Sequences & Series
Arithmetic progression sum condition
nta_pyq_2025_apr
Grade 12
Question:
Let a, a, a$\ldots be i_n a_n A.P. such that$$\sum 1 2 3 12$k=1$$a2k-1$= - 72 5$a_{1}$, a_{1}$$$\ne 0. If$$$\sum n$k=1$$ak = 0$, then n is$:
Step-by-Step Solution
Key Concept: Represent the AP as$a,a+d,\ldots$and translate the given summation into equations in$a,d$.
Let$a = a$, common$difference = d$1$(1)$$a_{1} +$$a_{3}$$+ a_{5}$+$\ldots$$\$ldots +$a_{23}$$= - 72 5 a 12 72$[$2a + 11$\times 2d] = - a 2 5 72$12a + 132d$= - a 5$132a + 132$\times $5d = 0$a = -5 d n ($2a +$(n - 1)$d) = 0$\Rightarrow -10$d + nd - d = 0$2$n = 11$
Correct Answer: 1