Find the sum:1 + 3 + 5 + 7 + . . . + 199

1.	Find the sum:1 + 3 + 5 + 7 + . . . + 199
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. 1 + 3 + 5 + 7 + . . . + 199 which is an A.P Where, a = 1 ,d = 3 – 1 = 2 and last term (a_n= l) = 199 We know that, a_n = a + (n – 1)d So, 199 = 1 + (n – 1)2 199 = 1 + 2n – 2 n = \(\frac{200}{2}\) = 100 Now, for the sum of these 100 terms S₁₀₀= \(\frac{100}{2}\) [1 + 199] = 50(200) = 10000 Hence, the sum of terms of the given series is 10000.

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. 1 + 3 + 5 + 7 + . . . + 199 which is an A.P

Where, a = 1 ,d = 3 – 1 = 2 and last term (a_n= l) = 199

We know that, a_n = a + (n – 1)d

So,

199 = 1 + (n – 1)2

199 = 1 + 2n – 2

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

Now, for the sum of these 100 terms

S₁₀₀= \(\frac{100}{2}\) [1 + 199]

= 50(200)

= 10000

Hence, the sum of terms of the given series is 10000.

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