Find the sum:25 + 28 + 31 + . . . + 100

1.	Find the sum:25 + 28 + 31 + . . . + 100
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, 25 + 28 + 31 + . . . + 100 which is an A.P Where, a = 25 ,d = 28 – 25 = 3 and last term (a_n= l) = 100 We know that, a_n = a + (n – 1)d So, 100 = 25 + (n – 1)(3) 100 = 25 + 3n – 3 n = \(\frac{(100 \,–\, 22)}{3}\) = 26 Now, for the sum of these 26 terms S₁₀₀= \(\frac{26}{2}\) [24 + 100] = 13(124) = 1625 Hence, the sum of terms of the given series is 1625.

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, 25 + 28 + 31 + . . . + 100 which is an A.P

Where, a = 25 ,d = 28 – 25 = 3 and last term (a_n= l) = 100

We know that, a_n = a + (n – 1)d

So,

100 = 25 + (n – 1)(3)

100 = 25 + 3n – 3

n = \(\frac{(100 \,–\, 22)}{3}\) = 26

Now, for the sum of these 26 terms

S₁₀₀= \(\frac{26}{2}\) [24 + 100]

= 13(124)

= 1625

Hence, the sum of terms of the given series is 1625.

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