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Given that the sequence formed by all the numbers between 9 and 95 which leaves remainder 1 when divided by 3.
AP: 10,13,16,19,22,...94
Let a be the first term and d be the common difference.
Here the first term a = 10.
Common difference d = 3.
Last term an = 94.
We know that number of n terms of an AP an = a + (n - 1) * d
94 = 10 + (n - 1) * 3
94 = 10 + 3n - 3
94 = 3n + 7
94 - 7 = 3n
87 = 3n
n = 87/3
n = 29.
Now,
Middle term = n + 1/(2)
= 29 + 1/(2)
= 30/2
= 15.
Now,
15th term a15 = a + (15 - 1) * d
= 10 + 14 * 3
= 10 + 42
= 52.
Therefore the Middle term = 52.
We know that sum of n terms of an AP sn = n/2(2a + (n - 1) * d)
sum of first 14 terms s14 = 14/2(2(10) + (14 - 1) * 3)
= 7(20 + 13 * 3)
= 413.
Therefore the Sum of first 14 terms = 413
Now,
Sum of the last 14 terms = S29 - S15
= 29/2(2 * 10 + (29 - 1) * 3) - 15/2(2 * 10 + (15 - 1) * 3)
= 29/2(20 + 84) - 15/2(20 + 42)
= 29/2 * 104 - 15/2 * 62
= 1043.
Therefore the Sum of last 14 terms = 1043.
Hope this helps!
AP: 10,13,16,19,22,...94
Let a be the first term and d be the common difference.
Here the first term a = 10.
Common difference d = 3.
Last term an = 94.
We know that number of n terms of an AP an = a + (n - 1) * d
94 = 10 + (n - 1) * 3
94 = 10 + 3n - 3
94 = 3n + 7
94 - 7 = 3n
87 = 3n
n = 87/3
n = 29.
Now,
Middle term = n + 1/(2)
= 29 + 1/(2)
= 30/2
= 15.
Now,
15th term a15 = a + (15 - 1) * d
= 10 + 14 * 3
= 10 + 42
= 52.
Therefore the Middle term = 52.
We know that sum of n terms of an AP sn = n/2(2a + (n - 1) * d)
sum of first 14 terms s14 = 14/2(2(10) + (14 - 1) * 3)
= 7(20 + 13 * 3)
= 413.
Therefore the Sum of first 14 terms = 413
Now,
Sum of the last 14 terms = S29 - S15
= 29/2(2 * 10 + (29 - 1) * 3) - 15/2(2 * 10 + (15 - 1) * 3)
= 29/2(20 + 84) - 15/2(20 + 42)
= 29/2 * 104 - 15/2 * 62
= 1043.
Therefore the Sum of last 14 terms = 1043.
Hope this helps!
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