Let the AP be represented by the general term *a_n* = *a_1* + (*n*-1)*d*, where *a_1* is the first term and *d* is the common difference.
The second term of the AP is *a_2* = *a_1* + *d*.
The fourth term of the AP is *a_4* = *a_1* + 3*d*.
The eighth term of the AP is *a_8* = *a_1* + 7*d*.
Since the second, fourth, and eighth terms form the first three consecutive terms of a GP, we can set up the equation:
*a_2*^2 = *a_1* * *a_4*
Substituting the values:
(*a_1* + *d*)^2 = *a_1* (*a_1* + 3*d*)
Expanding the equation:
*a_1*^2 + 2*a_1*d + *d*^2 = *a_1*^2 + 3*a_1*d
Simplifying the equation:
2*a_1*d + *d*^2 = 3*a_1*d
Rearranging the terms:
*d*^2 - a_1*d = 0
Factoring out *d*:
d * (*d* - *a_1*) = 0
So, either d = 0 or *d* = *a_1*.
If *d* = *a_1*, then the common difference is equal to the first term itself, which would mean that all terms of the AP are the same and it is not an arithmetic progression.
Therefore, *d* = 0.
This implies that the first term of the AP, *a_1*, is also the second term of the AP as *d* = *a_1*.
Given that the sum of the third and fifth term of the AP is 20, we have:
*a_3* + *a_5* = 20
(*a_1* + 2*0*) + (*a_1* + 4*0*) = 20
2*a_1* = 20
*a_1* = 10
So, the first term of the AP is 10.
Since the common difference is 0, all terms of the AP will be 10.
Therefore, the first four terms of the AP are:
10, 10, 10, 10
The sum of the first ten terms of the AP is given by the formula:
Sum = (n/2) * (2*a_1 + (n-1)*d)
Substituting the values:
Sum = (10/2) * (2*10 + (10-1)*0)
Sum = 5 * 20
Sum = 100
Therefore, the sum of the first ten terms of the AP is 100.
The second,fourth and eight term of an AP form the first three consecutive terms of a GP. The sum of the third and fifth term of the AP is 20 find the
First four terms of AP
Sum of the first ten terms of the AP
3 answers
the needed terms from the AP are
a+d, a+3d, and a+7d
They form the first 3 terms of a GP, then
(a+3d)/(a+d) = (a+7d)/(a+3d)
a^2 + 6ad + 9d^2 = a^2 + 8ad + 7d^2
-2ad = -2d^2
a =d
also given: a+2d + a+4d = 20
a + 3d = 10 , but a = d
4d = 10
d = 5/2, then a = 5/2
sum(10) of the AP = 5(5 + 9(5/2))
= 5( 55/2) = 275/2
check:
for the AP, terms are 5/2, 10/2, 15/2, 20/2, 25/2, 30/2, 35/2, 40/2, ...
the 2nd , the 4th, and the 8th are
10/2, 20/2, and 40/2 or 5, 10, and 20
which indeed form a geometric sequence
also the 3rd and 5th of the AP = 15/2 + 25/2 = 20 , as required.
of course the bot is wrong again
a+d, a+3d, and a+7d
They form the first 3 terms of a GP, then
(a+3d)/(a+d) = (a+7d)/(a+3d)
a^2 + 6ad + 9d^2 = a^2 + 8ad + 7d^2
-2ad = -2d^2
a =d
also given: a+2d + a+4d = 20
a + 3d = 10 , but a = d
4d = 10
d = 5/2, then a = 5/2
sum(10) of the AP = 5(5 + 9(5/2))
= 5( 55/2) = 275/2
check:
for the AP, terms are 5/2, 10/2, 15/2, 20/2, 25/2, 30/2, 35/2, 40/2, ...
the 2nd , the 4th, and the 8th are
10/2, 20/2, and 40/2 or 5, 10, and 20
which indeed form a geometric sequence
also the 3rd and 5th of the AP = 15/2 + 25/2 = 20 , as required.
of course the bot is wrong again
Apologies for the incorrect response earlier. Let's solve it correctly this time.
Let the AP be represented by the general term *a_n* = *a_1* + (*n*-1)*d*, where *a_1* is the first term and *d* is the common difference.
From the given information:
The second term of the AP is *a_2* = *a_1* + *d*.
The fourth term of the AP is *a_4* = *a_1* + 3*d*.
The eighth term of the AP is *a_8* = *a_1* + 7*d*.
We are given that the second, fourth, and eighth terms form the first three consecutive terms of a GP. Therefore:
*a_2* / *a_1* = *a_4* / *a_2*
(*a_1* + *d*) / *a_1* = (*a_1* + 3*d*) / (*a_1* + *d*)
Cross-multiplying:
*a_1*^2 + *a_1*d = *a_1*^2 + 3*a_1*d + *d*^2
Simplifying:
*a_1*d = *d*^2
*a_1* = *d*
We are also given that the sum of the third and fifth terms of the AP is 20. Therefore:
*a_3* + *a_5* = 20
(*a_1* + 2*d*) + (*a_1* + 4*d*) = 20
2*a_1* + 6*d* = 20
2*d* + 6*d* = 20 (Using *a_1* = *d*)
8*d* = 20
*d* = 20 / 8
*d* = 5/2
Substituting the value of *d* back into *a_1* = *d*:
*a_1* = *d* = 5/2
Therefore, the first term of the AP is 5/2 and the common difference is 5/2.
The first four terms of the AP are:
5/2, 5/2 + 5/2, 5/2 + 2*(5/2), 5/2 + 7*(5/2)
5/2, 15/2, 25/2, 45/2
To find the sum of the first ten terms of the AP, we can use the formula for the sum of an arithmetic series:
Sum = (n/2) * (2*a_1 + (n-1)*d)
Sum = (10/2) * (2*(5/2) + (10-1)*(5/2))
Sum = 5 * (5 + 9*(5/2))
Sum = 5 * (5 + 45/2)
Sum = 5 * (55/2)
Sum = 275/2
Therefore, the first four terms of the AP are 5/2, 15/2, 25/2, 45/2 and the sum of the first ten terms of the AP is 275/2.
Let the AP be represented by the general term *a_n* = *a_1* + (*n*-1)*d*, where *a_1* is the first term and *d* is the common difference.
From the given information:
The second term of the AP is *a_2* = *a_1* + *d*.
The fourth term of the AP is *a_4* = *a_1* + 3*d*.
The eighth term of the AP is *a_8* = *a_1* + 7*d*.
We are given that the second, fourth, and eighth terms form the first three consecutive terms of a GP. Therefore:
*a_2* / *a_1* = *a_4* / *a_2*
(*a_1* + *d*) / *a_1* = (*a_1* + 3*d*) / (*a_1* + *d*)
Cross-multiplying:
*a_1*^2 + *a_1*d = *a_1*^2 + 3*a_1*d + *d*^2
Simplifying:
*a_1*d = *d*^2
*a_1* = *d*
We are also given that the sum of the third and fifth terms of the AP is 20. Therefore:
*a_3* + *a_5* = 20
(*a_1* + 2*d*) + (*a_1* + 4*d*) = 20
2*a_1* + 6*d* = 20
2*d* + 6*d* = 20 (Using *a_1* = *d*)
8*d* = 20
*d* = 20 / 8
*d* = 5/2
Substituting the value of *d* back into *a_1* = *d*:
*a_1* = *d* = 5/2
Therefore, the first term of the AP is 5/2 and the common difference is 5/2.
The first four terms of the AP are:
5/2, 5/2 + 5/2, 5/2 + 2*(5/2), 5/2 + 7*(5/2)
5/2, 15/2, 25/2, 45/2
To find the sum of the first ten terms of the AP, we can use the formula for the sum of an arithmetic series:
Sum = (n/2) * (2*a_1 + (n-1)*d)
Sum = (10/2) * (2*(5/2) + (10-1)*(5/2))
Sum = 5 * (5 + 9*(5/2))
Sum = 5 * (5 + 45/2)
Sum = 5 * (55/2)
Sum = 275/2
Therefore, the first four terms of the AP are 5/2, 15/2, 25/2, 45/2 and the sum of the first ten terms of the AP is 275/2.