Question

In an arithmetic progression, the first term is 2 and the sum of its first five terms is one-fourth of the sum of the next five terms. The 20th term of that progression is

• A. -112
• B. -120
• C. -96
• D. 116

Hint:

When the problem involves the "next \(n\) terms", calculate their sum as: \[ S_{2n}-S_n \] instead of writing every term separately.

Answer 1

The Correct Option is A

Step 1: Find the sum of the first five terms.
Given first term \(a=2\) and common difference \(d\). Using \[ S_n=\frac{n}{2}[2a+(n-1)d] \] \[ S_5=\frac{5}{2}[4+4d] \] \[ S_5=10+10d \]

Step 2: Find the sum of the next five terms.
The sum of terms \(6\) to \(10\) is \[ S_{10}-S_5 \] Now, \[ S_{10}=\frac{10}{2}[4+9d] \] \[ S_{10}=20+45d \] Therefore, \[ S_{10}-S_5=(20+45d)-(10+10d) \] \[ =10+35d \]

Step 3: Use the given condition to find \(d\).
Given that the sum of the first five terms is one-fourth of the sum of the next five terms. \[ 10+10d=\frac14(10+35d) \] Multiplying by 4, \[ 40+40d=10+35d \] \[ 5d=-30 \] \[ d=-6 \]

Step 4: Find the 20th term.
The \(n^{th}\) term of an AP is \[ a_n=a+(n-1)d \] Hence, \[ a_{20}=2+19(-6) \] \[ =2-114 \] \[ =-112 \] {-112}