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Question
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Solution 1
The given sequence is 2, 4, 6, 8 ... 344, 346, 348
Now, surely the above-mentioned range is in an A.P. or Arithmetic Progression. (Because the difference between two consecutive numbers in the given series is always constant.)
The above sequence is an A.P. with
a = t1 = 2,
d = t2 - t1
= 4 - 2
= 2
tn = 348
Since tn = a + (n - 1) d
∴ 348 = 2 + (n - 1) 2
∴ 348 = 2 - 2n + 2
∴ 348 = 2 + 2n - 2
∴ 348 = 2n
∴ n = `348/2`
∴ n = 174
Thus, the number of terms (n) = 174.
Now, Sn = `n/2` [2a + (n - 1)d]
∴ S174 = `174/2` [2(2) + (174 - 1)2]
= 87 [4 + (173)2]
= 87 [4 + 346]
= 87 × 350
= 30450
Hence, the sum of all even numbers between 1 and 350 is 30450.
Solution 2
All even numbers between 1 and 350,
2, 4, 6, 8 ... 344, 346, 348
Given sequence is an A.P.
a = 2, d = 2, tn = 348
tn = a + (n - 1)d
348 = 2 + (n - 1)2
348 - 2 = (n - 1)2
`346/2` = n - 1
173 + 1 = n
n = 174
t1 = 2, tn = 348
Sn = `n/2[t_1+t_n]`
S174 = `174/2[2+348]`
S174 = 87 × 350
S174 = 30450
Hence, the sum of all even numbers between 1 and 350 is 30450.
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