find the sum of first 15 multiples of 999
Answers
Answer:
Hi there!
Hi there!The positive integers which are multiple of 8 are:
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 terms
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series.
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series. Now,
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series. Now,First term, a = 8
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series. Now,First term, a = 8Common difference, d = 8
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series. Now,First term, a = 8Common difference, d = 8Number of terms, n = 15
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series. Now,First term, a = 8Common difference, d = 8Number of terms, n = 15Now, sum of first 15 terms of AP = (n/2) × {2a + (n-1)d}
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series. Now,First term, a = 8Common difference, d = 8Number of terms, n = 15Now, sum of first 15 terms of AP = (n/2) × {2a + (n-1)d} = (15/2) × {2 × 8 + (15-1)8}
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series. Now,First term, a = 8Common difference, d = 8Number of terms, n = 15Now, sum of first 15 terms of AP = (n/2) × {2a + (n-1)d} = (15/2) × {2 × 8 + (15-1)8} = (15/2) × (16 + 14 × 8)
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series. Now,First term, a = 8Common difference, d = 8Number of terms, n = 15Now, sum of first 15 terms of AP = (n/2) × {2a + (n-1)d} = (15/2) × {2 × 8 + (15-1)8} = (15/2) × (16 + 14 × 8) = (15/2) × (16 + 112)
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series. Now,First term, a = 8Common difference, d = 8Number of terms, n = 15Now, sum of first 15 terms of AP = (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
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series. Now,First term, a = 8Common difference, d = 8Number of terms, n = 15Now, sum of first 15 terms of AP = (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
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series. Now,First term, a = 8Common difference, d = 8Number of terms, n = 15Now, sum of first 15 terms of AP = (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
Hi there!The positive integers which are multiple of 8 are:8, 16, 24... to 15 termsIt forms an arithmetic series. Now,First term, a = 8Common difference, d = 8Number of terms, n = 15Now, sum of first 15 terms of AP = (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 = 960Cheers!
Answer:
multiples of 999: 999, 1998, 2997, 3996,................,14985
here
a = 999
d = 999
n = 15
an = 14985
according to question
Sn = n/2{a+an}
Sn = 15/2[999+14985]
Sn = [15984/2]15
Sn = 119880 ans
Step-by-step explanation: