First slide

Arithmetic progression

Question

If $a_{1}, a_{2}, a_{3}, \dots, a_{24}$ are in arithmetic progression and $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$ then $a_{1} + a_{2} + a_{3} + \dots + a_{23} + a_{24}$ is equal to

Moderate

A

909
B

75
C

750
D

900

Solution

Given that,
$a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$
$\begin{array}{l} \Rightarrow (a_{1} + a_{24}) + (a_{5} + a_{20}) + (a_{10} + a_{15}) = 225 \\ \Rightarrow 3 (a_{1} + a_{24}) = 225 \\ \Rightarrow a_{1} + a_{24} = 75 \end{array}$
[ $∵$ in an AP the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of the first and last term]
$\begin{array}{l} ∴ a_{1} + a_{2} + \dots + a_{24} = \frac{n}{2} (a + l) = \frac{24}{2} (a_{1} + a_{24}) \\ = 12 \times 75 = 900 \end{array}$ [from eq.(i)]

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