find the sum of first 15 terms of an A.P whose 1st and last are -4 and 46 respectively.
last term is term(n)
= a + (n-1)d
= 46
so -4 + nd - d = 46
d(n-1) = 50
sum(15) = (15/2)(2a + (n-1)d )
= (15/2)(-8 + 50) = 315
To find the sum of the first 15 terms of an arithmetic progression (A.P), we need to use the formula for the sum of an A.P.
The formula for the sum (S) of an A.P is given by:
S = (n/2) * (a + l)
Where:
n is the total number of terms
a is the first term
l is the last term
In this case, we are given:
a = -4 (first term)
l = 46 (last term)
n = 15 (number of terms)
Now, let's substitute these values into the formula:
S = (15/2) * (-4 + 46)
First, we calculate the value inside the parentheses:
-4 + 46 = 42
Now we substitute this value into the formula:
S = (15/2) * 42
To multiply a fraction by a whole number, we can multiply the numerator by the whole number and keep the denominator the same:
S = (15 * 42) / 2
15 * 42 = 630
S = 630 / 2
Dividing 630 by 2 gives us:
S = 315
Therefore, the sum of the first 15 terms of the given arithmetic progression is 315.
Sn = n/2 (a1 + an) = 15/2(-4 + 46) = 15/2 * 42 = 315
I was going to use the formula
sum(n) = (n/2)(first + last), but then hesitated because we would
assume that there were 15 terms and term(15) = 46
So I went with the more general sum of terms formula
which contained d(n-1) as a constant of 50, so we got the same result.
however, suppose we had 26 terms, then term(26) = 46, and d = 2 , a = -4
sum(15) = (15/2)(-8 + 25(2)) = 315
suppose we have 21 terms then term(21) = 46 and d = 2.5
sum(15) = (15/2)(-8 + 20(2.5)) = 315
interesting