1 Crore+ students have signed up on EduRev. Have you? 
Ques 1: Find the sum of the following APs:
(i) 2, 7, 12, ..., to 10 terms.
(ii)  37,  33,  29, ..., to 12 terms.
(iii) 0.6, 1.7, 2.8, ..., to 100 terms.
(iv) , ..., to 11 terms.
Sol:
Formula for Sum of an AP
(i) Here, a = 2
d = 7  2 = 5
n = 10
Since, S_{n} = n/2 [2a + (n  1) d]
∴ S_{10} = [2 × 2 + (10  1) × 5]
⇒ S_{10} = 5 [4 + 9 × 5]
⇒ S_{10 }= 5 [49] = 245
Thus, the sum of first 10 terms is 245.
(ii) We have:
a =  37
d =  33  ( 37) = 4
n = 12
∴ S_{n} = n/2 [2a + (n  1) d]
⇒ S_{12} = 12/2 [2 ( 37) + (12  1) × 4]
= 6 [ 74 + 11 × 4]
= 6 [ 74 + 44]
= 6 × [ 30] =  180
Thus, sum of first 12 terms = 180.
(iii) Here, a = 0.6
d = 1.7  0.6 = 1.1
n = 100
∴ S_{n} = n/2 [2a + (n  1) d]
S_{100} = 100/2 [2 (0.6) + (100  1) × 1.1]
= 50 [1.2 + 99 × 1.1]
= 50 [1.2 + 108.9]
= 50 [110.1]
= 5505
Thus, the required sum of first 100 terms is 5505.
(iv) Here,
a = 1/15
d =
n = 11
∴ S_{n} = n/2 [2a + (n  1) d]
S_{11} =
Thus, the required sum of first 11 terms = 33/20.
Ques 2: Find the sums given below:
(i)
(ii) 34 + 32 + 30 + ... + 10
(iii)  5 + ( 8) + ( 11) + ... + ( 230)
Sol: (i) Here, a = 7
l = 84
Let n be the number of terms
∴ T_{n} = a + (n  1) d
⇒
⇒
⇒ n = 22 + 1 = 23
Now,
⇒
Thus, the required sum =
(ii) Here,a = 34
d = 32  34 =  2
l = 10
Let the number of terms be n
∴ T_{n} = 10
Now T_{n} = a + (n  1) d
⇒ 10 = 34 + (n  1) × ( 2)
⇒ (n  1) × ( 2) = 10  34 =  24
⇒
⇒ n = 13
⇒
Now,
⇒
=
=
=
= 13 × 22 = 286
OR
S_{13} = n/2 (a + l)
Thus, the required sum is 286.
(iii) Here, a =  5
d =  8  ( 5) =  3
l =  230
Let n be the number of terms.
∴ T_{n} =  230
⇒  230 =  5 + (n  1) × ( 3)
⇒ (n  1) × ( 3) =  230 + 5 =  225
⇒ n  1 = 225/3 = 75
⇒ n = 75 + 1 = 76
Now,
= 38 × ( 235)
=  8930
∴ The required sum =  8930.
Ques 3: In an AP:
(i) given a = 5, d = 3, a_{n} = 50, find n and S_{n}.
(ii) given a = 7, a_{13} = 35, find d and S_{13}.
(iii) given a_{12} = 37, d = 3, find a and S_{12}.
(iv) given a_{3} = 15, S_{10} = 125, find d and a_{10}.
(v) given d = 5, S_{9} = 75, find a and a_{9}.
(vi) given a = 2, d = 8, S_{n} = 90, find n and a_{n}.
(vii) given a = 8, a_{n} = 62, S_{n} = 210, find n and d.
(viii) given a_{n} = 4, d = 2, S_{n} =  14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
Sol: (i) Here, a = 5, d = 3 and an = 50 = l
∵ a_{n} = a + (n  1) d
∴ 50 = 5 + (n  1) × 3
⇒ 50  5 = (n  1) × 3
⇒ (n  1) × 3 = 45
⇒ (n  1) = 45/3 =15
⇒ n = 15 + 1 = 16
Now S_{n} = n/2 (a + l)
= 16/2 (5 + 50)
= 8 (55) = 440
Thus, n = 16 and S_{n} = 440
(ii) Here, a = 7 and a_{13} = 35 = l
∴ a_{n} = a + (n  1) d
⇒ 35 = 7 + (13  1) d
⇒ 35  7 = 12d
⇒ 28 = 12d
⇒ d = 28/12 = 7/3
Now, using
S_{n} = n/2 (a + l)
S_{13} = 13/2 (7 + 35)
S_{n} = 273 and d = 7/3
(iii) Here, a_{12} = 37 = l and d = 3
Let the first term of the AP be ‘a’.
Now a_{12} = a + (12  1) d
⇒ 37 = a + 11d
⇒ 37 = a + 11 × 3
⇒ 37 = a + 33
⇒ a = 37  33 = 4
Now, S_{n} = n/2 (a + l)
⇒ S_{12} = 12/2 (4 + 37)
⇒ S_{12} = 6 × (41) = 246
Thus, a = 4 and S_{12} = 246
(iv) Here, a_{3} = 15 = l
S_{10} = 125
Let first term of the AP be ‘a’ and the common difference = d
∴ a_{3} = a + 2d
⇒ a + 2d = 15 ...(1)
Again S_{n} = n/2 [2a + (n  1) d]
⇒ S_{10} = 10/2 [2a + (10  1) d]
⇒ 125 = 5 [2a + 9d]
⇒ 2a + 9d = 125/5 = 25
⇒ 2a + 9d = 25 ...(2)
Multiplying (1) by 2 and subtracting (2) from it, we get
2 [a + 2d = 15]  [2a + 9d = 25]
⇒ 2a + 4d  2a  9d = 30  25
⇒  5d = 5
⇒ d =5/5 = 1
∴ From (1), a + 2 ( 1) = 15 ⇒ a = 15 + 2 ⇒ a = 17
Now, a_{10} = a + (10  1) d
= 17 + 9 × ( 1)
= 17  9 = 8
Thus, d =  1 and a_{10} = 8
(v) Here, d = 5, S_{9} = 75
Let the first term of the AP is ‘a’.
∴ S_{9} = 9/2 [2a + (9  1) × 5]
⇒ 75 = 9/2 [2a + 40]
⇒
⇒ 50/3 = 2a + 40
⇒
⇒
Now, a_{9} = a + (9  1) d
=
=
Thus,
(vi) Here, a = 2, d = 8 and S_{n} = 90
∵ S_{n} = n/2 [2a + (n  1) d]
∴ 90 = n/2 [2 × 2 + (n  1) × 8]
⇒ 90 × 2 = 4n + n (n  1) × 8
⇒ 180 = 4n + 8n^{2}  8n
⇒ 180 = 8n^{2}  4n
⇒ 45 = 2n^{2}  n
⇒ 2n^{2}  n  45 = 0
⇒ 2n^{2}  10n + 9n  45 = 0
⇒ 2n (n  5) + 9 (n  5) = 0
⇒ (2n + 9) (n  5) = 0
∴ Either
or n  5 = 0 ⇒ n = 5
But is not required. ∴n = 5
Now, a_{n} = a + (n  1) d
⇒ a_{5} = 2 + (5  1) × 8
= 2 + 32 = 34
Thus, n = 5 and a_{5} = 34.
(vii) Here, a = 8, a_{n} = 62 = l and S_{n} = 210
Let the common difference = d
Now, S_{n} = 210
⇒ 210 = n/2 (a + l)
⇒ 210 = n/2 (8 + 62) =
∴ n = 210/35 = 6
Again a_{n} = a + (n  1) d
⇒ 62 = 8 + (6  1) × d
⇒ 62  8 = 5d
⇒ 54 = 5d ⇒ d = 54/5
Thus, n = 6 and d = 54/5 .
(viii) Here, a_{n} = 4, d = 2 and S_{n} =  14
Let the first term be ‘a’.
∵ a_{n} = 4
∴ a + (n  1) 2 = 4
⇒ a + 2n  2 = 4
⇒ a = 4  2n + 2
⇒ a = 6  2n ...(1)
Also S_{n} =  14
⇒ n/2 (a + l) =  14
⇒ n/2 (a + 4) =  14
⇒ n (a + 4) =  28 ...(2)
Substituting the value of a from (1) into (2),
n [6  2n + 4] =  28
⇒ n [10  2n] =  28
⇒ 2n [5  n] =  28
⇒ n (5  n) =  14 [Dividing throughout by 2]
⇒ 5n  n^{2} + 14 = 0
⇒ n^{2}  5n  14 = 0
⇒ n^{2}  7n + 2n  14 = 0
⇒ n (n  7) + 2 (n  7) = 0
⇒ (n  7) (n + 2) = 0
∴ Either n  7 = 0 ⇒ n = 7
or n + 2 = 0 ⇒ n =  2
But n cannot be negative,
∴ n = 7
Now, from (1), we have
a = 6  2 × 7 ⇒ a =  8
Thus, a =  8 and n = 7
(ix) Here, a = 3, n = 8 and S_{n} = 192
Let the common difference = d.
∵ S_{n} = n/2 [2a + (n  1) d]
∴ 192 = 8/2 [2 (3) + (8  1) d]
⇒ 192 = 4 [6 + 7d]
⇒ 192 = 24 + 28d
⇒ 28d = 192  24 = 168
⇒ d = 168/28 =6
Thus, d = 6.
(x) Here, l = 28 and S_{9} = 144
Let the first term be ‘a’.
Then S_{n} = n/2 (a + l)
⇒ S_{9} = 9/2 (a + 28)
⇒ 144 = 9/2 (a + 28)
⇒
⇒ a = 32  28 = 4
Thus, a = 4.
Ques 4: How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?
Sol: Here, a = 9
d = 17  9 = 8
S_{n} = 636
∵ S_{n} = n/2 [2a + (n  1) d] = 636
∴ n/2 [(2 × 9) + (n  1) × 8] = 636
⇒ n [18 + (n  1) × 8] = 1272
⇒ n (8n + 10) = 1272
⇒ 8n^{2} + 10n  1272 = 0
⇒ 4n^{2} + 5n  636 = 0
⇒ 4n^{2} + 53n  48n  636 = 0
⇒ n (4n + 53)  12 (4n + 53) = 0
⇒ (n  12) (4n + 53) = 0
⇒ n = 12 and
Rejecting , we have n = 12.
Ques 5: The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Sol: Here, a = 5
l = 45 = T_{n}
S_{n} = 400
∵ T_{n} = a + (n  1) d
∴ 45 = 5 + (n  1) d
⇒ (n  1) d = 45 – 5
⇒ (n  1) d = 40 ...(1)
Also S_{n }= n/2 (a + l)
⇒ 400 = n/2 (5 + 45)
⇒ 400 × 2 = n × 50
⇒
From (1), we get
(16  1) d = 40
⇒ 15d = 40
⇒ d = 40/15 = 8/3
53 videos403 docs138 tests
