Introduction / Context:
This problem tests your understanding of arithmetic progressions, specifically how to use known terms to find the common difference and then any other term. Such questions are standard in quantitative aptitude sections of competitive exams.
Given Data / Assumptions:
- The sequence is an arithmetic progression (A.P.).
- T3 = −13 (the 3rd term).
- T8 = 2 (the 8th term).
- We need to find the 14th term T14.
Concept / Approach:
For an arithmetic progression with first term a and common difference d, the nth term is given by: Tn = a + (n − 1) * d. We can form two equations using T3 and T8, solve for a and d, and then substitute into the formula for T14.
Step-by-Step Solution:
Let a be the first term and d be the common difference.
T3 = a + 2d = −13.
T8 = a + 7d = 2.
Subtract the first equation from the second: (a + 7d) − (a + 2d) = 2 − (−13).
This gives 5d = 15, so d = 3.
Substitute d = 3 into a + 2d = −13: a + 2 * 3 = −13.
Thus a + 6 = −13, so a = −19.
Now find T14: T14 = a + 13d = −19 + 13 * 3.
Compute 13 * 3 = 39, so T14 = −19 + 39 = 20.
Verification / Alternative check:
You can quickly check by listing some terms. T3 = −13, T4 = −10, and adding 3 repeatedly should eventually reach T8 = 2 and T14 = 20. This matches the algebraic solution, confirming the result.
Why Other Options Are Wrong:
Option a (23), option b (17), option d (26), and option e (11) do not match the arithmetic progression generated from a = −19 and d = 3. When you compute T14, only 20 is consistent with the derived formula.
Common Pitfalls:
Learners often confuse the position index n in Tn with the term value. Another error is to mix up formulas for A.P. with those for geometric progressions. Always remember that in an A.P. the difference between consecutive terms is constant and use Tn = a + (n − 1)d.
Final Answer:
Therefore, the 14th term of the arithmetic progression is 20.
Discussion & Comments