Answer:
[tex]\text{Solution:}\\\text{First term}(a)=2\\\text{Sum of first five terms = }\dfrac{1}{4}\times\text{Sum of next five terms}\\[/tex]
[tex]\text{Now let }d\text{ denote the common difference and }n\text{ denote the }\bold{number\ of\ terms.}[/tex]
[tex]\text{Then, Sum of first } n \text{ terms is given by:}\\\\\text{S}_n=\dfrac{n}{2}[2a+(n-1)d][/tex]
[tex]\therefore\ \text{Sum of first five terms = }\dfrac{5}{2}[2(2)+(5-1)d]=\dfrac{5}{2}[4+4d]\\\text{or, S}_5=10+10d[/tex]
[tex]\text{Now for the next series of five terms, the sixth term will be the first term.}\\\therefore\ \text{Sum of next five terms(S}_t)=\dfrac{5}{2}[2(a+4d)+(5-1)d]\\\\\text{or, S}_t=\dfrac{5}{2}[2(2+5d)+4d]=\dfrac{5}{2}[4+10d+4d]=\dfrac{5}{2}[4+14d]\\\\\text{or, S}_t=5(2+7d)\\\text{or, S}_t=10+35d[/tex]
[tex]\text{According to question,}\\\text{S}_5=\dfrac{1}{4}\text{S}_t\\\text{or, }10+10d=\dfrac{1}{4}(10+35d)\\\text{or, }40+40d=10+35d\\\text{or, }5d=-30\\\text{or, }d=-6[/tex]
[tex]\text{Now the 20th term(t}_{20})=a+19d=2+19(-6)=112[/tex]
You will get this after seeing the posted figure!
The arithmetic progression under given conditions, we used the formulas for the nth term and sum of n terms of an A.P., solving for the common difference and then for the 20th term, confirming it to be -112.
To solve the problem of finding the 20th term of an arithmetic progression (A.P.) where the first term is 2.
And the sum of the first five terms is one-fourth of the sum of the next five terms, we can start by using the formulas for the nth term of an A.P.
And the sum of the first n terms of an A.P.
The nth term is given by a_n = a_1 + (n-1)d, and the sum of the first n terms is S_n = \(\frac{n}{2}\)[2a_1 + (n-1)d], where a_1 is the first term, d is the common difference, and n is the term number.
Since the first term a_1 = 2, the sum of the first five terms is S_5 = \(\frac{5}{2}\)(2\times2 + (5-1)d), and the sum of the next five terms (i.e., the sum of terms 6 to 10) is S_{5 to 10} = S_{10} - S_{5}.
Given that S_5 = \frac{1}{4}S_{5 to 10}, we can set up an equation S_{5} = \frac{1}{4}(S_{10} - S_{5}), substitute the sums' expressions, and solve it to find the common difference d.
Once d is found, calculating the 20th term involves substituting n = 20, a_1 = 2, and the found value of d into the nth term formula.
Performing these steps leads us to find that d = -7 and subsequently that the 20th term, a_20, is indeed -112.
