Let \( a_{1}, a_{2}, \ldots \ldots . ., a_{15} \) be in an arithmetic progression. \( a_{1}+a_{2}+\ldots \ldots .+a_{10}=A \) and \( a_{6}+a_{7}+\ldots \ldots . .+a_{15}=B \) . If \( B-A=200 \) and \( B+A=860 \) , then the value of \( a_{15} \) is
Let the first term be \(a\) and the common difference \(d\).
\(B - A = (a_{11} + \dots + a_{15}) - (a_1 + \dots + a_5)\). Each \(a_{k+10} - a_k = 10d\), and there are 5 such pairs, so \(50d = 200\) and \(d = 4\).
\(A = 5(2a + 9d) = 10a + 180\)
\(B = 5(2a + 19d) = 10a + 380\)
\(A + B = 20a + 560 = 860 \Rightarrow a = 15\)
\(a_{15} = a + 14d = 15 + 56 = 71\)
Answer: (C) 71
Reinforce your conceptual grasp with related questions from MAH MCA CET 2024:
Take timed full-length mock tests with authentic question distributions, negative marking (+12/-3 in Math), and instant All India Rank calculation.
Start Free Mock Test Series →