Is this correct? Arithmetic sum and sequence $100 \sum n = 51 7 n$ $sum_(n=51)^100 7n$

Question

$100 \sum n = 51 7 n$ $sum_(n=51)^100 7n$

$S_{n} = \frac{50}{2} (357 + 700) = = 26, 425$ $S_n=50/2 (357+700)= =26,425$
$a_{100} = 7 (51) + 7 (100 - 51) = 700$ $a_100=7(51)+7(100−51)=700$

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Answer 1 · 2018-04-19T13:10:49

See below.

You need the sum from $n = 51$ $n=51$ so:

$100 \sum n = 1 7 n - 50 \sum n = 1 7 n$ $sum_(n=1)^(100)7n-sum_(n=1)^(50)7n$

$7 \cdot (100 \sum n = 1 n - 50 \sum n = 1 n)$ $7*(sum_(n=1)^(100)n-sum_(n=1)^(50)n)$

$7 \cdot (\frac{100}{2} (2 + 99) - \frac{50}{2} (2 + 49))$ $7*(100/2(2+99)-50/2(2+49))$

$7 \cdot (5050 - 1275)$ $7*(5050-1275)$

$7 \cdot (3775) = 26425$ $7*(3775)=26425$

$n_{100}$ $n_100$

First term is 7 and common difference is 7:

$n t h$ $nth$ term

$a + (n - 1) d$ $a+(n-1)d$

$n_{100} = 7 + (99) \cdot 7 = 700$ $n_100=7+(99)*7=700$

So you are correct, or we are both wrong.