1st term of an a.p is twice its common diff,if the sum of the 1st 10 terms of the a.p is 130 find the 1st term of the a.p
1st term of an a.p is twice its common difference means:
a1 = 2 d
Sum of the first n terms of an A.P.
Sn = n ( a1 + an ) / 2
In this case:
S10 = 10 ∙ ( a1 + a10 ) / 2
130 = 10 ∙ ( a1 + a10 ) / 2
Since:
a10 = a1 + 9 d
130 = 10 ∙ ( a1 + a1 + 9 d ) / 2
130 = ( 10 / 2 ) ∙ ( 2 a1 + 9 d )
130 = 5 ∙ ( 2 ∙ 2 d + 9 d )
130 = 5 ∙ ( 4 d + 9 d )
130 = 5 ∙ 13 d
130 = 65 d
65 d = 130
d = 130 / 65
d = 2
a1 = 2 d
a1 = 2 ∙ 2
a1 = 4
Let's assume the first term of the arithmetic progression (AP) is 'a' and the common difference is 'd'.
Given that the first term is twice the common difference, we can write:
a = 2d ----(1)
We know the formula to find the sum of the first 'n' terms of an AP is:
Sn = (n/2) * [2a + (n-1)d] ----(2)
Given that the sum of the first 10 terms is 130, we can substitute the values into equation (2):
130 = (10/2) * [2a + (10-1)d]
Simplifying further:
130 = 5(2a + 9d) ----(3)
Since we have two equations (equations (1) and (3)) with two variables (a and d), we can solve them simultaneously to find their values.
Using equation (1), we substitute the value of a in equation (3):
130 = 5(2(2d) + 9d)
130 = 5(4d + 9d)
130 = 5(13d)
Dividing both sides by 5:
26 = 13d
Simplifying further:
d = 26/13
d = 2
Now substitute the value of 'd' into equation (1) to find 'a':
a = 2d
a = 2(2)
a = 4
Therefore, the first term of the arithmetic progression is 4.
To find the first term of an arithmetic progression (AP), where the sum of the first 10 terms is given, we need to use the formula for the sum of an AP.
The formula for the sum of an AP is:
S = (n/2)(2a + (n-1)d)
Where:
S is the sum of the terms of the AP
n is the number of terms
a is the first term of the AP
d is the common difference between the terms
In this case, given that the sum of the first 10 terms (S) is 130, we can substitute the known values into the formula and solve for the first term (a).
130 = (10/2)(2a + (10-1)d)
Simplifying further,
130 = 5(2a + 9d)
Now, we are given that the first term (a) is twice the common difference (d). So we can substitute 2d for a in the equation.
130 = 5(2(2d) + 9d)
Simplifying again,
130 = 5(4d + 9d)
130 = 5(13d)
130 = 65d
Dividing both sides of the equation by 65, we get:
d = 2
Now, substitute this value of d back into the equation for the first term (a):
a = 2d = 2(2) = 4
Therefore, the first term of the AP is 4.