If the sixth term of an arithmetic progression (A.P) is 37 and the sum of the six term is 147, find the first term, common difference, sum of the first fifteen terms
Just use the formulas you know. You have only learned a few rules so far, so the problems will have to use the few you already have.
a + 5d = 37
6/2 (2a+5d) = 147
so, a=12, d=5, and
S15 = 15/2 (2*12 + 14*5) = 705
In an arithmetic progression nth term:
an = a1 + ( n - 1 ) d
a1= initial term
d = common difference
a6 = a1 + ( 6 - 1 ) d = 37
a1 + 5 d = 37
Sum of n terms in A.P:
Sn = n ∙ [ 2 a1 + ( n - 1) ∙ d ] / 2
In this case n = 6
S6 = 6 ∙ [ 2 a1 + ( 6 - 1) ∙ d ] / 2 = 147
3 ∙ [ 2 a1 + 5 d ] = 147
6 a1 + 15 d = 147
Now you have system of two equations:
a1 + 5 d = 37
6 a1 + 15 d = 147
___________
Try to solve it.
The solutons are:
a1 = 12 , d = 5
Sum of the first fifteen terms:
Sn = n ∙ [ 2 a1 + ( n - 1) ∙ d ] / 2
S15 = 15 ∙ [ 2 ∙ 12 + ( 15 - 1) ∙ 5 ] / 2
S15 = 15 ∙ [ 24 + 14 ∙ 5 ] / 2
S15 = 15 ∙ ( 24 + 70 ) / 2
S15 = 15 ∙ 94 / 2
S15 = 1410 / 2
S15 = 705
Check of results.
Your A.P is:
12 , 17 , 22 , 27 , 32 , 37 , 42 , 47 , 52 , 57 , 62 , 67 , 72 , 77 , 82
Sum of first 6 terms in:
S6 = 12 + 17 + 22 + 27 + 32 + 37 = 147
S15 = 12 + 17 + 22 + 27 + 32 + 37 + 42 + 47 + 52 + 57 + 62 + 67 + 72 + 77 + 82 = 705
To find the first term, common difference, and the sum of the first fifteen terms, we can use the formulas for arithmetic progressions.
1. Finding the first term (a):
We know that the sixth term of the A.P is 37. The formula to find the nth term of an A.P is:
an = a + (n-1)d
Let's substitute the value of the sixth term (an = 37) and the position of the sixth term (n = 6) into the formula:
37 = a + (6-1)d
37 = a + 5d
2. Finding the common difference (d):
We still have one equation with two variables (a and d). To solve for d, we need more information.
3. Finding the sum of the first six terms (S6):
Given that the sum of the six terms is 147, we can use the formula for the sum of an A.P up to the nth term:
Sn = (n/2)(2a + (n-1)d)
Substituting the values of S6 (147) and n (6) into the formula:
147 = (6/2)(2a + (6-1)d)
147 = 3(2a + 5d)
49 = 2a + 5d
Now we have two equations:
37 = a + 5d
49 = 2a + 5d
4. Solving for a and d:
We can solve this system of equations by subtracting the first equation from the second equation to eliminate d:
(2a + 5d) - (a + 5d) = 49 - 37
a = 12
Plugging this value of a back into the first equation:
37 = 12 + 5d
37 - 12 = 5d
25 = 5d
d = 5
Now we have found the first term (a = 12) and the common difference (d = 5).
5. Finding the sum of the first fifteen terms (S15):
We can use the formula for the sum of an A.P up to the nth term again:
Sn = (n/2)(2a + (n-1)d)
Substituting the values of n (15), a (12), and d (5) into the formula:
S15 = (15/2)(2(12) + (15-1)(5))
S15 = (15/2)(24 + 14 × 5)
S15 = (15/2)(24 + 70)
S15 = (15/2)(94)
S15 = (15 × 94)/2
S15 = 705
Therefore, the sum of the first fifteen terms is 705.