Updated on: 2026-03-31 | Author: Aarti Kulkarni

Algebra (Math 1) Chapter 3 Arithmetic Progression Solutions

Q1. 1, 6, 11, 16 ...... Find the 18 th term of this A.P.

The given A.P. is 1, 6, 11, 16, ......

Here, a = 1, d = 6 – 1 = 5
Since tn = a + (n – 1)d,
t18 = 1 + (18 – 1)(5)

= 1 + 17(5)

= 1 + 85

= 86

∴ 18th term of the given A.P. is 86.

Q2. 1, 7, 13, 19 ...... find 18 th term of this A.P.

The given A.P. is 1, 7, 13, 19, ......

Here, a = 1, d = 7 – 1 = 6
Since tn = a + (n – 1)d,
t18 = 1 + (18 – 1)(6)

= 1 + 17(6)

= 1 + 102

= 103

∴ 18th term of the given A.P. is 103.

Q3. 12, 16, 20, 24, ...... Find 25 th term of this A.P.

The given A.P. is 12, 16, 20, 24, ......

Here, a = 12, d = 16 – 12 = 4
Since tn = a + (n – 1)d
t25 = 12 + (25 – 1)(4)

= 12 + 24(4)

= 12 + 96

= 108

∴ 25th term of the given A.P. is 108.

Q4. A merchant borrows ₹ 1000 and agrees to repay its interest ₹ 140 with principal in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 10. Find the amount of the first instalment

The installments are in A.P.

Amount repaid in 12 instalments (S12)

= Amount borrowed + total interest

= 1000 + 140

∴ S12 = 1140
Number of instalments (n) = 12

Each instalment is less than the preceding one by ₹ 10.

∴ d = – 10

Now, Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`

∴ S12 = `12/2 [2"a" + (12 - 1)(- 10)]`
∴ 1140 = 6[2a + 11(– 10)]
∴ 1140 = 6(2a – 110)
∴ `1140/6` = 2a – 110
∴ 190 = 2a – 110
∴ 2a = 300
∴ a = `300/2` = 150
∴ The amount of first instalment is ₹ 150.

Q5. Choose the correct alternative answer for the following sub question 1, 4, 7, 10, 13, ... Next two terms of this A.P. are ______

16, 19

Q6. Choose the correct alternative answer for the following sub question Find d of an A.P. whose first two terms are – 3 and 4

7

Q7. Choose the correct alternative answer for the following sub-question Find t 3 = ? in an A.P. 9, 15, 21, 27, ...

21

Explanation:

A.P. 9, 15, 21, 27, ...(Given)

t1 = 9, t2 = 15, t3 = 21, t4 = 27
∴ t3 = 21

Q8. Choose the correct alternative answer for the following sub question For an A.P. 5, 12, 19, 26, … a = ?

5

Q9. Choose the correct alternative answer for the following sub-question If the third term and fifth term of an A.P. are 13 and 25 respectively, find its 7 th term

37

Q10. Choose the correct alternative answer for the following sub question In an A.P., 0, – 4, – 8, – 12, ... find t 2 = ?

– 4

Q11. Choose the correct alternative answer for the following sub question. In an Arithmetic Progression 2, 4, 6, 8, ... the common difference d is ______

2

Q12. Choose the correct alternative answer for the following sub questions A set of numbers where the numbers are arranged in a definite order, like the natural numbers, is called a ______

sequence

Q13. Choose the correct alternative answer for the following sub question. What is the common difference of the sequence 0, – 4, – 8, – 12?

– 4

Q14. Common difference, d = ? for the given A.P., 7, 14, 21, 28 ........ Activity :- Here t 1 = 7, t 2 = 14, t 3 = 21, t 4 = `square` t 2 − t 1 = `square` t 3 – t 2 = 7 t 4 – t 3 = `square` Therefore, common difference d = `square`

Here t1 = 7, t2 = 14, t3 = 21, t4 = 28
t2 − t1 = 14 – 7 = 7
t3 – t2 = 7
t4 – t3 = 28 – 21 = 7
Therefore, common difference d = 7

Q15. Decide whether 301 is term of given sequence 5, 11, 17, 23, ..... Activity :- Here, d = `square`, therefore this sequence is an A.P. a = 5, d = `square` Let n th term of this A.P. be 301 t n = a + (n – 1) `square` 301 = 5 + (n – 1) × `square` 301 = 6n – 1 n = `302/6` = `square` But n is not positive integer. Therefore, 301 is `square` the term of sequence 5, 11, 17, 23, ......

Here, d = 11 – 5 = 6, therefore this sequence is an A.P.
a = 5, d = 6

Let nth term of this A.P. be 301

tn = a + (n – 1) d
∴ 301 = 5 + (n – 1) × 6
∴ 301 = 5 + 6n – 6
∴ 301 = 6n – 1
∴ 6n = 302
∴ n = `302/6` = `151/3`

But n is not positive integer.

Therefore, 301 is not the term of sequence 5, 11, 17, 23, ......

Q16. Decide whether the following sequence is an A.P. or not. 3, 5, 7, 9, 11 ........

The given sequence is 3, 5, 7, 9, 11 ........

Here, t1 = 3, t2 = 5, t3 = 7, t4 = 9
∴ t2 – t1 = 5 – 3 = 2
t3 – t2 = 7 – 5 = 2
t4 – t3 = 9 – 7 = 2
∴ t2 – t1 = t3 – t2 = 2 = d = constant

The difference between two consecutive terms is constant.

∴ The given sequence is an A.P.

Q17. Decide whether the given sequence 2, 4, 6, 8,… is an A.P.

Here, t1 = 2, t2 = 4, t3 = 6
∴ t2 − t1 = 4 − 2 = 2
t3 − t2 = 6 − 4 = 2
∴ t2 − t1 = t3 − t2 = ... = 2 = constant

The difference between two consecutive terms is constant.

∴ The given sequence is an A.P.

Q18. Decide whether the given sequence 24, 17, 10, 3, ...... is an A.P.? If yes find its common term (t n )

The given sequence is 24, 17, 10, 3, ......

Here, t1 = 24, t2 = 17, t3 = 10, t4 = 3
∴ t2 – t1 = 17 – 24 = – 7
t3 – t2 = 10 – 17 = – 7
t4 – t3 = 3 – 10 = – 7
∴ t2 – t1 = t3 – t2 = …= – 7 = d = constant

The difference between two consecutive terms is constant.

∴ The given sequence is an A.P.
tn = a + (n – 1)d

= 24 + (n – 1)(– 7)

= 24 – 7n + 7

= 31 – 7n

Q19. Determine the sum of first 100 terms of given A.P. 12, 14, 16, 18, 20, ...... Activity :- Here, a = 12, d = `square`, n = 100, S 100 = ? S n = `"n"/2 [square + ("n" - 1)"d"]` S 100 = `square/2 [24 + (100 - 1)"d"]` = `50(24 + square)` = `square` = `square`

Here, a = 12, d = 14 - 12 = 2, n = 100, S100 = ?

Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`

S100 = `100/2 [24 + (100 - 1)"d"]`

= 50[24 + 99(2)]

= 50(24 + 198)

= 50(222)

= 11100

Q20. Find 27 th and n th term of given A.P. 5, 2, – 1, – 4, ......

The given A.P. is 5, 2, – 1, – 4, ......

Here, a = 5, d = 2 – 5 = – 3
Since tn = a + (n – 1)d,
tn = 5 + (n – 1)(– 3)
∴ tn = 5 – 3n + 3
∴ tn = 8 – 3n
∴ t27 = 8 – 3(27)

= 8 – 81

= – 73

∴ nth term of the given A.P. is 8 – 3n and 27th term is – 73.

Q21. Find a and d for an A.P., 1, 4, 7, 10,.........

Here, t1 = 1, t2 = 4
∴ First term (a) = t1 = 1 and
Common difference (d) = t2 – t1

= 4 – 1

= 3

Q22. Find common difference of an A.P., 0.9, 0.6, 0.3 ......

Here, t1 = 0.9, t2 = 0.6
∴ Common difference = t2 – t1

= 0.6 – 0.9

= -0.3

Q23. Find d if t 9 = 23 व a = 7

t9 = 23 ......[Given]
∴ 7 + (9 – 1)d = 23 ......[tn = a + (n – 1)d]
∴ 7 + 8d = 23
∴ 8d = 16
∴ d = `16/8`

= 2

Q24. Find first four terms of an A.P., whose first term is 3 and common difference is 4.

First term (a) = 3, Common difference (d) = 4 ......[Given]
t1 = a = 3
t2 = t1 + d = 3 + 4 = 7
t3 = t2 + d = 7 + 4 = 11
t4 = t3 + d = 11 + 4 = 15
∴ The first four terms of A.P. are 3, 7, 11, 15.

Q25. Find first four terms of the sequence t n = n + 2

tn = n + 2 ......[Given]
t1 = 1 + 2 = 3
t2 = 2 + 2 = 4
t3 = 3 + 2 = 5
t4 = 4 + 2 = 6
∴ The first four terms of the sequence are 3, 4, 5, 6.

Q26. Find first term of the sequence t n = 2n + 1

tn = 2n + 1 ......[Given]
∴ t1 = 2(1) + 1

= 2 + 1

= 3

∴ The first term of the sequence is 3.

Q27. Find S 10 if a = 6 and d = 3

a = 6 and d = 3 ......[Given]

Since Sn = `"n"/2[2"a" + ("n" - 1)"d"]`,

S10 = `10/2[2(6) + (10 - 1)(3)]`

= 5[12 + 9(3)]

= 5(12 + 27)

= 5(39)

= 195

Q28. Find t 21 , if S 41 = 4510 in an A.P.

For an A.P., let a be the first term and d be the common difference.

S41 = 4510 ......[Given]

Since Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`,

S41 = `41/2 [2"a" + (41 - 1)"d"]`

∴ 4510 = `41/2 (2"a" + 40"d")`
∴ 4510 = `41/2 xx 2 ("a" + 20"d")`
∴ 4510 = 41(a + 20d)
∴ a + 20d = `4510/41`
∴ a + 20d = 110 .....(i)
Now, tn = a + (n – 1)d
∴ t21 = a + (21 – 1)d

= a + 20d

∴ t21 = 110 ......[From (i)]

Q29. Find t 5 if a = 3 आणि d = −3

tn = a + (n – 1)d
∴ t5 = 3 + (5 – 1)(– 3)

= 3 + 4(– 3)

= 3 – 12

= – 9

Q30. Find the first terms and common difference of an A.P. whose t 8 = 3 and t 12 = 52.

For an A.P., let a be the first term and d be the common difference.

t8 = 3 and t12 = 52 ......[Given]
Since tn = a + (n – 1)d,
t8 = a + (8 – 1)d
∴ 3 = a + 7d
i.e., a + 7d = 3 ......(i)
Also, t12 = 52
∴ a + (12 – 1)d = 52
∴ a + 11d = 52 ......(ii)

Subtracting equation (i) from (ii), we get

a + 11d = 52
a + 7d = 3

– – –

4d = 49
∴ d = `49/4`

Substituting d = `49/4` in equation (i), we get

`"a" + 7(49/4)` = 3

∴ `"a" + 343/4` = 3
∴ a = `3 - 343/4`

= `(-331)/4`

∴ The first term and common difference of A.P. are `(-331)/4` and `49/4` respectively.

Q31. Find the next 4 terms of the sequence `1/6, 1/4, 1/3`. Also find S n .

The given sequence is `1/6, 1/4, 1/3`

The above sequence is an A.P.

∴ a = `1/6`
d = `1/4 -1/6 = ((6 - 4)/(6 xx 4))= 2/24 = 1/12`

The next four terms of the sequence are

t4 = t3 + d = `1/3 + 1/12 = 5/12`
t5 = t4 + d = `5/12 + 1/12 = 6/12 = 1/2`
t6 = t5 + d = `1/2 + 1/12 = 7/12`
t7 = t6 + d = `7/12 + 1/12 = 8/12 = 2/3`

Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`

= `"n"/2 [2(1/6) + ("n" - 1)(1/12)]`

= `"n"/2 (1/3 + 1/12 "n" - 1/12)`

= `"n"/2("n"/12 + 1/4)`

∴ Sn = `("n"("n" + 3))/24`

Q32. Find the sum of first 1000 positive integers. Activity :- Let 1 + 2 + 3 + ........ + 1000 Using formula for the sum of first n terms of an A.P., S n = `square` S 1000 = `square/2 (1 + 1000)` = 500 × 1001 = `square` Therefore, Sum of the first 1000 positive integer is `square`

Let 1 + 2 + 3 + ........ + 1000

Using formula for the sum of first n terms of an A.P.,

Sn = `"n"/2 ("t"_1 + "t"_"n")`

S1000 = `1000/2 (1 + 1000)`

= 500 × 1001

= 500500

Therefore, Sum of the first 1000 positive integer is 500500.

Q33. Find the sum of natural numbers between 1 to 140, which are divisible by 4. Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136 Here d = 4, therefore this sequence is an A.P. a = 4, d = 4, t n = 136, S n = ? t n = a + (n – 1)d `square` = 4 + (n – 1) × 4 `square` = (n – 1) × 4 n = `square` Now, S n = `"n"/2["a" + "t"_"n"]` S n = 17 × `square` S n = `square` Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.

The numbers from 1 to 140 which are divisible by 4 are 4, 8, 12, ... 140

This sequence is an A.P. with a = 4, d = 8 – 4 = 4, tn = 140
But, tn = a + (n − 1)d
∴ 140 = 4 + (n − 1)4
∴ 140 - 4 = (n − 1)4
∴ 136 = (n − 1)4
∴ `136/4` = n − 1
∴ 34 + 1 = n
∴ n = 35

Now, Sn = `n/2` [2a + (n − 1)d]

∴ S35 = `35/2`[2 × 4 + (35 − 1)4]

= `35/2` [8 + (34)4]

= `35/2`[8 + 136]

= `35/2` × 144

= 35 × 72

∴ S35 = 2520
∴ The sum of all numbers from 1 to 140 which are divisible by 4 is 2520.

Q34. Find the sum of numbers between 1 to 140, divisible by 4

The numbers between 1 to 140 divisible by 4 are

4, 8, 12, ......, 140

The above sequence is an A.P.

∴ a = 4, d = 8 - 4 = 4

Let the number of terms in the A.P. be n.

Then, tn = 140
Since tn = a + (n – 1)d,
140 = 4 + (n – 1)(4)
∴ 140 - 4 = (n – 1)(4)
∴ 136 = (n – 1)(4)
∴ `136/4` = n - 1
∴ 34 + 1 = n
∴ n = 35

Now, Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`

∴ S35 = `35/2 [2xx4+(35-1)4]`

= `35/2[8+(34)4]`

= `35/2[8+136]`

= `35/2xx144`

= 35 × 72

S35 = 2520
∴ The sum of numbers between 1 to 140, which are divisible by 4 is 2520.

Q35. Find the sum of odd natural numbers from 1 to 101

The odd natural numbers from 1 to 101 are

1, 3, 5, ....., 101

The above sequence is an A.P.

∴ a = 1, d = 3 – 1 = 2

Let the number of terms in the A.P. be n.

Then, tn = 101
Since tn = a + (n – 1)d,
101 = 1 + (n – 1)(2)
∴ 101 = 1 + 2n – 2
∴ 101 = 2n – 1
∴ 102 = 2n
∴ n = `102/2` = 51

Now, Sn = `"n"/2 ("t"_1 + "t"_"n")`

∴ S51 = `51/2 (1 + 101)`

= `51/2 (102)`

= 51 × 51

= 2601

∴ The sum of odd natural numbers from 1 to 101 is 2601.

Q36. Find the sum of three-digit natural numbers, which are divisible by 4

The three-digit natural numbers divisible by 4 are

100, 104, 108, ......, 996

The above sequence is an A.P.

∴ a = 100, d = 104 – 100 = 4

Let the number of terms in the A.P. be n.

Then, tn = 996
Since tn = a + (n – 1)d,
996 = 100 + (n – 1)(4)
∴ 996 = 100 + 4n – 4
∴ 996 = 96 + 4n
∴ 996 – 96 = 4n
∴ 4n = 900
∴ n = `900/4` = 225

Now, Sn = `"n"/2 ("t"_1 + "t"_"n")`

∴ S225 = `225/2 (100 + 996)`

= `225/2 (1096)`

= 225 × 548

= 123300

∴ The sum of three digit natural numbers, which are divisible by 4 is 123300.

Q37. Find t n if a = 20 आणि d = 3

tn = a + (n – 1)d

= 20 + (n – 1)3

= 20 + 3n – 3

= 17 + 3n

Q38. Find two terms of the sequence t n = 3n – 2

tn = 3n – 2 ......[Given]
∴ t1 = 3(1) – 2

= 3 – 2

= 1

t2 = 3(2) – 2

= 6 – 2

= 4

∴ The first two terms of the sequence are 1 and 4.

Q39. First four terms of an A.P., are ______ whose first term is –2 and the common difference is –2.

First, four terms of an A.P., are – 2, – 4, – 6, – 8 whose first term is –2 and the common difference is –2.

Explanation:

Let the first four terms of an A.P are a, a + d, a + 2d and a + 3d

Given that the first term is – 2 and difference is also – 2, then the A.P would be:

– 2, (– 2 – 2), [– 2 + 2 (– 2)], [– 2 + 3(– 2)]

= – 2, – 4, – 6, – 8

Q40. For an A.P., If t 1 = 1 and t n = 149 then find S n . Activitry :- Here t 1 = 1, t n = 149, S n = ? S n = `"n"/2 (square + square)` = `"n"/2 xx square` = `square` n, where n = 75

Here, t1= 1, tn = 149, Sn = ?

Sn = `"n"/2 ("t"_1 + "t"_"n")`

= `"n"/2 (1 + 149)`

= `"n"/2 xx 150`

= 75 n, where n = 75

Q41. For an A.P., t 4 = 12 and its common difference d = – 10, then find t n

t4 = 12, d = – 10 .....[Given]
Since tn = a + (n – 1)d,
t4 = a + (4 – 1)(– 10)
∴ 12 = a + 3(– 10)
∴ 12 = a – 30
∴ a = 42
tn = a + (n – 1)d

= 42 + (n – 1)(– 10)

= 42 – 10n + 10

= 52 – 10n

Q42. How many two-digit numbers are divisible by 5? Activity :- Two-digit numbers divisible by 5 are, 10, 15, 20, ......, 95. Here, d = 5, therefore this sequence is an A.P. Here, a = 10, d = 5, t n = 95, n = ? t n = a + (n − 1) `square` `square` = 10 + (n – 1) × 5 `square` = (n – 1) × 5 `square` = (n – 1) Therefore n = `square` There are `square` two-digit numbers divisible by 5

Two-digit numbers divisible by 5 are, 10, 15, 20, ......, 95.

Here, d = 5, therefore this sequence is an A.P.
Here, a = 10, d = 5, tn = 95, n = ?
tn = a + (n − 1) d
∴ 95 = 10 + (n – 1) × 5
∴ 95 – 10 = (n – 1) × 5
∴ 85 = (n – 1) × 5
∴ `85/5` = (n – 1)
∴ 17 = (n – 1)
∴ n = 17 + 1
Therefore n = 18

There are 18 two-digit numbers divisible by 5

Q43. If ₹ 3900 will have to be repaid in 12 monthly instalments such that each instalment being more than the preceding one by ₹ 10, then find the amount of the first and last instalment

The instalments are in A.P.

Amount repaid in 12 instalments (S12) = 3900
Number of instalments (n) = 12

Each instalment is more than the preceding one by ₹ 10.

∴ d = 10

Now, Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`

∴ S12 = `12/2[2"a" + (12 - 1)(10)]`
∴ 3900 = 6[2a + 11(10)]
∴ 3900 = 6(2a + 110)
∴ `3900/6` = 2a + 110
∴ 650 = 2a + 110
∴ 2a = 540
∴ a = `540/2` = 270
tn = a + (n – 1)d
∴ t12 = 270 + (12 – 1)(10)

= 270 + 11(10)

= 270 + 110

= 380

∴ Amount of the first instalment is ₹ 270 and that of the last instalment is ₹ 380.

Q44. If a = 6 and d = 10, then find S 10

a = 6 and d = 10 ......[Given]

Since Sn = `"n"/2[2"a" + ("n" - 1)"d"]`,

S10 = `10/2 [2(6) + (10 - 1)(10)]`

= 5[12 + 9 (10)]

= 5(12 + 90)

= 5(102)

= 510

Q45. If t n = 2n – 5 is the nth term of an A.P., then find its first five terms

tn = 2n – 5 ......[Given]
∴ t1 = 2(1) – 5 = 2 – 5 = – 3
t2 = 2(2) – 5 = 4 – 5 = – 1
t3 = 2(3) – 5 = 6 – 5 = 1
t4 = 2(4) – 5 = 8 – 5 = 3
t5 = 2(5) – 5 = 10 – 5 = 5
∴ The first five terms of A.P. are – 3, – 1, 1, 3, 5.

Q46. In a ‘Mahila Bachat Gat’, Sharvari invested ₹ 2 on first day, ₹ 4 on second day and ₹ 6 on third day. If she saves like this, then what would be her total savings in the month of February 2010?

Amount invested by Sharvari in the month of February 2010 are as follows:

2, 4, 6, .....

The above sequence is an A.P

∴ a = 2, d = 4 – 2 = 2

Number of days in February 2010,

n = 28

Now, Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`

∴ S28 = `28/2 [2(2) + (28 - 1)(2)]`

= 14[4 + 27(2)]

= 14(4 + 54)

= 14(58)

= 812

∴ Total savings of Sharvari in the month of February 2010 is ₹ 812.

Q47. In an A.P., a = 10 and d = −3 then find its first four terms

a = 10, d = – 3 ......[Given]
t1 = a = 10
t2 = t1 + d = 10 – 3 = 7
t3 = t2 + d = 7 – 3 = 4
t4 = t3 + d = 4 – 3 = 1
∴ The first four terms of A.P. are 10, 7, 4, 1.

Q48. In an A.P. a = 2 and d = 3, then find S 12

a = 2 and d = 3 ......[Given]

Since Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`,

S12 = `12/2 [2(2) + (12 - 1)(3)]`

= 6[4 + 11(3)]

= 6(4 + 33)

= 6(37)

= 222

Q49. In an A.P. a = 4 and d = 0, then find first five terms

a = 4 and d = 0 ......[Given]
t1 = a = 4
t2 = t1 + d = 4 + 0 = 4
t3 = t2 + d = 4 + 0 = 4
t4 = t3 + d = 4 + 0 = 4
t5 = t4 + d = 4 + 0 = 4
∴ The first five terms of A.P. are 4, 4, 4, 4, 4.

Q50. In an A.P. t 10 = 57 and t 15 = 87, then find t 21

For an A.P., let a be the first term and d be the common difference.

t10 = 57 and t15 = 87 ......[Given]
Since tn = a + (n – 1)d
t10 = a + (10 – 1)d
∴ 57 = a + 9d
i.e., a + 9d = 57 ......(i)
Also, t15 = 87
∴ a + (15 – 1)d = 87
∴ a + 14d = 87 ......(ii)

Subtracting equation (i) from (ii), we get

a + 14d = 87
a + 9d = 57

– – –

5d = 30
∴ d = `30/5` = 6
Substituting d = 6 in equation (i), we get
a + 9(6) = 57
∴ a + 54 = 57
∴ a = 57 – 54 = 3
t21 = a + (21 – 1)d

= a + 20d

= 3 + 20(6)

= 3 + 120

∴ t21 = 123

Q51. Kalpana saves some amount every month. In the first three months, she saves ₹ 100, ₹ 150, and ₹ 200 respectively. In how many months will she save ₹ 1200? Activity :- Kalpana’s monthly saving is ₹ 100, ₹ 150, ₹ 200, ......, ₹ 1200 Here, d = 50. Therefore this sequence is an A.P. a = 100, d = 50, t n = `square`, n = ? t n = a + (n – 1) `square` `square` = 100 + (n – 1) × 50 `square/50` = n – 1 n = `square` Therefore, she saves ₹ 1200 in `square` months.

Kalpana’s monthly saving is ₹ 100, ₹ 150, ₹ 200, ......, ₹ 1200

Here, d = 50. Therefore this sequence is an A.P.
a = 100, d = 50, tn = 1200, n = ?
tn = a + (n – 1) d
∴ 1200 = 100 + (n – 1) × 50
∴ 1200 – 100 = (n – 1) × 50
∴ 1100 = (n – 1) × 50
∴ `1100/50` = n – 1
∴ 22 = n – 1
n = 23
Therefore, she saves ₹ 1200 in 23 months.

Q52. Merry got a job with salary ₹ 15000 per month. If her salary increases by ₹ 100 per month, how much would be her salary after 20 months?

Merry’s monthly salaries form an A.P.

With first term (a) = ₹ 15000 and common difference (d) = ₹ 100

Number of months (n) = 20
tn = a + (n – 1)d
∴ t20 = 15000 + (20 – 1)(100)

= 15000 + 19(100)

= 15000 + 1900

= 16900

∴ Merry’s salary after 20 months will be ₹ 16,900.

Q53. Shubhankar invested in a national savings certificate scheme. In the first year he invested ₹ 500, in the second year ₹ 700, in the third year ₹ 900 and so on. Find the total amount that he invested in 12 years

Amount invested by Shubhankar in the national savings certificate scheme is as follows:

500, 700, 900, ......

The above sequence is an A.P.

∴ a = 500, d = 700 – 500 = 200, n = 12

Now, Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`

∴ S12 = `12/2 [2(500) + (12 - 1)(200)]`

= 6[1000 + 11(200)]

= 6(1000 + 2200)

= 6(3200)

= 19200

∴ The total amount invested by Shubhankar is ₹ 19,200.

Q54. Sum of first 55 terms in an A.P. is 3300, find its 28 th term.

For an A.P., let a be the first term and d be the common difference.

S55 = 3300 ......[Given]

We know that,

Since Sn = `"n"/2`[2a + (n – 1)d]

∴ S55 = `55/2`[2a + (55 – 1) d]
∴ 3300 = `55/2`[2a + 54d]
∴ 3300 = `55/2` × 2[a + 27d]
∴ 3300 = 55[a + 27d]
∴ a + 27d = `3300/55`
∴ a + 27d = 60 ......(i)
Now, tn = a + (n – 1)d
∴ t28 = a + (28 – 1)d
∴ t28 = a + 27d
∴ t28 = 60 .......[From (i)]
∴ 28th term of A.P. is 60.

Q55. t 19 = ? for the given A.P., 9, 4, −1, −6 ........ Activity :- Here a = 9, d = `square` t n = a + (n − 1)d t 19 = 9 + (19 − 1) `square` = 9 + `square` = `square`

Here a = 9, d = 4 - 9 = − 5
tn = a + (n − 1)d
t19 = 9 + (19 − 1) − 5

= 9 + 18(-5)

= 9 + − 90

= − 81

Q56. The n th term of an A.P. 5, 8, 11, 14, ...... is 68. Find n = ?

The given A.P. is 5, 8, 11, 14, ......

Here, a = 5, d = 8 – 5 = 3, tn = 68
Since tn = a + (n – 1)d,
68 = 5 + (n – 1)(3)
∴ 68 = 5 + 3n – 3
∴ 68 = 2 + 3n
∴ 66 = 3n
∴ n = `66/3`
∴ n = 22

Q57. t n = 2n − 5 in a sequence, find its first two terms

tn = 2n − 5 .......[Given]
∴ t1 = 2(1) – 5

= 2 – 5

= – 3

t2 = 2(2) – 5

= 4 – 5

= – 1

∴ The first two terms of the sequence are – 3 and – 1.

Q58. What is the sum of an odd numbers between 1 to 50?

The odd numbers between 1 to 50 are

1, 3, 5, ......, 49

The above sequence is an A.P.

∴ a = 1
∴ d = 5 – 3 = 2

Let the number of terms in the A.P. be n.

Then, tn = 49
Since tn = a + (n – 1)d,
49 = 1 + (n – 1)(2)
∴ 49 = 1 + 2n – 2
∴ 49 = 2n - 1
∴ 49 + 1 = 2n
∴ n =`50/2`
∴ n = 25

Now, Sn = `"n"/2 ("t"_1 + "t"_"n")`

∴ S25 = `25/2 (1 + 49)`

= 12.5 × 50

= 625

∴ The sum of odd numbers between 1 to 50 is 625.

Q59. Which term of following A.P. is −940. 50, 40, 30, 20 ........ Activity :- Here a = `square`, d = `square`, t n = −940 According to formula, t n = a + (n − 1)d −940 = `square` n = `square`

Here a = 50, d = 40 − 50 = −10, tn = −940
According to formula, tn = a + (n − 1)d
−940 = 50 + (n − 1)(−10)
∴ – 940 = 50 – 10n + 10
∴ – 940 = 60 – 10n
∴ 10n = 60 + 940
∴ 10n = 1000
∴ n = `1000/10`
n = 100

Q60. Write the formula of the sum of first n terms for an A.P.

Answer

Sum of the first n terms is given by

Sn = `"n"/2[2"a" + ("n" - 1)"d"]`,

where a = first term,
d = common difference

OR

Sn = `"n"/2("t"_1 + "t"_"n")`,

where t1 = first term,
tn = last term

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