Q1. How many twodigit numbers are divisible by 7?
Sol. Let 'n' two digit numbers are divisible by 7.
So, the numbers are: 14, 21, 28, 35, 42, 49 ....................98.
This series form an A.P.
∴ a = 14,d = 7,and an = 98
Now, using an = a + (n  1) d, we get
98 = 14 + (n  1) 7
⇒ 98 = 14 + 7n  7
⇒ 98 = 7 + 7n or 7n
= 98  7 = 91 ⇒ n = 91/7 = 13
Thus, required twodigit numbers is 13
Q2. If the numbers x  2, 4x  1 and 5x + 2 are in A.P. Find the value of x.
Sol. ∵ x  2, 4x  1 and 5x + 2 are in A.P.
∴ (4x  1)  (x  2) = (5x + 2)  (4x  1)
⇒ 3x + 1 = x + 3
⇒ 2x = 2 ⇒ x = 1
Q3. Which term of the A.P. 4, 9, 14, ..... is 109?
Sol. Let 109 is the nth term,
∴ Using T_{n} = a + (n  1) d, we have:
109 = 4 + (n  1) 5
[∵ a = 4 and d = 9  4 = 5]
⇒
⇒ n = 21 + 1 = 22
Thus, the 22nd term is 109.
Q4. If a, (a  2) and 3a are in A.P. then what is the value of a?
Sol. ∵ a, (a  2) and 3a are in A.P.
∴ (a  2)  a = 3a  (a  2)
⇒ a  2  a = 3a  a + 2
⇒  2 = 2a + 2
⇒ 2a =  2  2 =  4
⇒ a = 4/2 = 2
Thus, the required value of a is  2.
Q5. How many terms are there in the A.P.?
7, 10, 13, ....., 151
Sol. Here, a = 7, d = 10  7 = 3
Let there are nterms.
∴ T_{n} = a + (n  1) d
⇒ T_{51} = 7 + (n  1) × 3
⇒
⇒ 144/3 = n − 1 ⇒ n = 48 + 1 = 49
i.e., n = 49
Q6. Which term of the A.P. 72, 63, 54, ..... is 0?
Sol. Here, a = 72
d = 63  72 =  9
Let nth term of this A.P. be 0
∴ T_{n} = a + (n  1) d
⇒ 72 + (n  1) × ( 9) = 0
⇒ (n  1) = 72/9 = 8
⇒ n = 8 + 1 = 9
Thus the 9th term of the A.P. is 0.
Q7. The first term of an A.P. is 6 and its common difference is  2. Find its 18th term.
Sol. Using T_{n} = a + (n  1) d, we have:
T_{18} = 6 + (18  1) × ( 2)
= 6 + 17 × ( 2)
= 6  34 =  28
Thus, the 18th term is  28.
Q8. The 4th term of an A.P. is 14 and its 12th term 70. What is its first term
Sol. Let the first term = a
If ‘d’ is the common difference,
Then T_{4} = a + 3d = 14 ...(1)
And T_{12} = a + 11d = 70 ...(2)
Subtracting (1) from (2),
a + 11d  a  3d = 70  14
⇒ 8d = 56 ⇒ d = 56/8 = 7
∴ From (1), a + 3 (7) = 14
⇒ a + 21 = 14
⇒ a = 14  21 = ( 7)
Thus, the first term is  7.
Q9. Which term of A.P. 5, 2,  1,  4 ..... is  40?
Sol. Here, a = 5
d = 2  5 =  3
Let nth term be  40
∴ T_{n} = a + (n  1) d
⇒  40 = 5 + (n  1) × ( 3)
⇒ n = 15 + 1 = 16 i.e.,
i.e., The 16th term of the A.P. is  40.
Q10. What is the sum of all the natural numbers from 1 to 100?
Sol.
We have:
1, 2, 3, 4, ....., 100 are in an A.P. such that
a = 1 and l = 100
∴ S_{n} = n/2 [a + l]
⇒ S_{100} =100/2 [1 + 100]
= 50 × 101 = 5050.
Q11. For an A.P., the 8th term is 17 and the 14th term is 29. Find its common difference.
Sol. Let the common difference = d and first term = a
∴ T_{8} = a + 7d = 17 ...(1)
T_{14} = a + 13d = 29 ...(2)
Subtracting (1) from (2), we have:
a + 13d  a  7d = 29  17
⇒ 6d = 12
⇒ d = 12/6 = 2
∴ The required common difference = 2.
Q12. If the first and last terms of an A.P. are 10 and  10. How many terms are there? Given that d =  1.
Sol. Let the required number of terms is n and
1st term a = 10
nth term T_{n} =  10
Let common difference be d then using,
T_{n} = a + (n  1) d, we have:
 10 = 10 + (n  1) × ( 1)
⇒  10 = 10  n + 1
⇒  n + 1 =  10  10 =  20
⇒  n =  20  1 =  21
⇒ n = 21
13. The nth term of an A.P. is (3n  2) find its first term.
Sol. ∵ T_{n} = 3n  2
∴ T_{1} = 3 (1)  2 = 3  2 = 1
⇒ First term = 1
Q14. The nth term of an A.P. is (2n  3) find the common difference.
Sol. Here, T_{n} = 2n  3
∴ T_{1} = 2 (1)  3 =  1
T_{2} = 2 (2)  3 = 1
∴ d = T_{2}  T_{1} = 1  ( 1) = 2
Thus the common difference is 2.
Q15. If the nth term of an A.P. is (7n  5). Find its 100th term.
Sol. Here, T_{n} = 7n  5
∴ T_{1} = 7 (1)  5 = 2
T_{2} = 7 (2)  5 = 9
∴ a = 2
and d = T_{2}  T_{1}
= 9  2 = 7
Now T_{100} = 2 + (100  1) 7
[using T_{n} = a + (n  1) d]
= 2 + 99 × 7
= 2 + 693 = 695.
Q16. Find the sum of first 12 terms of the A.P. 5, 8, 11, 14, ...... .
Sol. Here, a = 5
d = 8  5 = 3
n = 12
Using S_{n} = n/2 [2 (a) + (n  1) d]
we have: S_{12} = 12/2 [2 (5) + (12  1) × 3]
= 6 [10 + 33]
= 6 × 43 = 258
Q17. Write the common difference of an A.P. whose nth term is 3n + 5.
Sol. T_{n} = 3n + 5
∴ T_{1} = 3 (1) + 5 = 8
T_{2} = 3 (2) + 5 = 11
⇒ d = T_{2}  T_{1}
= 11  8 = 3
Thus, the common difference = 3.
Q18. Write the value of x for which x + 2, 2x, 2x + 3 are three consecutive terms of an A.P.
Sol. Here,
T_{1} = x + 2
T_{2} = 2x
T_{3} = 2x + 3
For an A.P., we have:
∴ 2x  (x + 2) = 2x + 3  2x
⇒ 2x  x  2 = 2x + 3  2x
⇒ x  2 = 3
⇒ x = 3 + 2 = 5
Thus, x = 5
Q19. What is the common difference of an A.P. whose nth term is 3 + 5n?
Sol. ∵ T_{n} = 3 + 5_{n}
∴ T_{1} = 3 + 5 (1) = 8
And T_{2} = 3 + 5 (2) = 13
∵ d = T_{2}  T_{1}
∴ d = 13  8 = 5
Thus, common difference = 5.
Q20. For what value of k, are the numbers x, (2x + k) and (3x + 6) three consecutive terms of an A.P.?
Sol. Here, T_{1} = x, T_{2} = (2x + k) and T_{3} = (3x + 6)
For an A.P., we have
T_{2}  T_{1} = T_{3}  T_{2}
i.e., 2x + k  x = 3x + 6  (2x + k)
⇒ x + k = 3x + 6  2x  k
⇒ x + k = x + 6  k
⇒ k + k = x + 6  x
⇒ 2k = 6
⇒ k = 6/2 = 3
Q21. For what value of k, will k + 9, 2k  1 and 2k + 7 are consecutive terms of an A.P.?
Sol. ∵ T_{1} = k + 9, T_{2} = 2k  1 and T_{3} = 2k + 7
For an A.P., T_{2}  T_{1} = T_{3}  T_{2}
∴ (2k  1)  (k + 9) = (2k + 7)  (2k  1)
⇒ 2k  1  k  9 = 2k + 7  2k + 1
⇒ k  10 = 8
⇒ k = 8 + 10
⇒ k = 18
Q22. If 4/5, a, 2 are three consecutive terms of an A.P., then find the value of a?
Sol. Here,
T_{1} = 4/5
T_{2} = a
T_{3} = 2
∵ For an A.P.,
T_{2}  T_{1} = T_{3}  T_{2}
∴
⇒
⇒ 2a = 14/5
⇒
Thus, a = 7/5
Q23. For what value of p are 2p  1, 7 and 3p three consecutive terms of an A.P.?
Sol. Here,
T_{1 }= 2p  1
T_{2} = 7
T_{3} = 3p
∵ For an A.P., we have:
T_{2}  T_{1} = T_{3}  T_{2}
⇒ 7  (2p  1) = 3p  7
⇒ 7  2p + 1 = 3p  7
⇒  2p  3p =  7  1  7
⇒  5p =  15
⇒ p = 15/5 = 3
Thus, p = 3
Q24. For what value of p are 2p + 1, 13 and 5p  3 three consecutive terms of an A.P.?
Sol. Here,
T_{1 }= 2p + 1
T_{2} = 13
T_{3} = 5p  3
For an A.P., we have:
T_{2}  T_{1} = T_{3}  T_{2}
⇒ 13  (2p + 1) = 5p  3  13
⇒ 13  2p  1 = 5p  16
⇒  2p + 12 = 5p  16
⇒  2p  5p =  16  12 =  28
⇒  7p =  28
⇒
∴ p = 4
Q25. The nth term of an A.P. is 7  4n. Find its common difference.
Sol. ∵ T_{n} = 7  4n
∴ T_{1} = 7  4 (1) = 3
T_{2} = 7  4 (2) =  1
∴ d = T_{2}  T_{1}
= ( 1)  3 =  4
Thus, common difference =  4
Q26. The nth term of an A.P. is 6n + 2. Find the common difference.
Sol. Here, T_{n} = 6n + 2
∴ T_{1} = 6 (1) + 2 = 8
T_{2} = 6 (2) + 2 = 14
⇒ d = T_{2}  T_{1} = 14  8 = 6
∴ Common difference = 6.
Q27. Write the next term of the A.P. √8, √18 √32 ......
Sol.
Here,
∴
Now, d = T_{2} − T_{1}
∴ T_{4} = a + 3d
Thus, the next term of the A.P. is 5√2 or
Q28. The first term of an A.P. is p and its common difference is q. Find the 10th term.
Sol. Here, a = p and d = q
∵ T_{n} = a + (n  1) d
∴ T_{10} = p + (10  1) q
= p + 9q
Thus, the 10th term is p + 9q.
Q29. (a) Find the next term of the A.P.
(b) Find the tenth term of the sequence
Sol. (a) Here,
Now, d = T_{2} − T_{1}
Now, using T_{n} = a + (n  1) × d, we have
T_{4} = a + 3d
Thus, the next term = √32.
(b) Here, T_{10} = a + 9d
Q30. Which term of the A.P.:
21, 18, 15, ..... is zero?
Sol. Here, a = 21
d = 18  21 =  3
Since Tn = a + (n  1) d
⇒ 0 = 21 + (n  1) × ( 3)
⇒  3 (n  1) =  21
⇒ (n  1) = 21/3 = 7
⇒ n = 7 + 1 = 8
Thus, the 8th term of this A.P. will be 0.
Q31. Which term of the A.P.:
14, 11, 8, ..... is  1?
Sol. Here, a = 14
d = 11  14 =  3
Let the nth term be ( 1)
∴ Using T_{n} = a + (n  1) d, we get
 1 = 11 + (n  1) × ( 3)
⇒  1  14 =  3 (n  1)
⇒  15 =  3 (n  1)
∴ n  1 = 15/3 = 5
⇒ n = 5 + 1 = 6
Thus, 1 is the 6th term of the A.P.
Q32. The value of the middlemost term (s) of the AP : –11, –7, –3, ...49.
Sol. ∵ a = –11, an = 49 and d = (–7) – (–11) = 4
∴ a_{n} = a + (n – 1)d
⇒ 49 = –11 + (n – 1) × 4
⇒ n = 16
Since, n is an even number
∴ There will be two middle terms, which are:
or 8th and 9th
Now, a_{8} = a + (8 – 1)d
= –11 + 7 × 4 = 17
a_{9} = a + (9 – 1)d
= –11 + 8 × 4 = 21
Thus, the values of the two middlemost terms are 17 and 21.
118 videos463 docs105 tests

1. What is an arithmetic progression? 
2. How is the nth term of an arithmetic progression calculated? 
3. Can an arithmetic progression have negative terms? 
4. How can we find the sum of the first n terms of an arithmetic progression? 
5. Can the common difference of an arithmetic progression be zero? 
118 videos463 docs105 tests


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