Find the sum:3 + 11 + 19 + . . . + 803

1.	Find the sum:3 + 11 + 19 + . . . + 803
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. 3 + 11 + 19 + . . . + 803 which is an A.P Where, a = 3 ,d = 11 – 3 = 8 and last term (a_n= l) = 803 We know that, a_n = a + (n – 1)d So, 803 = 3 + (n – 1)8 803 = 3 + 8n – 8 n = \(\frac{808}{8}\) = 101 Now, for the sum of these 101 terms S₁₀₁= \(\frac{101}{2}\) [3 + 803] = \(\frac{101(806)}{2}\) = 101 x 403 = 40703 Hence, the sum of terms of the given series is 40703.

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. 3 + 11 + 19 + . . . + 803 which is an A.P

Where, a = 3 ,d = 11 – 3 = 8 and last term (a_n= l) = 803

We know that, a_n = a + (n – 1)d

So,

803 = 3 + (n – 1)8

803 = 3 + 8n – 8

n = \(\frac{808}{8}\) = 101

Now, for the sum of these 101 terms

S₁₀₁= \(\frac{101}{2}\) [3 + 803]

= \(\frac{101(806)}{2}\)

= 101 x 403

= 40703

Hence, the sum of terms of the given series is 40703.

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