The 10th and 15th terms of an a.p are -5 and -7 and the half, respectively. What is the sum of the first 20 terms
We know that the 10th term of the arithmetic progression is -5 and the 15th term is -7.
Using the formula for the nth term of an arithmetic progression, we can write the equations:
a + 9d = -5 ...(1)
a + 14d = -7 ...(2)
where a is the first term and d is the common difference.
Subtracting equation (1) from equation (2), we get:
( a + 14d ) - ( a + 9d ) = -7 -(-5)
5d = -2
d = -2/5
Substituting the value of d in equation (1), we get:
a + 9(-2/5) = -5
a - 18/5 = -5
a = -5 + 18/5
a = -25/5 + 18/5
a = ( -25 + 18 ) / 5
a = -7/5
So, the first term (a) is -7/5 and the common difference (d) is -2/5.
The formula for the sum of the first n terms of an arithmetic progression is given by:
S_n = n/2(2a + (n-1)d)
Substituting the values, we have:
S_20 = 20/2(2(-7/5) + (20-1)(-2/5))
= 10(-14/5 + 19(-2/5))
= 10(-14/5 + (-38/5))
= 10(-14 - 38)/5
= 10(-52)/5
= -520/5
= -104
Therefore, the sum of the first 20 terms is -104. Answer: \boxed{-104}.
To find the sum of the first 20 terms of the arithmetic progression (AP), we need to first find the common difference (d) and the first term (a) of the AP.
Given that the 10th term is -5 and the 15th term is -7, we can use the formula for the nth term of an AP:
an = a + (n - 1) * d
Substituting the values:
-5 = a + (10 - 1) * d
-7 = a + (15 - 1) * d
Simplifying these two equations, we get:
-5 = a + 9d (Equation 1)
-7 = a + 14d (Equation 2)
Now, let's find the common difference (d) by subtracting Equation 1 from Equation 2:
-7 - (-5) = (a + 14d) - (a + 9d)
-2 = 5d
d = -2/5
Now, we can substitute the value of d back into Equation 1 to find the first term (a):
-5 = a + 9 * (-2/5)
-5 = a - 18/5
To get the common denominator, multiply each term by 5:
-25/5 = 5a - 18/5
-25 = 5a - 18
-25 + 18 = 5a
-7 = 5a
a = -7/5
So, the first term (a) is -7/5 and the common difference (d) is -2/5.
Now that we know a and d, we can find the sum of the first 20 terms using the formula:
Sn = (n/2) * (2a + (n - 1) * d)
Substituting the values:
S20 = (20/2) * [2(-7/5) + (20 - 1) * (-2/5)]
= 10 * [-14/5 + 19 * (-2/5)]
Simplifying,
S20 = 10 * [-14/5 - 38/5]
= 10 * [-52/5]
= -520/5
= -104
Therefore, the sum of the first 20 terms of the arithmetic progression is -104.
To find the sum of the first 20 terms of an arithmetic progression (AP), we'll need to know the common difference (d) or two terms in the AP.
Given that the 10th term (a₁₀) is -5 and the 15th term (a₁₅) is -7, we can find the common difference (d) as follows:
a₁₀ = a₁ + (10 - 1)d
-5 = a₁ + 9d
a₁₅ = a₁ + (15 - 1)d
-7 = a₁ + 14d
Now we can solve these two equations to find the common difference (d):
-5 = a₁ + 9d
-7 = a₁ + 14d
Subtracting the first equation from the second equation, we get:
-7 - (-5) = (a₁ + 14d) - (a₁ + 9d)
-2 = 14d - 9d
-2 = 5d
Dividing both sides by 5, we find:
d = -2/5
Now that we have the common difference (d), we can find the first term (a₁) by substituting it into one of the previous equations. Let's use the first equation:
-5 = a₁ + 9(-2/5)
-5 = a₁ - (18/5)
-5 + 18/5 = a₁
(-25/5 + 18/5) = a₁
-7/5 = a₁
So, the first term (a₁) is -7/5, and the common difference (d) is -2/5.
To find the sum of the first 20 terms, we'll use the formula for the sum of an AP:
Sn = (n/2)[2a₁ + (n - 1)d]
where Sn is the sum of the first n terms. Plugging in the values:
n = 20
a₁ = -7/5
d = -2/5
Sn = (20/2)[2(-7/5) + (20 - 1)(-2/5)]
Sn = 10[-14/5 + 19(-2/5)]
Sn = 10[-14/5 + (-38/5)]
Sn = 10[-52/5]
Sn = -10 * 52/5
Sn = -520/5
Sn = -104
Therefore, the sum of the first 20 terms in the arithmetic progression is -104.