The sum of the first 6 term of AP is 21if the 7th term is 3 times the sum of the 3rd and 4th term .Find the first term and common difference.
6/2 (2a+5d) = 21
a+6d = 3(a+2d + a+3d)
so you have
a = -9
d = 5
In an Arithmetic Progression:
an = a + ( n - 1 ) d
where
a = a1 = the initial term
an = the nth term
d = the common difference of successive members
The sum of the first n term:
Sn = n / 2 [ 2 a + ( n - 1 ) d ]
Ginen conditions:
S6 = n / 2 [ 2 a + ( n - 1 ) d ] = 21
6 / 2 [ 2 a + ( 6 - 1 ) d ] = 21
3 ( 2 a + 5 d ) = 21
Divide both sides by 3
2 a + 5 d = 7
The 7th term is 3 times the sum of the 3rd and 4th term means:
a7 = 3 ( a3 + a4 )
Since:
a3 = a + 2 d , a4 = a + 3 d , a7 = a + 6 d
a7 = 3 ( a3 + a4 )
a + 6 d = 3 ( a + 2 d + a + 3 d )
a + 6 d = 3 ( 2 a + 5 d )
a + 6 d = 6 a + 15 d
Subract 15 d d to both sides
a - 9 d = 6 a
Subract a to both sides
- 9 d = 5 a
5 a = - 9 d
Now you must solve system of two equations:
2 a + 5 d = 7
5 a = - 9 d
Try that.
The solution is:
a = - 9 , d = 5
Check result:
a1 = - 9
a2 = - 9 + 5 = - 4
a3 = - 4 + 5 = 1
a4 = 1 + 5 = 6
a5 = 6 + 5 = 11
a6 = 11 + 5 = 16
a7 = 16 + 5 = 21
The sum of the first 6 term:
- 9 + ( - 4 ) + 1 + 6 +11 +16 = - 9 - 4 + 1 + 6 +11 +16 = 21
Correct.
a7 = 3 ( a3 + a4 )
21 = 3 ( 1 + 6 )
21 = 3 ∙ 7
Correct.
To solve this problem, let's first find the sum of the first 6 terms of the arithmetic progression (AP).
The sum of an arithmetic progression is given by the formula: Sn = (n/2)(a + l), where Sn is the sum of the first n terms, a is the first term, and l is the last term.
Given that the sum of the first 6 terms is 21, we have:
21 = (6/2)(a + l)
Simplifying, we get: 21 = 3(a + l)
Now, let's find the 7th term of the AP.
It is given that the 7th term is 3 times the sum of the 3rd and the 4th term. Let's assume that the common difference is d.
The 3rd term is a + 2d, and the 4th term is a + 3d. Therefore, the sum of the 3rd and 4th term is (a + 2d) + (a + 3d).
Given that the 7th term is 3 times this sum, we have:
7th term = 3((a + 2d) + (a + 3d))
Now, we can use the fact that the 7th term is also given by the formula for the n-th term of an arithmetic progression:
7th term = a + (7-1)d
= a + 6d
Equating the two expressions for the 7th term, we have:
a + 6d = 3((a + 2d) + (a + 3d))
Simplifying, we get:
a + 6d = 3(2a + 5d)
Further simplification leads to:
a + 6d = 6a + 15d
Rearranging the terms, we get:
6d - 15d = 6a - a
Simplifying, we have:
-9d = 5a
From the given information, we have two equations:
21 = 3(a + l)
-9d = 5a
Now, we can solve these equations simultaneously to find the values of the first term (a) and the common difference (d).
By substituting the second equation into the first equation, we get:
21 = 3(a + l)
-9d = 5(-9d)
21 = 3(a - 9d)
Dividing both sides by 3, we have:
7 = a - 3d
Now, we have two equations:
7 = a - 3d
-9d = 5a
Let's solve these equations together to find the values of a and d.
Multiplying the first equation by 5 and the second equation by 3 to eliminate the common terms, we have:
35 = 5a - 15d
-27d = 15a
Now, subtract the second equation from the first equation, we get:
35 - (-27d) = 5a - 15d - 15a
Simplifying, we have:
35 + 27d = -10a - 15d
Rearranging, we get:
10a = -15d - 27d - 35
Further simplifying, we have:
10a = -42d - 35
Finally, dividing both sides by 10, we get:
a = (-42d - 35)/10
From the given equations, we can substitute the value of a into the second equation to find the value of d.
-9d = 5((-42d - 35)/10)
Simplifying, we have:
-9d = (-210d - 175)/10
Cross-multiply and simplify, we get:
-90d = -210d - 175
Add 210d to both sides, we have:
120d = -175
Dividing both sides by 120, we get:
d = -175/120
Simplifying further, we get:
d = -7/5
Now substitute the value of d back into the first equation to find the value of a.
7 = a - 3(-7/5)
Simplifying, we have:
7 = a + 21/5
Rearranging, we have:
a = 7 - 21/5
Further simplifying, we have:
a = 35/5 - 21/5
a = 14/5
Therefore, the first term (a) is 14/5 and the common difference (d) is -7/5.