. is 3, 7, 11, 15, …, 147
11If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is (a + c)(b + c − 2a) / 2(b − a).
First term = a, common difference d = b − a, last term tn = c
c = a + (n − 1)(b − a)
∴ n − 1 = c − ab − a
∴ n = c − ab − a + 1 = b + c − 2ab − a
Sn = n2[first term + last term] = n2(a + c)
= (a + c)(b + c − 2a)2(b − a)
Hence proved.
12If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)
Sp = Sq
p2[2a + (p − 1)d] = q2[2a + (q − 1)d]
∴ 2ap + p(p − 1)d = 2aq + q(q − 1)d
∴ 2a(p − q) + d[p2 − q2 − (p − q)] = 0
∴ (p − q)[2a + (p + q − 1)d] = 0
Since p ≠ q: 2a + (p + q − 1)d = 0
Sp+q = p + q2[2a + (p + q − 1)d] = p + q2 × 0 = 0
Hence the sum of the first (p + q) terms is zero.
13If m times the mth term of an A.P. is equal to n times nth term then show that the (m + n)th term of the A.P. is zero.
m × tm = n × tn
m[a + (m − 1)d] = n[a + (n − 1)d]
∴ a(m − n) + d[m2 − n2 − (m − n)] = 0
∴ (m − n)[a + (m + n − 1)d] = 0
For different terms m ≠ n, so a + (m + n − 1)d = 0
But tm+n = a + (m + n − 1)d ∴ tm+n = 0
Hence the (m + n)th term is zero.
14₹ 1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. Complete the following activity.
Simple interest = P × R × N100
Simple interest after 1 year = 1000 × 10 × 1100 = 100
Simple interest after 2 years = 1000 × 10 × 2100 = 200
Simple interest after 3 years = 1000 × 10 × 3100 = 300
The simple interest for 4, 5, 6 years will be 400, 500, 600 respectively.
The interests 100, 200, 300, … form an A.P. From this d = 100 and a = 100.
tn = a + (n − 1)d
t20 = 100 + (20 − 1) × 100
t20 = 2000
Simple interest after 20 years = ₹ 2000
