Find the sum of all integers between 84 and 719, which are multiples of 5.

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Solution

Multiples of 5 between 84 and 719 are: 85,90,95........715.

Therefore, a=85, d=5, an=715.

a + (n-1)d = an.

Therefore, 85 + (n-1)5 = 715.

Implies, (n-1)5 = 630.

Implies, n-1 = 126.

Therefore, n = 127.

$S_{n} = \frac{n}{2} (2 a + (n - 1) d)$

$S_{n} = \frac{127}{2} (2 \times 85 + (127 - 1) 5)$

$S_{n} = \frac{127}{2} \times 800$

$S_{n} = 127 \times 400$

$S_{n} = 50800$

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Find the sum of all integers between 84 and 719, which are multiples of 5.

Multiples of 5 between 84 and 719 are: 85,90,95........715. Therefore, a=85, d=5, an=715. a + (n-1)d = an. Therefore, 85 + (n-1)5 = 715. Implies, (n-1)5 = 630. Implies, n-1 = 126. Therefore, n = 127. Sn=n2(2a+(n−1)d) Sn=1272(2×85+(127−1)5) Sn=1272×800 Sn=127×400 Sn=50800