Find the sum of the first twelve multiples of 7.

 first 12 multiples of 7 are = 7 , 14 .21 ..............84.

a = 7

d = 7

L = 84

n = 12

sn = n 2(2a + (n-1 ) d )

= 6 ( 14 + 77 )

= 6 (91 )

= 546

 

 

therefore ans is 546

  • 1

First twelve multiples of 7 are - 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, and 84.

Here, a = 7, d = a2 - a1 = 14 - 7 = 7, and n = 12

Sn = n / 2 {2a + (n-1) d}

S12 = 12 / 2 {2 * 7 +(12 - 1) 7}

      => 6 { 14 + (13 * 7)}

      => 6 {14 + 91}

      => 6 * 105

S12  => 630

  • -1

@vivek ur answer is wrong

  • 0

First 12 multiples of 7 are = 7 , 14 .21 ..............84.

Here,

a = 7

d = 7

an = 84

n = 12

sn = n / 2 ( a + an  )

= 6 ( 7 + 84 )

= 6 (91 )

= 546

 

 

therefore ans is 546

Cheers!

  • 8

n = 12

a=7

d=7

Sn=n/2 [2a + (n-1)d]

=12/2 [2(7) + (12-1)7]

=6 [14 + 77]

=6 [91]

=546

  • 3

 thX a lot//......For ur kindest immediate help guyZzz..... @all

  • 0
first 12 multiples of  7 are= 7,14,21.......

a=7
d=7
n=12

s12= n/2{2a +(n-1)d}
s12= 12/2{2x7+(12-1)7}
    = 6(14+77)
   = 6(91)
   =546

answer-546 
 
  • 0
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