Let a_1,a_2, ,a_2024 be an Arithmetic Progression such that a_1+(a_5+a_10+a_15+ +a_2020)+a…

Question

Let ${a}_{1},{a}_{2},\ldots ,{a}_{2024}$ be an Arithmetic Progression such that ${a}_{1}+({a}_{5}+{a}_{10}+{a}_{15}+\ldots +{a}_{2020})+{a}_{2024}=2233$. Then ${a}_{1}+{a}_{2}+{a}_{3}+\ldots +{a}_{2024}$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 11132

Accepted Answer

${a}_{1}+{a}_{5}+{a}_{10}+\ldots \ldots +{a}_{2020}+{a}_{2024}=2233$

In an A.P. the sum of terms equidistant from ends is equal.

$\begin{gathered} {a}_{1}+{a}_{2024}={a}_{5}+{a}_{2020}={a}_{10}+{a}_{2015}\ldots \ldots \ \Rightarrow 203 \ \Rightarrow 203({a}_{1}+{a}_{2024})=2233 \end{gathered}$

Hence,

$\begin{gathered} {S}_{2024}=\frac{2024}{2}({a}_{1}+{a}_{2024}) \ = 1012	imes 11 \ = 11132 \end{gathered}$

Solutions · 1