In an AP the first term is 2 and the sum of the first five terms is one fourth of the sum of the next five terms. Show that the 20th term is -112.
nth term is a+(n-1)d
sum of 1st n terms = n/2(T1+Tn)
a = 2
sum of 1st 5 terms = 5/2 (2 + 2+4d) = 10+10d
6th term = 2+5d
sum of terms 6-10 = 5/2 (2+5d + 2+9d) = 10+35d
10+10d = 1/4 (10+35d)
40+40d = 10 + 35d
5d = -30
d = -6
sequence is 2 -4 -10 -16 -22 -28 -34 -40 -46 -52
sum of T1..T5 = -50
sum of T6..T10 = -200
(copied From above)
20th Term = a +(n-1)d
= 2+19 X -6
= 2 - 114
= -112
2ab/a² b²
nth term is a+(n-1)d
sum of 1st n terms = n/2(T1+Tn)
a = 2
sum of 1st 5 terms = 5/2 (2 + 2+4d) = 10+10d
6th term = 2+5d
sum of terms 6-10 = 5/2 (2+5d + 2+9d) = 10+35d
10+10d = 1/4 (10+35d)
40+40d = 10 + 35d
5d = -30
d = -6
sequence is 2 -4 -10 -16 -22 -28 -34 -40 -46 -52
sum of T1..T5 = -50
sum of T6..T10 = -200
To solve this problem, we need to understand the properties of an Arithmetic Progression (AP) and the given conditions.
An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. Let's denote the common difference by 'd'.
We are given that the first term of the AP is 2. Let's assume the common difference is 'd', so the second term would be 2 + d, the third term would be 2 + 2d, and so on.
Now let's analyze the given condition: the sum of the first five terms is one fourth of the sum of the next five terms.
The sum of the first five terms can be written as:
Sum_first_five = 2 + (2 + d) + (2 + 2d) + (2 + 3d) + (2 + 4d) = 10 + 10d
The sum of the next five terms can be written as:
Sum_next_five = (2 + 5d) + (2 + 6d) + (2 + 7d) + (2 + 8d) + (2 + 9d) = 10 + 35d
According to the given condition, we have:
Sum_first_five = (1/4) * Sum_next_five
Substituting the values, we get:
10 + 10d = (1/4) * (10 + 35d)
Simplifying the equation, we have:
40 + 40d = 10 + 35d
Bringing like terms together, we get:
40d - 35d = 10 - 40
Simplifying further, we have:
5d = -30
Dividing both sides by 5, we get:
d = -6
Now that we have found the common difference 'd' as -6, we can find the 20th term of the AP.
The general formula to find the nth term of an AP is given by:
nth_term = First_term + (n - 1) * common difference
Using the formula, we have:
20th_term = 2 + (20 - 1) * (-6)
= 2 + 19 * (-6)
= 2 - 114
= -112
Therefore, the 20th term of the given AP is -112.