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Question
Find the sum of all multiples of 9 lying between 300 and 700.
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Solution
The multiples of 9 lying between 300 and 700 are 306, 315,………, 693.
This is an AP with a = 306,d = 9 and l = 693.
Suppose these are n terms in the AP. Then,
an = 693
⇒ 306+ (n-1) × 9 = 639 [an = a + (n-1) d]
⇒ 9n + 297 = 693
⇒ 9n = 693 - 297 = 396
⇒ n=44
∴ Required sum = `44/2 (306 = 693) [s_n = n/2 (a+1)]`
= 22× 999
= 21978
Hence, the required sum is 21978.
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