sum of 1st 15 multiples of 8
Answers
Answer:
hi there, it's aishik
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The first 8 multiples of 8 are
8, 16, 24, 32, 40, 48, 56,64
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S15 = ?
Sn = n/2 [2a + (n - 1)d]
S15 = 15/2 [2(8) + (15 - 1)8]
= 15/2[6 + (14) (8)]
= 15/2[16 + 112]
= 15(128)/2
= 15 × 64
= 960
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Answer;
The first 8 multiples of 8 are
8, 16, 24, 32, 40, 48, 56,64
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S15 = ?
Sn = n/2 [2a + (n - 1)d]
S15 = 15/2 [2(8) + (15 - 1)8]
= 15/2[6 + (14) (8)]
= 15/2[16 + 112]
= 15(128)/2
= 15 × 64
= 960