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Question
Find the sum of first 15 multiples of 8.
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Solution 1
The multiples of 8 are
8, 16, 24, 32…
These are in an A.P., having the first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S15 =?
`S_n = n/2[2a+(n-1)d]`
= `15/2[2(8)+(15-1)8]`
= `15/2[16+14(8)]`
= `15/2(16+112)`
= `(15(128))/2`
= 15 × 64
= 960
Therefore, the sum of the first 15 multiples of 8 is 960.
Solution 2
Multiples of 8 are: 8, 16, 24, 32, ........... , Which form an A.P. with first term, a = 8 and common difference, d = 8
∵ Sum of nth term of A.P.
Sn = `n/2[2a + (n - 1)d]`
∴ S15 = `15/2 [2 xx 8 + (15 - 1) xx 8]`
= `15/2 [16 + 112]`
= `15/2 xx 128`
= 15 × 64
= 960
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