Question

Q1. 1, 4, 6, 5, 11, 6...... Find the sum first 100 terms?​

Answer 1

Answer:

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Answer 2

$The \: above \: series \: can \: be \: rewritten \: as \\ </p><p>(1 + 6 + 11 +........50 \: terms) + (4 + 5 + 6 +....50 \: terms) \\ = \frac{50}{2} (2 \times 1 + (50 - 1) \times 5) + \frac{50}{2} (2 \times 4 + (50 - 1) \times 1) \\ = 25 \times (2 + 245) + 25 \times (8 + 49) \\ = 25 \times (247 + 57) \\ = 25 \times 304 \\ = 7600$

Formula used :-

$\small\bold\red{sum \: of \: n \: terms \: = \frac{n}{2}(2a + (n - 1) \times d) } \\ \small\bold\red{a \: = first \: term} \\ \small\bold\red{d = common \: difference} \\ \small\bold\red{n \: = number \: of \: terms}$