The given sequence is an arithmetic progression with common difference d = 4 and first term a = 3.
Sum of n terms of an A.P. is given by: Sn = n/2 x [2a + (n-1)d]
We need to find sum of 11 terms so n = 11.
Putting value of n, a and d we get:
S11 = 11/2 x [2x3 + (11 - 1) x 4]
S11 = 11/2 x [6 + 40]
S11 = 11/2 x 46 = 11 x 23 = 253
3 + 11 + 11 + 11 + 12 + 13 + 15 + 15 + 16 + 17 + 19 = 143
No it doesn't. That sum is 100.
11
The total sum is 110
There are 3 terms in the given expression of 6a+9b+15
The sum of addends 11 and 15 is: 11 + 15 = 26
15 x 11
(5 x 15) + (6 x 15) = 11 x 15 = 165
Expressed as a proper fraction in its simplest form, 1/3 + 2/5 = 11/15 or eleven fifteenths.1/3 + 2/55/15 + 6/1511/155/15 + 6/15 = 11/15
They sum up to 15
3 + 11 + 11 + 11 + 12 + 13 + 15 + 15 + 16 + 17 + 19 = 143
No it doesn't. That sum is 100.
11
the sum of -14 + 29 is 15
11+4=15
15 + 13 + 11 = 39
15 + 14 + 14 + 11 + 2x + x = 75 Collect like terms 54 + 3x = 75 Subtract 54 3x = 21 Divide by 3 x = 7 ■