1.

Find the sum of series 4+12+20+28+ +100terms

Answer»

Given ,

FIRST term (a) = 4

Common difference (d) = 8

Last term (an) = 100

We know that , the NTH term of an AP is given by

\boxed{ \sf{ a_{<klux>N</klux>} = a + (n - 1)d }}

THUS ,

\implies \tt 100 = 4 + (n - 1)8

\implies \tt 96 = (n - 1)8

\implies \tt 12 = n - 1

\implies \tt n = 13

Now , the sum of first n terms of an AP is given by

\boxed { \sf{ S_{n} = \frac{n}{2}(a + a_{n}) }}

Thus ,

\implies \tt sum = \frac{13}{2} (4 + 100)

\implies \tt sum =13 \times 52

\implies \tt sum =<klux>676</klux>

★ The sum of given series is 676


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