Question

Find the sum of all multiples of 7 lying between 500 and 900.

Class 10

Arithmetic Progressions

Answer 1

Find the first term:

500 ÷ 7 = 71 remainder 3 = 72 (round up)

First term = 72 x 7 = 504

Find the last term:

900 ÷ 7 = 128 remainder 4 = 128 (round down)

last term = 128 x 7 = 896

Find the number of terms

an = a1 + ( n - 1) d

896 = 504 + (n - 1)7

896 = 504 + 7n - 7

896 = 497 + 7n

7n = 896 - 497

7n = 399

n = 57

Find the sum:

sn = n/2 (a1 + an)

s57 = 57/2 (504 + 896) = 39900

Answer: The sum is 39900

Answer 2

sum of all multiples of 7 lying between 500 and 900.
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Solution :
A.P : 504, 511, 518.......896
a=504
d = 7
an = 896
Sn = n/2 (a+an)

n = ?

an = a + (n-1)d
896 = 504 + (n-1) 7
896-504 = (n-1)7
7 (n-1) = 392
(n-1) = 392/7
= 56
n = 57

Sn = 57/2 (504+896)
= 57/2 × 1400
57 × 700 = 39,900