Find the sum:2 + 4 + 6 + . . . + 200

1.	Find the sum:2 + 4 + 6 + . . . + 200
Answer» We know that the sum of terms for an A.P is given by S_n = \(\frac{n}{2}\)[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms Or S_n = \(\frac{n}{2}\)[a + l] Where; a = first term for the given A.P. ;l = last term for the given A.P Given series. 2 + 4 + 6 + . . . + 200 which is an A.P Where, a = 2 ,d = 4 – 2 = 2 and last term (a_n= l) = 200 We know that, a_n = a + (n – 1)d So, 200 = 2 + (n – 1)2 200 = 2 + 2n – 2 n = \(\frac{200}{2}\) = 100 Now, for the sum of these 100 terms S₁₀₀= \(\frac{100}{2}\) [2 + 200] = 50(202) = 10100 Hence, the sum of terms of the given series is 10100.

Answer»

We know that the sum of terms for an A.P is given by

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Or S_n = \(\frac{n}{2}\)[a + l]

Where; a = first term for the given A.P. ;l = last term for the given A.P

Given series. 2 + 4 + 6 + . . . + 200 which is an A.P

Where, a = 2 ,d = 4 – 2 = 2 and last term (a_n= l) = 200

We know that, a_n = a + (n – 1)d

So,

200 = 2 + (n – 1)2

200 = 2 + 2n – 2

n = \(\frac{200}{2}\) = 100

Now, for the sum of these 100 terms

S₁₀₀= \(\frac{100}{2}\) [2 + 200]

= 50(202)

= 10100

Hence, the sum of terms of the given series is 10100.

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