The sum of the first 5 terms of an AP is 40 and the 7th term is equal to 4 times the second term, find the first term (a), common difference (d) and the sum of the second five terms.
4. The sum of the first 5 terms of an AP is 40 and the 7th term is equal to 4 times the second term, find the first term (a), common difference (d) and the sum of the second five terms.
why post it twice? Using your formulas, you see that
5/2 (2a+4d) = 40
a+6d = 4(a+d)
Now you can find a and d, and thus the sum of the 2nd five terms is the sum of the first 10, less the sum of the first 5:
10/2 (2a+9d) - 40
To find the first term (a) and the common difference (d) of an arithmetic progression (AP), we can use the information given. Let's solve step by step:
Step 1: Finding the first term (a) and common difference (d)
The sum of the first 5 terms of the AP is given as 40. The formula to find the sum of the first n terms of an AP is given by:
Sum (Sn) = (n/2) * [2a + (n - 1) * d]
Using the given information, we have:
40 = (5/2) * [2a + (5 - 1) * d]
Simplifying further:
80 = 10a + 10d
8 = a + d --- (Equation 1)
Step 2: Finding the relation between the 2nd term and the 7th term
The 7th term is equal to 4 times the second term. Setting up the equation:
7th term = 4 * 2nd term
Using the formula for the nth term of an AP:
a + (n - 1) * d = 4 * (a + d)
Simplifying further:
a + 6d = 4a + 4d
3a = 2d --- (Equation 2)
Step 3: Solving the equations
We have two equations:
Equation 1: 8 = a + d
Equation 2: 3a = 2d
We can substitute Equation 2 into Equation 1:
8 = a + (3a/2)
Multiplying both sides by 2:
16 = 2a + 3a
16 = 5a
Dividing by 5:
a = 16/5
Substituting the value of a back into Equation 1:
8 = (16/5) + d
Multiplying both sides by 5 to remove the fraction:
40 = 16 + 5d
24 = 5d
d = 24/5
So, the first term (a) is 16/5 and the common difference (d) is 24/5.
Step 4: Finding the sum of the second five terms
The sum of the second five terms can be found using the formula for the sum of an AP:
Sum (Sn) = (n/2) * [2a + (n - 1) * d]
For the second five terms, n = 5 and a = 16/5, d = 24/5:
Sum (S5) = (5/2) * [2 * (16/5) + (5 - 1) * (24/5)]
= (5/2) * [32/5 + 4 * (24/5)]
= (5/2) * [32/5 + 96/5]
= (5/2) * [128/5]
= (5/2) * 128/5
= 64
Therefore, the sum of the second five terms is 64.