Sequences and Series
- The sum of 10 terms of the arithmetic series is 390. If the third term of the series is 19, find the first term
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Here , S10 = 390 and first term ( a ) = ?
Let the first term of A.P. be ‘a’ and the common difference be ‘d’.∴ Sn = n [2a + (n - 1)d] 2 ⇒ 390 = 10 [2a + (10 - 1)d] 2
⇒ 390 = 5 (2a + 9d)⇒ 2a + 9d = 390 = 78 .....(i) 5
Again, third term = 19
[ ∴ Tn = a + (n – 1)d]
⇒ a + 2d = 19 .... (ii)
By equation (i) – 2 × (ii),
Correct Option: A
Here , S10 = 390 and first term ( a ) = ?
Let the first term of A.P. be ‘a’ and the common difference be ‘d’.∴ Sn = n [2a + (n - 1)d] 2 ⇒ 390 = 10 [2a + (10 - 1)d] 2
⇒ 390 = 5 (2a + 9d)⇒ 2a + 9d = 390 = 78 .....(i) 5
Again, third term = 19
[ ∴ Tn = a + (n – 1)d]
⇒ a + 2d = 19 .... (ii)
By equation (i) – 2 × (ii),
2a + 9d – 2a – 4d = 78 – 38
⇒ 5d = 40⇒ d = 40 = 8 5
From equation (ii),
a + 2 × 8 = 19
⇒ a = 19 – 16 = 3
- Given 2² + 4² + 6² + .......+ 40² = 11480, then the value of 1² + 2² + 3² + ......+20² is :
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Given in question , 2² + 4² + 6² + .... + 40² = 11480
⇒ ( 2 . 1 )² + ( 2 . 2 )² + ( 2 . 3 )² + .... + ( 2 . 20 )² = 11480
⇒ 1².2² + 2².2² + 3².2² + ...+ 20².2² = 11480
⇒ 2² (1² + 2² + 3² + .... + 20²) = 11480
⇒ 4 [ 1² + 2² + 3² + .... + 20² ] = 11480Correct Option: A
Given in question , 2² + 4² + 6² + .... + 40² = 11480
⇒ ( 2 . 1 )² + ( 2 . 2 )² + ( 2 . 3 )² + .... + ( 2 . 20 )² = 11480
⇒ 1².2² + 2².2² + 3².2² + ...+ 20².2² = 11480
⇒ 2² (1² + 2² + 3² + .... + 20²) = 11480
⇒ 4 [ 1² + 2² + 3² + .... + 20² ] = 11480Required answer = 11480 = 2870 4
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If 1² + 2² + 3² + ..... + p² = p(p + 1)(2p + 1) , then 1² + 3² + 5² + ..... + 17² is equal to : 6
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As we know that ,
1² + 2² + 3² + .... + p² = p(p + 1)(2p + 1) 6
∴ 1² + 3² + 5² + .... + 17² = (1² + 2² + 3² + ... + 17²) – (2² + 4² + ...+ 16²)
Required answer = (1² + 2² + 3² + ... + 17²) – 4 (1² + 2² + ... + 8²)Required answer = 17(17 + 1)(34 + 1) - 4 × 8(8 + 1)(16 + 1) 6 6
Correct Option: D
As we know that ,
1² + 2² + 3² + .... + p² = p(p + 1)(2p + 1) 6
∴ 1² + 3² + 5² + .... + 17² = (1² + 2² + 3² + ... + 17²) – (2² + 4² + ...+ 16²)
Required answer = (1² + 2² + 3² + ... + 17²) – 4 (1² + 2² + ... + 8²)Required answer = 17(17 + 1)(34 + 1) - 4 × 8(8 + 1)(16 + 1) 6 6 Required answer = 17 × 18 × 35 - 4 × 8 × 9 × 17 6 6
Required answer = 1785 – 816 = 969
- If 7 times the seventh term of an Arithmetic Progression (AP) is equal to 11 times its eleventh term, then the 18th term of the AP will be
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We know that , nth term of an arithmetic progression = an = a + (n – 1) d
Here , n = 7 and 11
∴ a7 = a + (7 – 1) d = a + 6d
a11 = a + (11 – 1) d = a + 10d
According to the question,
7 a7 = 11 a11
⇒ 7 (a + 6d) = 11 (a + 10d)
⇒ 7a + 42d = 11a + 110 d
⇒ 11a – 7a = 42d – 110dCorrect Option: B
We know that , nth term of an arithmetic progression = an = a + (n – 1) d
Here , n = 7 and 11
∴ a7 = a + (7 – 1) d = a + 6d
a11 = a + (11 – 1) d = a + 10d
According to the question,
7 a7 = 11 a11
⇒ 7 (a + 6d) = 11 (a + 10d)
⇒ 7a + 42d = 11a + 110 d
⇒ 11a – 7a = 42d – 110d
⇒ 4a = – 68d
⇒ a = – 17d .... (i)
∴ a18 = a + (18 – 1)d = a + 17d = –17d + 17d = 0
- (45 + 46 + 47 + .... + 113 + 114 + 115) is equal to
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We can write , (45 + 46 + 47 + ....... + 114 + 115) = (1 + 2 + 3 +..... + 115) – (1 + 2 + 3 + .... + 44)
∵ 1 + 2 + 3 + ..... + n = n ( n + 1) 
2 Required answer = 115 × (115 + 1) - 44 × (44 + 1) 2 2
Correct Option: C
We can write , (45 + 46 + 47 + ....... + 114 + 115) = (1 + 2 + 3 +..... + 115) – (1 + 2 + 3 + .... + 44)
∵ 1 + 2 + 3 + ..... + n = n ( n + 1) 2 Required answer = 115 × (115 + 1) - 44 × (44 + 1) 2 2 Required answer = 115 × 116 - 44 × 45 2 2
Required answer = 115 × 58 – 22 × 45
Required answer = 6670 – 990 = 5680
